Oleh: Marsigit

Koyber:

Wahai Capteu, aku telah menerima pesan agar disampaikan kepadamu.

Capteu:

Pesan apakah itu dan dari siapa?

Koyber:

Pesan dari Monasiu

Capteu:

Siapakah Monasiu itu?

Koyber:

Dia adalah penguasa kita. Dia yang berhak mematikan dan menghidupkan kita setiap saat. Tiadalah daya upaya kita tanpa dirinya.

Capteu:

Baiklah sebutkanlah apa pesan dia kepadaku.

Koyber:

Pertama, dia menuliskan selembar surat pertama yang isinya permintaan agar engkau menyimpannya.

Kedua, dia membuat dokumen agar engkau mencetakya.

Ketiga, dia membuat rumus-rumus, tetapi meminta agar engkau menghitungnya.

Keempat, dia menunjuk dokumen, dan ,meminta agar engkau menyalinnya.

Kelima, dia menulis surat dan memeinta agar engkau mengirimkan ke temannya.

Keenam, dia membutuhkan gambar, dan meminta agar engkau menyimpannya.

Ketujuh, dia ingin pidato, dan meminta agar engkau menyimpannya.

Kedelapan, dia menunjuk beberapa dokumen, dan meminta agar engkau menghapuskannya.

Kesembilan, dia menunjuk beberapa dokumen, dan meminta agar engkau memindahkannya.

Kesepuluh, dia menunjuk suatu dokumen, dan meminta agar engkau mengeditnya.

Capteu:

Baiklah, semua akan segera saya kerjakan satu persatu dengan kecepatan tinggi dan hasil yang memuaskan.

Capteu:

Wahai haradiso, simpanlah semua dokumen-dokumen itu termasuk perintah-perintahku. Simpan pula semua rencana-rencanaku.

Capteu:

Wahai starago romao, ini ada beberapa dokumen penting dari monasiu. Maka simpanlah dalam dirimu. Tetapi jangan lupa, simpanlah semua perintah-perintahnya, data-datanya, dan tetaplah engkau disitu menjaganya sampai dia mencabut kekuatanmu.

Capteu:

Wahai starago ramoa, ini ada beberapa dokumen penting dari monasiu. Maka simpanlah dalam dirimu. Tetapi jangan lupa, simpanlah semua perintah-perintahnya, data-datanya, dan tetaplah engkau disitu menjaganya dan mempertahankannya, jika perlu sampai titik darah penghabisan.

Capteu:

Wahai paranto, ini ada beberapa dokumen penting dari monasiu. Maka cetak segeralah dokumen-dokumen itu tanpa keragu-raguanmu.

Capteu:

Wahai contorola, kendalikan semua perintah-perintahku. Pastikanlah bahwa semua perintah-perintahku dilaksanakan dengan baik.

Capteu:

Wahai manatora, bukalah jendelamu agar engkau mereka dapat melihat semua diri kita ini. Biarkan mereka mengetahui pekerjaan kita. Jangan engkau tutup-tutupi semua simpanan kita ini. Maka sebenar-benar fungsi dirimu adalah untuk memberi penerangan, sampai monasiu mencabut semua kekuatanmu.

Capteu:

Wahai falasa, engkau adalah anak kesayangan dari haradiso. Maka dengan keterbatasan dirimu, engkau masih mampu berperan sebagi haradiso. Maka simpanlah beberapa dokumen penting dan yang siap digunakan.

Capteu:

Wahai intoranot, sampaikan pesan-pesan dari monasiu. Ketahuilah bahwa pesan-pesan itu sesungguhnya bagi teman-temannya yang jauh.

Capteu:

Wahai manatora, sampaikan berita kepada monasiu bahwa semua perintah-perintahnya telah aku laksanakan.

Monatora:

Wahai monasiu. Perkenankanlah aku menyampaikan pesan dari capteu bahwa semua pesan dan perintahmu sudah dilaksanakannya. Maka semua nasib-nasibku itu terserahlah anda.

Monasiu:

Wahai semuanya, sangat senanglah diriku mendengarkan engkau semua telah melaksanakan pekerjaanmu dengan sebaik-baiknya. Tetapi ketahuilah bahwa aku tidaklah bisa menjaga dirimu setiap saat. Oleh karena itu sekarang engkau semua aku perkenankan untuk beristirahat. Dan untuk itu maka akan segera aku cabutlah semua kekuatanmu itu.

Orang tua berambut putih:

Wahai mahasiswaku semua. Inilah sebenar-benar teka-teki bagimu. Maka tugas-tugasmu adalah untuk menjawab siapakah monasiu. Siapakah manatora. Siapakah capteu. Siapakah intoranot. Siapakah falasa. Siapakah haradiso. Siapakah contorola. Siapakah paranto. Siapakah starago romao. Siapakah starago ramoa. Dan siapakah koyber. Sebenar-benar jawabanmu ditunggu di kolom comment pada elegi oleh Pak Marsigit.

## Friday, March 20, 2009

## Saturday, March 7, 2009

### Power and Root

Dear all students,

I try a new method to share the reference of English for Mathematics

My last posting is about Power and Root. You may download the file by at the following

Click here for Power and Root

Thanks,

Marsigit

I try a new method to share the reference of English for Mathematics

My last posting is about Power and Root. You may download the file by at the following

Click here for Power and Root

Thanks,

Marsigit

## Saturday, February 28, 2009

### PLANE

By Marsigit

Basic Competency

To identify the properties of triangle in term of its sides and angles

To identify the properties of rectangle, square, trapezium, parallelogram, rhombus and kite

To determine the perimeter and area of triangle and rectangle and apply them to solve mathematical problems.

To construct triangle, altitude line, interior angle bisector line, median line, and perpendicular bisector line.

Two dimensional shapes such as triangle, rectangle, and square are easily found in your surroundings. For example, the front side of the roof of a building can be seen as a triangle shape and tennis court can be in the form of rectangular. There are still other two dimensional shapes, for example, parallelogram, trapezium, kite and rhombus.

Can you find the examples of each of those geometrical shapes from your surroundings.

What will you learn in this chapter?

Triangle

Rectangle

Triangle

Understanding Triangle

Triangle is a plane figure with three straight sides. You can construct a triangle by following methods:

Connecting Three Non-Collinear Points to Each Other.

Consider the following figure.

[..gambar 3 titik dan segitiga.. ]

Points A, B and C above do not lie in one straight line (non-collinear). If you connect those three points each other, you will get a triangle ABC.

Note:

Triangle is usually symbolized with “ ∆” . So, you can notate triangle ABC as ∆ABC.

Connecting The Ends Points of A Line Segment to A Certain Point Which not Lies In The Segment Line.

[.......gambar ...]

Let ¯AB is a line segment. Construct point C not in the line segment ¯AB . If you connect point C to the ends of line segment ¯AB you will get a triangle as it is shown in the above figure.

Dividing a Square or Rectangle Through Its Diagonal

Consider the following square ABCD and rectangle EFGH

If you divide the square or the rectangle through its diagonals, you will get two congruence triangles.

[gambar ]

Classifying Triangles

Three types of triangles are distinguished as follows:

Type of triangles in term of the measure of their angles

Type of triangles in term of the length of their sides

Type of triangles in term of the length of their sides and the measure of their angles

Classifying Triangles by Angles

In term of the measure of their angles there are three types of triangles i.e. acute triangle, right triangle and obtuse triangle.

A triangle is acute if all three of its angles are acute

A triangle is obtuse if one of their angles is obtuse

A triangle is right if one of their angles is right

[Gambar 3 buah Segitiga]

Note:

An angle A is right if A = 90°

An angle A is acute if 0° < A < 90°;

An angle A is obtuse if 90° < A < 180°.

Classifying Triangles by Sides

In term of the length of their sides there are three types of triangles i.e. isosceles triangle, equilateral triangle, and scalene triangle.

A triangle is isosceles if two sides are equal

A triangle is equilateral if all three sides are equal

A triangle is scalene if no two of its sides are equal in length

[gambar 3 buah segitiga]

Classifying Triangles by Angles and Sides

In term of the length of their sides and the measure of their angles, there are some types of triangles as follows:

A right isosceles triangle is a triangle in which one of their angles is right and two sides are equal

An acute isosceles triangle is a triangle in which all three of its angles are acute and two sides are equal

An obtuse isosceles triangle is a triangle in which one of their angles is obtuse and two sides are equal

A right scalene triangle is a triangle in which one of their angles is right and no two of its sides are equal in length

An acute scalene triangle is a triangle in which all three of its angles are acute and no two of its sides are equal in length

An obtuse scalene triangle is a triangle in which one of their angles is obtuse and no two of its sides are equal in length

Example

Describe the type of the following triangles in term of the measure of their angles

[gambar 3 buah segitiga]

Solution:

The first triangle is right

The second triangle is obtuse

The third triangle is acute

Exercise

Describe the type of the following triangles in term of the measure of their angles

[Gambar 3 buah segitiga]

Consider the following square!

Construct both diagonals of that square. Then describe four triangles that you found!

[Gambar persegi]

Construct three type of triangles as follows:

An acute isosceles triangle

An obtuse triangle

An equilateral triangle

An obtuse isosceles triangle

3. Constructing Triangles

a. Constructing Equilateral Triangles

The series of diagrams below shows the different stages of construction.

Let, you will construct Equilateral Triangle ∆ABC with the length of its side is 4 cm.

Construct 4 cm line segment ¯AB using pencil and ruler

At point A, set compass to a wide of line segment ¯AB to produce an arch

At point B, set compass to a wide of line segment ¯AB to produce other arch. You find that the second arch intersects with the first arch. Call the intersection point as C.

Connect point C to point A and to point B

[Gambar Lukisan Segitiga]

Exercise

Construct the following equilateral triangles!

∆ABC with the length of the sides is 5 cm

∆PQR with the length of the sides is 6.5 cm

∆ABC with the length of the sides is 4.5 cm

Construct equilateral triangle ABC in which the line segment ¯AB is as a base in the following positions!

[Gambar alas segitiga]

b. Constructing Isosceles Triangles

Followings are the stages of constructing isosceles triangles

Let you will construct isosceles triangle PQR with the length of the sides PR = QR =

cm.

Construct line segment ¯PQ . The length of this segment is arbitrary, i.e. PQ = 2 cm

At point P, set compass to a wide of 3.5 cm to produce an arch.

At point Q, set compass to a wide of 3.5 cm to produce other arch. You find that the second arch intersects with the first arch. Call the intersection point as R.

Connect point R to point P and to point Q

[Gambar lukisan segitiga samakaki]

Note:

A triangle is isosceles if two sides are equal

Exercise

Construct the following isosceles triangles!

∆ABC, AB = AC = 6 cm

∆KLM, KL = KM = 5 cm

b. Constructing Scalene Triangles

If you want to construct any triangle, make sure that the length of the sides should comply with triangular inequalities, i.e. the length of each side is less than the sum of the lengths of the other two sides.

Let, the length of AB, BC, and AC are consecutively c unit of length, a unit of length and b unit of length. If ABC is a triangle, it should comply with the following triangular inequalities:

a + b > c,

a + c > b, and

b + c > a.

In other words, you can construct a scalene triangle if the length of each side is less than the sum of the lengths of the other two sides.

Followings are the stages of constructing isosceles triangles

Let, you are asked to construct ∆ABC, AB = 4 cm, AC = 6 cm, and BC = 5 cm.

Construct 6 cm length of line segment ¯AC

Construct the side of AB by setting compass at point A to a wide of 4 cm to produce an arch.

Construct the side of BC by setting compass at point B to a wide of 5 cm to produce another arch. Call the intersection point of two arches as B.

Connect point B to point A and point C.

. The length of this segment is arbitrary, i.e. PQ = 2 cm

At point P, set compass to a wide of 3.5 cm to produce an arch.

At point Q, set compass to a wide of 3.5 cm to produce other arch. You find that the second arch intersects with the first arch. Call the intersection point as R.

Connect point R to point P and to point Q

[Gambar lukisan segitiga sembarang]

Exercise

Is it possible for you to construct the triangles with the length of the sides as follows? If it is possible, construct them!

PQ = 3,5 cm; PR = 5 cm; and QR = 4 cm

PQ = 3,5 cm; PR = 6 cm; and QR = 8 cm

PQ = 3,5 cm; PR = 6 cm; and QR = 5 cm

AB = 4 cm; BC = 7 cm; and AC = 5 cm

AB = 5 cm; BC = 9 cm; and A = 8 cm

4.Lines inTriangles

There are specific lines in a certain triangle:

Altitude of triangle

Bisector of triangle

Median of triangle and perpendicular bisector of triangle

Altitude of Triangles

The altitude of a triangle is the line passing through one vertex of the triangle, perpendicular to the line including the side opposite this angle.

The stages of constructing altitude of triangle:

Construct triangle ABC

At point A, set compass to a certain wide so that the resulting arch intersects with the opposite side of angle A i.e. side BC. Call the intersection point as P and Q.

At point P, set compass to a similar wide with that of procedure 2 to produce an arch.

At point Q, set compass to a similar wide with that of procedure 3 to produce another arch. Call the intersection point of the arches as M.

Connect point A to point M.

Line segment AM intersects with line segment BC at point D.

Line segment AD is called the altitude of triangle ABC.

[Gambar lukisan garis tinggi)

Exercise

Construct the altitude of the following triangles!

[Gambar 3 buah segitiga]

Bisector of Triangles

Bisector of a triangle is the line that divides the certain angle of a given triangle into smaller angles of equal size.

The stages of constructing bisector of triangle:

Construct a triangle ABC

At point A, set compass to a certain wide so that the resulting arch intersects with the side AB in D and with the side AC in E

At point D, set compass to the similar wide with that of procedure 2 to produce an arch.

At point E, set compass to the similar wide with that of procedure 3 to produce another arch. Call the intersection point of the arches as M.

Connect point A to point M.

Line segment AM intersects with line segment BC at point T.

Line segment AT is called the bisector of triangle ABC at A.

Median and Perpendicular Bisector of Triangles

Perpendicular bisector of triangle is a line passing through the mid-point and perpendicular to a certain side of a given triangle. A median of a triangle is a line passing through one vertex of the triangle and the midpoint of the side opposite this angle.

[Gambar garis sumbu dan garis berat]

In order to construct the median CD you need first to construct perpendicular bisector PQ to indicate the mid-point of AB, i.e. D. After you can indicate point D, you then are to construct the median CD.

The stages of constructing perpendicular bisector and median of a triangle:

Construct a triangle ABC

At point A, set compass to a certain wide so that the resulting arch intersects with the side AB

At point B, set compass to the similar wide with that of procedure 2 to produce another arch so that it intersects with the first arch in P and Q.

Connect point P to point Q

Line segment PQ is perpendicular bisector of line segment AB

Call it D for the intersection between PQ and AB

Connect point D to point C

Line segment CD is the median of triangle ABC passing through C.

Exercise

Construct all bisectors, perpendicular bisectors and medians of the following triangles!

[Gambar 3 biah segitiga]

Perimeter and Area of Triangles

Determining the Perimeter of Triangles

Perimeter of a triangle is the sum of its three sides.

[Gambar keliling segitiga]

Perimeter (P) of triangle ABC is formulated as P = AB + AC + BC

Example

Determine the perimeter of the following triangle!

[Gambar segitiga]

P = AB + AC + BC

= 10 + 4 + 8

= 22

Hence, the perimeter of ABC is 22 cm

The perimeter of a triangle-like garden is 60 m. The length of the two sides of the garden are 15 m and 28 m. Determine the length of its other side!

Solution:

Let, the length of the unknown side is b, then

P = b + 15 + 28

60 = b + 15 + 28

= b + 43

b = 60-43 = 17

Hence, the length of the other side of the garden is 17 m.

Exercise

Find out the perimeter of the following triangles!

16 cm, 10 cm, and 20 cm

30 cm, 45 cm, and 35 cm

75 cm, 100 cm, and 120 cm

The perimeter of a triangle is 55 cm. The length of the two sides are 14 cm and 16 cm. Determine the length of the third side!

Determining the Area of Triangles

At any triangle there are base and height. Each side of triangle can be its base. The height of triangle is the straight line perpendicular to the base, passing through the opposite apex.

Followings are the examples of triangles with their base and height.

[Gambar tiga segitiga dengan alas dan garis tinggi]

You can determine the area of triangle using the following formula

Area of triangle = ½ x base x height

Example

Determine the area of triangle in which its base is 8 cm and the height is 6 cm!

Solution:

[Gambar segitiga yang dimaksud]

Area of triangle = ½ x base x height = ½ x 8 x 6 = 24

Hence, the area of triangle is 24 cm2

Exercise

Determine the area of the following triangles!

[Gambar segitiga 1 sd 5]

Angles of Triangles

The Sum of the Angles in a Triangle

The sum of the three angles of a triangle is 180°. Consider the following figures:

[Gambar guntingan segitiga]

Tear off each angle, and then put the three angles together like the following figure.

You noticed that the sum of the three angles is 180°.

Example

Determine the measure of angles in the following triangle!

[Gambar segitiga siku-siku]

Solution:

Triangle ABC is the right triangle at A. Therefore, ∠ CAB=90°

On the other hand we have

∠ CAB+∠ABC+ ∠BCA=180° then

90° + 2 x° + x° = 180°

90° + 3 x° = 180°

3 x° = 90°

x° = 30°

Hence the measure of each angle of triangle ABC are:

∠ CAB=90° , ∠ BCA=x°=30° , and ∠ ABC=2x°=2(30°)=60°

The comparison among angles of a triangles is 2 : 3 : 4

Determine the measure of each angle!

Solution:

Let, the measure of the angles are 2x°, 3x° and 4x°.

On the other hand we have the sum of the three angles of a triangle is 180°, then

2x°+ 3x°+ 4x° = 180°

9x° = 180°

x° = (180°)/9=20°

Hence, 2x°=2(20°)=40°; 3x°=3 (20°)=60°; 4x°=4(20°)= 80°

Thus, the measure of the angles are 40°, 60° and 80°.

Exercise

Determine the value of y in the following triangles!

[Gambar Segitiga a, b c]

Then, determine also the measure of each angle!

Proportion of angles is 5: 2: 3

Determine the angles.

Interior and Exterior Angles of Triangles

In the following triangle, ∠ BAC,∠ BCA,and ∠ ABC are called interior angles of triangle ABC, while ∠ ACD is called exterior angle of triangle ABC.

[Gambar segitiga]

An exterior angle is equal to the sum of the two interior angles that do not share a side

with the exterior angle.

You know that the sum of the three angles of a triangle is 180°.

or , ∠ CAB+ ∠ ACB+ ∠ ABC= 180°

Due to ∠ ACB shares a side with ∠ ACD, then ∠ ACB+ ∠ ACD= 180°

Hence, ∠ ACD= ∠ BAC+ ∠ ABC

Example

Consider the following figure

[Gambar segitiga]

Determine the measure of ∠ ACD and the measure of ∠ ACB,

when ∠ ABC= 80° and ∠ CAB= 35°!

Solution:

The sum of the three angles of a triangle is 180°, therefore,

∠ CAB+ ∠ ACB+ ∠ ABC= 180°

= 35°+80°=115°

Due to ∠ ACD= ∠ CAB+∠ ACB, hence

∠ ACB= 180°- ∠ ACD

= 180°-115°=65°

Exercise

Triangle ABC, ∠ ABC= 75° and ∠ ACB= 55°. Determine:

∠ BAC;

Exterior angle of A;.

Exterior angle of B.

Determine the exterior angles of P in the following triangles!

[Gambar segitiga]

Quadrilateral

Rectangle

Rectangle is a quadrilateral with two pairs of parallel sides, four right angles, opposite sides equal in length, equal diagonals bisecting one another. Consider the following rectangle ABCD.

[Gambar persegi panjang]

The elements of rectangle consist of length, width, and diagonal.

AB and CD of the rectangle ABCD are called the length

AD and BC of the rectangle ABD are called the width

AC and BD of the rectangle ABD are called diagonal

Followings are the properties of rectangle.

Opposite sides of rectangle are equal and parallel

Each interior angle of rectangle is a right angle

Diagonals of rectangle are equal

Diagonals of rectangle have the same midpoint

Let, a rectangle has its length p and width l then:

Perimeter (K) of the rectangle is K = 2 (p + l)

Area (L) of the rectangle is L = p x l

Example

Draw a rectangle ABCD in which AD = 4 cm and CD = 6 cm!

Is the length of AB equal to CD?

Is the length of AD equal to BC?

Determine the perimeter and the area of the rectangle ABCD!

Solution:

The following is a drawing of rectangle ABCD

[Gambar persegi panjang ABCD]

The length of AB is equal to the length of CD because the opposite sides of rectangle are equal

The length of AD is equal to the length of BC because the opposite sides of rectangle are equal

The length of rectangle ABCD is 6 cm. The width of rectangle ABCD is 4 cm. Therefore,

we can determine the perimeter and the area as:

K = 2 (p + l)

= 2 (6 + 4)

= 2 (10)

= 20

L = p x l

= 6 x 4

= 24

Hence, the perimeter and the area of rectangle are 20 cm and 24 cm2

Exercise

Complete the following table of rectangle!

[Tabel ]

Mr Karto has a field in the form of rectangle. The length of the field is twice as the width.

If the perimeter of Mr. Karto’s fiels is 48 m, determine:

the length and the width of the field;

the area of the field;

if the field of Mr. Karto produces 5 kg casava for each m2

, how many kg casava will produced by Mr. Karto’s field totally?

Square

Square is two-dimensional figure with four straight sides, whose four interior angles are right angles (90°), and whose four sides are of equal length. Consider the following square ABCD.

[Gambar persegi]

The elements of square are:

AB, BC, CD, and AD are the sides of square ABCD.

AC and BD are diagonals of square ABCD.

The properties of square:

all four sides of a square are of equal length

the diagonals of a square are the bisectors

the diagonals of a square are perpendicular bisector of each other.

Let, a square has its side of s cm length:

The perimeter of the square K = 4 s

The area of the square L = s2

Example

The side of a square is 16 cm. Then, each angle is torn off 2 cm. Determine the perimeter and the area of the torn off square!

Solution

The following figure is resulted from the torn off square of 16 cm length of side.

[Gambar bangun persegi yang telah terpotong]

You can indicate the area of the last figure by subtracting the area of the square before it is torn off by the total area of torn off parts.

The area of square before it is torn off is 16 cm x 16 cm = 256 cm2

The area of each torn off part is 2 cm x 2 cm = 4 cm2

Therefore, the total area of torn off parts = 4 x 4 cm2 = 16 cm2

The area of figure after it is torn off = 256 – 16 = 240 cm2

The perimeter of the figure after it is torn off is:

K = 4 x (2 + 12 + 2)

= 4 x 16

= 64 cm

Exercise

Complete the following table!

[ Tabel soal}

A bricklayer is to set the square marbles of the size 20 cm x 20 cm each, to fit together without leaving any space of a rectangle-flour with the size of 4 m x 3 m.

Calculate the number of marbles to tessellate the flour.

Determine the area of the flour to be tessellated.

[Gambar lantai]

Parallelogram

We can form the parallelogram by combining two congruence triangles.

The followings are the elements of parallelogram.

AB, BC, CD and AD are the sides of parallelogram

AC and BD are the diagonals of parallelogram

AB is the base of parallelogram

t is the height of parallelogram

Note:

Triangles which have the exact same size and shape are congruence

[Gambar jajaran genjang]

The properties of parallelogram are:

The opposite sides are equal and parallel to each other

The opposite angles are equal

The sum of the consecutive angles is 1800

The diagonals are perpendicular bisector to each other

We can determine the perimeter of parallelogram by adding up all sides.

The perimeter of parallelogram, K = 2 (AB + BC)

[Gambar jajaran genjang]

The area of parallelogram can be found by the following formula.

The area of parallelogram, L = base x height

= a x t

Example

The length of AB and AD of a parallelogram ABCD are consecutively 12 cm and 7 cm. The height of parallelogram is 5 cm. Determine the perimeter and the area of parallelogram ABCD!

Solution:

You can draw the parallelogram ABCD as follow.

[Gambar jajaran genjang]

Hence, the perimeter of the parallelogram is 38 cm and the area is 60 cm2

Exercise

Complete the following table of parallelogram

[Tabel jajarangenjang]

Let, PQRS is a parallelogram in which its diagonals intersect at point O.

Indicate two couples of parallel sides of the parallelogram PQRS!

Indicate two couples of obtuse triangle in the parallelogram PQRS!

Indicate two couples of equal angles of the parallelogram PQRS!

Consider the following parallelogram

[Gambar jajarangenjang]

Calculate the value of x.

And then, calculate the length of AB.

Calculate the area of the parallelogram.

Trapezoid

Trapezoid is quadrilateral figure with two parallel sides. or bases, of unequal length.

The elements of trapezoid are lower base, upper base, height and legs. Consider the following trapezoid figure.

[Gambar Trapesium]

AB, BC, CD, and AD are the sides of trapezoid

AB has a special name, i.e. the lower base of trapezoid

CD has a special name, i.e. the upper base of trapezoid

AD and BC have a special name, i.e. the legs of trapezoid

Line t is called the height of trapezoid

Type of Trapezoid Can be formed from The Figure

Right Trapezoid A rectangle and a right triangle

gambar

A square and a right triangle

gambar

Isosceles Trapezoid A rectangle and two congruence right triangles

gambar

A square and two congruence right triangles

gambar

Scalene Trapezoid Some plane figures

gambar

The properties of Trapezoid

[Gambar trapesium]

Right Trapezoid

It has exactly two right angles, i.e. ∠ BAD and ∠ ADC

∠ BAD+∠ ADC=1800

∠ ABC+∠ BCD= 1800

[Gambar trapesium]

Isosceles Trapezoid

∠ BAD= ∠ ABC

∠ ADC= ∠ BCD

∠ BAD+ ∠ ADC=1800

∠ ABC+ ∠ BCD=1800

The two diagonals are equal in length (AC = BD)

[Gambar trapesium]

Scalene Trapezoid

∠ BAD+∠ ADC=1800

∠ ABC+ ∠ BCD=1800

The perimeter and area of trapezoid are:

K = AB + BC + CD + AD

L = ½ x (AB + DC) x t

[Gambar trapasium]

Example

Consider the following figure

[Gambar trapesium]

Determine:

The perimeter of trapezoid ABCD

The area of trapezoid ABCD

Solution:

The perimeter of trapezoid can be determined by adding up all sides

K = AB + BC + CD + AD

= 8 + 5 + 5 + 4

= 22

Hence, the perimeter of trapezoid ABCD is 22 cm

Trapezoid ABCD is the right trapezoid. Therefore, the height of trapezoid is (t) equals AD.

L = ½ x (AB + CD) x t

= ½ x ( 8 + 5)) x 4

= ½ x 13 x 4

= 26

Hence, the area of trapezoid ABCD is 26 cm2

BAGIAN KE TIGA (Hal 160 sd 166)

Bagian Terakhir (Hal 167 -172: Soal-soal) menyusul

Exercise

Complete the angles of the following trapezoid

[Gambar 3 bua trapesium]

Trapezoid PQRS is isosceles trapezoid in which PQ // RS. The length of PQ is 15 cm, PR = 1/3 PQ, and the height of trapezoid PQRS is 12 cm. Determine the area of trapezoid PQRS!

The field of Mr. Arman has its square form with 30 m length of side. Mr. Arman then buys new field to extend the field on the left and on the right hand such that the ultimate form of Mr. Aman’s field is isosceles trapezoid as it is shown in the figure. Mr. Arman wish to fence the field, calculate how long the fence that Mr. Aman needs?

Kite

A kite can be formed from two congruence triangles by joining them through their bases.

[Gambar layang-layang]

The elements of a kite:

PQ, PS, QR and SR are called the sides of the kite

PR and QS are called diagonals of the kite

The properties of a kite are:

Two pairs of adjacent sides are equal

One of the diagonals is a perpendicular bisector of the other.

Two pairs of opposite angles are equal

The perimeter and the area of a kite are:

K = 2 (PQ + QR)

L = ½ x QS x PR

Note:

Axis of symmetry is a line producing a figure identical to the original or a mirror image of the original figure.

Example

Consider the following figure.

[Gambar layang-layang]

Determine:

the perimeter of the kite PQRS

the area of the kite PQRS

Solution:

K = 2(PQ + QR)

= 2(13 + 20)

= 2 (33)

= 66

Hence, the perimeter of the kite PQRS is 66 cm

L = ½ x QS x PR

= ½ x (12 + 12) x (5 + 16)

= ½ x 24 x 21

= 252

Hence, the area of the kite PQRS is 252 cm2

Exercise

Consider the following figure

[Gambar layang-layang]

Determine the unknown angles of the kite ABCD

Consider the following figure

[Gambar layang-layang]

Determine:

the perimeter of the kite PQRS

the area of the kite PQRS

Rhombus

Rhombus can be formed by combining two congruence isosceles triangles through its bases.

[Gambar belah ketupat]

The elements of rhombus:

AB, BC, CD, and AD are the sides of rhombus ABCD

AC and BD are the diagonals of rhombus ABCD

The properties of rhombus:

The four sides of rhombus are equal

The opposite angles of rhombus are equal

The diagonals are perpendicular bisector to each other

The perimeter and the area of rhombus are:

K = 4 s

L = ½ x AC x BD

Example

Consider the following figure

[Gambar belah ketuat]

Determine:

the perimeter of rhombus ABCD

the area of rhombus ABCD

Solution:

The side of rhombus ABCD is 15 cm

K = 4 s = 4 x 15 = 60

Hence, the perimeter of rhombus ABCD is 60 cm

We know that ½ AC = 12 cm and ½ BD = 9 cm, hence AC = 2 x 12 cm = 24 cm

and BD = 2 x 9 = 18 cm.

L = ½ x BD x AC

= ½ x 18 x 24

= 216

Hence, the area of rhombus ABCD is 216 cm2

Exercise

Complete the following table of rhombus

[Tabel belah ketupat]

A cassette-shelf is in the form of rhombus as it is shown by the following figure.

[Gambar rak kaset]

Determine the perimeter and the area of that cassette-shelf!

(Hint: to determine the length of OD, use the formula OD = ...)

EXERCISE 9

Chose the correct answer

The measure of the angles of a triangle are consecutively 2xo, (x + 40)o and (4x + 35)o.

The value of x is ...

55

40

35

15

Of the isosceles triangle ABC, the length AB = BC and ∠ ABC=30o. The measure of ∠ BAC is…

150o

120o

75o

60o

Of the isosceles triangle PQR, the length PR = QR and ∠ ABC=37,5o. The measure of

∠ PRQ is…

a.

b.

c.

d.

(lihat dok)

The perimeter of triangle ABC is 120 cm. If the proportion of the sides AB : BC : AC = 3: 4 : 5, the length of AB is...

a.

b.

c.

d.

(lihat dok)

The following ∆ABC can be constructed, except..

a.

b.

c.

d.

(lihat dok)

Consider the following figure

[Gambar segitiga]

The area of ∆KLM is ...

a.

b.

c.

d.

(lihat dok)

The area of a triangle is 135 cm2. The base of the triangle is 18 cm2. The height of the triangle is...

a.

b.

c.

d.

(lihat dok)

The area of ∆ABC in the following figure is..

a.

b.

c.

d.

(lihat dok)

The area of ∆ABC in the following figure is..

a.

b.

c.

d.

(lihat dok)

The length and the width of a rectangle are consecutively (3x + 4) cm and (2x + 4) cm. The perimeter of the rectangle is 36 cm. The value of x is ...

a.

b.

c.

d.

(lihat dok)

A yard has its form of rectangle with the size of 10 m x 8 m. In the middle of the yard is built a pond with the size of 4m x 6m. Therefore, the area of the yard out of the pond is..

a.

b.

c.

d.

(lihat dok)

The proportion of the length and the width of a rectangle is 2:5. If the area of the rectangle is 40 cm2, hence the width of the rectangle is..

a.

b.

c.

d.

(lihat dok)

A square yard has it sides to be planting with akasias tree. The distance between every two akasias is 3 m. If the side of the yard is 15 m, then the number of akasia tree is..

a.

b.

c.

d.

(lihat dok)

The length and the width of a rectangle are 30 cm and 18 cm. If you have a square PQRS in which its perimeter is ½ of the perimeter of the rectangle ABCD, then the side of the square PQRS is..

a.

b.

c.

d.

(lihat dok)

The perimeter of shaded-area in the following figure is...

[Gambar]

a.

b.

c.

d.

(lihat dok)

Consider the following figure.

[Gambar trapesium]

The area of trapezoid showing in the figure is...

a.

b.

c.

d.

(lihat dok)

The area of the following shaded-area is ...

a.

b.

c.

d.

(lihat dok)

Consider the following figure of a kite

[Gambar layang-layang]

If the perimeter of the kite is 240 cm and AD = ½ BC then the value of ½ x + 23 is...

a.

b.

c.

d.

(lihat dok)

[Gambar trapesium]

The area of shaded-area in the above figure is...

a.

b.

c.

d.

(lihat dok)

The perimeter of a rhombus is 52 cm. If the length of its diagonal AC = 10 cm then the area of rhombus ABCD is..

a.

b.

c.

d.

(lihat dok)

Solve the following problems correctly

The perimeter of ∆ABC is 35 cm. The length AB = BC. Determine the length of BC if the length of AC is 10 cm!

The perimeter of a rectangle is 62 cm. The width of the rectangle is 12 cm. Determine:

the length of the rectangle

the area of the rectangle

The length of one of the diagonals of rhombus is 10 cm. While the proportion of the length of both diagonals is 2 : 5. Determine the area of the rhombus!

Consider the following figure of a kite!

[Gambar layang-layang]

Determine the area of shaded-area of the kite if the length of Ac is 10 cm, the length of BD is 4 cm, and the proportion of AO and OC is 2:3!

The size of the flour if 12 m x 9 m. The flour will be tessellated using the marbles with the size of 30 cm x 30 cm each. Determine the number of marbles to be used to fit together without leaving any space of the flour.

Basic Competency

To identify the properties of triangle in term of its sides and angles

To identify the properties of rectangle, square, trapezium, parallelogram, rhombus and kite

To determine the perimeter and area of triangle and rectangle and apply them to solve mathematical problems.

To construct triangle, altitude line, interior angle bisector line, median line, and perpendicular bisector line.

Two dimensional shapes such as triangle, rectangle, and square are easily found in your surroundings. For example, the front side of the roof of a building can be seen as a triangle shape and tennis court can be in the form of rectangular. There are still other two dimensional shapes, for example, parallelogram, trapezium, kite and rhombus.

Can you find the examples of each of those geometrical shapes from your surroundings.

What will you learn in this chapter?

Triangle

Rectangle

Triangle

Understanding Triangle

Triangle is a plane figure with three straight sides. You can construct a triangle by following methods:

Connecting Three Non-Collinear Points to Each Other.

Consider the following figure.

[..gambar 3 titik dan segitiga.. ]

Points A, B and C above do not lie in one straight line (non-collinear). If you connect those three points each other, you will get a triangle ABC.

Note:

Triangle is usually symbolized with “ ∆” . So, you can notate triangle ABC as ∆ABC.

Connecting The Ends Points of A Line Segment to A Certain Point Which not Lies In The Segment Line.

[.......gambar ...]

Let ¯AB is a line segment. Construct point C not in the line segment ¯AB . If you connect point C to the ends of line segment ¯AB you will get a triangle as it is shown in the above figure.

Dividing a Square or Rectangle Through Its Diagonal

Consider the following square ABCD and rectangle EFGH

If you divide the square or the rectangle through its diagonals, you will get two congruence triangles.

[gambar ]

Classifying Triangles

Three types of triangles are distinguished as follows:

Type of triangles in term of the measure of their angles

Type of triangles in term of the length of their sides

Type of triangles in term of the length of their sides and the measure of their angles

Classifying Triangles by Angles

In term of the measure of their angles there are three types of triangles i.e. acute triangle, right triangle and obtuse triangle.

A triangle is acute if all three of its angles are acute

A triangle is obtuse if one of their angles is obtuse

A triangle is right if one of their angles is right

[Gambar 3 buah Segitiga]

Note:

An angle A is right if A = 90°

An angle A is acute if 0° < A < 90°;

An angle A is obtuse if 90° < A < 180°.

Classifying Triangles by Sides

In term of the length of their sides there are three types of triangles i.e. isosceles triangle, equilateral triangle, and scalene triangle.

A triangle is isosceles if two sides are equal

A triangle is equilateral if all three sides are equal

A triangle is scalene if no two of its sides are equal in length

[gambar 3 buah segitiga]

Classifying Triangles by Angles and Sides

In term of the length of their sides and the measure of their angles, there are some types of triangles as follows:

A right isosceles triangle is a triangle in which one of their angles is right and two sides are equal

An acute isosceles triangle is a triangle in which all three of its angles are acute and two sides are equal

An obtuse isosceles triangle is a triangle in which one of their angles is obtuse and two sides are equal

A right scalene triangle is a triangle in which one of their angles is right and no two of its sides are equal in length

An acute scalene triangle is a triangle in which all three of its angles are acute and no two of its sides are equal in length

An obtuse scalene triangle is a triangle in which one of their angles is obtuse and no two of its sides are equal in length

Example

Describe the type of the following triangles in term of the measure of their angles

[gambar 3 buah segitiga]

Solution:

The first triangle is right

The second triangle is obtuse

The third triangle is acute

Exercise

Describe the type of the following triangles in term of the measure of their angles

[Gambar 3 buah segitiga]

Consider the following square!

Construct both diagonals of that square. Then describe four triangles that you found!

[Gambar persegi]

Construct three type of triangles as follows:

An acute isosceles triangle

An obtuse triangle

An equilateral triangle

An obtuse isosceles triangle

3. Constructing Triangles

a. Constructing Equilateral Triangles

The series of diagrams below shows the different stages of construction.

Let, you will construct Equilateral Triangle ∆ABC with the length of its side is 4 cm.

Construct 4 cm line segment ¯AB using pencil and ruler

At point A, set compass to a wide of line segment ¯AB to produce an arch

At point B, set compass to a wide of line segment ¯AB to produce other arch. You find that the second arch intersects with the first arch. Call the intersection point as C.

Connect point C to point A and to point B

[Gambar Lukisan Segitiga]

Exercise

Construct the following equilateral triangles!

∆ABC with the length of the sides is 5 cm

∆PQR with the length of the sides is 6.5 cm

∆ABC with the length of the sides is 4.5 cm

Construct equilateral triangle ABC in which the line segment ¯AB is as a base in the following positions!

[Gambar alas segitiga]

b. Constructing Isosceles Triangles

Followings are the stages of constructing isosceles triangles

Let you will construct isosceles triangle PQR with the length of the sides PR = QR =

cm.

Construct line segment ¯PQ . The length of this segment is arbitrary, i.e. PQ = 2 cm

At point P, set compass to a wide of 3.5 cm to produce an arch.

At point Q, set compass to a wide of 3.5 cm to produce other arch. You find that the second arch intersects with the first arch. Call the intersection point as R.

Connect point R to point P and to point Q

[Gambar lukisan segitiga samakaki]

Note:

A triangle is isosceles if two sides are equal

Exercise

Construct the following isosceles triangles!

∆ABC, AB = AC = 6 cm

∆KLM, KL = KM = 5 cm

b. Constructing Scalene Triangles

If you want to construct any triangle, make sure that the length of the sides should comply with triangular inequalities, i.e. the length of each side is less than the sum of the lengths of the other two sides.

Let, the length of AB, BC, and AC are consecutively c unit of length, a unit of length and b unit of length. If ABC is a triangle, it should comply with the following triangular inequalities:

a + b > c,

a + c > b, and

b + c > a.

In other words, you can construct a scalene triangle if the length of each side is less than the sum of the lengths of the other two sides.

Followings are the stages of constructing isosceles triangles

Let, you are asked to construct ∆ABC, AB = 4 cm, AC = 6 cm, and BC = 5 cm.

Construct 6 cm length of line segment ¯AC

Construct the side of AB by setting compass at point A to a wide of 4 cm to produce an arch.

Construct the side of BC by setting compass at point B to a wide of 5 cm to produce another arch. Call the intersection point of two arches as B.

Connect point B to point A and point C.

. The length of this segment is arbitrary, i.e. PQ = 2 cm

At point P, set compass to a wide of 3.5 cm to produce an arch.

At point Q, set compass to a wide of 3.5 cm to produce other arch. You find that the second arch intersects with the first arch. Call the intersection point as R.

Connect point R to point P and to point Q

[Gambar lukisan segitiga sembarang]

Exercise

Is it possible for you to construct the triangles with the length of the sides as follows? If it is possible, construct them!

PQ = 3,5 cm; PR = 5 cm; and QR = 4 cm

PQ = 3,5 cm; PR = 6 cm; and QR = 8 cm

PQ = 3,5 cm; PR = 6 cm; and QR = 5 cm

AB = 4 cm; BC = 7 cm; and AC = 5 cm

AB = 5 cm; BC = 9 cm; and A = 8 cm

4.Lines inTriangles

There are specific lines in a certain triangle:

Altitude of triangle

Bisector of triangle

Median of triangle and perpendicular bisector of triangle

Altitude of Triangles

The altitude of a triangle is the line passing through one vertex of the triangle, perpendicular to the line including the side opposite this angle.

The stages of constructing altitude of triangle:

Construct triangle ABC

At point A, set compass to a certain wide so that the resulting arch intersects with the opposite side of angle A i.e. side BC. Call the intersection point as P and Q.

At point P, set compass to a similar wide with that of procedure 2 to produce an arch.

At point Q, set compass to a similar wide with that of procedure 3 to produce another arch. Call the intersection point of the arches as M.

Connect point A to point M.

Line segment AM intersects with line segment BC at point D.

Line segment AD is called the altitude of triangle ABC.

[Gambar lukisan garis tinggi)

Exercise

Construct the altitude of the following triangles!

[Gambar 3 buah segitiga]

Bisector of Triangles

Bisector of a triangle is the line that divides the certain angle of a given triangle into smaller angles of equal size.

The stages of constructing bisector of triangle:

Construct a triangle ABC

At point A, set compass to a certain wide so that the resulting arch intersects with the side AB in D and with the side AC in E

At point D, set compass to the similar wide with that of procedure 2 to produce an arch.

At point E, set compass to the similar wide with that of procedure 3 to produce another arch. Call the intersection point of the arches as M.

Connect point A to point M.

Line segment AM intersects with line segment BC at point T.

Line segment AT is called the bisector of triangle ABC at A.

Median and Perpendicular Bisector of Triangles

Perpendicular bisector of triangle is a line passing through the mid-point and perpendicular to a certain side of a given triangle. A median of a triangle is a line passing through one vertex of the triangle and the midpoint of the side opposite this angle.

[Gambar garis sumbu dan garis berat]

In order to construct the median CD you need first to construct perpendicular bisector PQ to indicate the mid-point of AB, i.e. D. After you can indicate point D, you then are to construct the median CD.

The stages of constructing perpendicular bisector and median of a triangle:

Construct a triangle ABC

At point A, set compass to a certain wide so that the resulting arch intersects with the side AB

At point B, set compass to the similar wide with that of procedure 2 to produce another arch so that it intersects with the first arch in P and Q.

Connect point P to point Q

Line segment PQ is perpendicular bisector of line segment AB

Call it D for the intersection between PQ and AB

Connect point D to point C

Line segment CD is the median of triangle ABC passing through C.

Exercise

Construct all bisectors, perpendicular bisectors and medians of the following triangles!

[Gambar 3 biah segitiga]

Perimeter and Area of Triangles

Determining the Perimeter of Triangles

Perimeter of a triangle is the sum of its three sides.

[Gambar keliling segitiga]

Perimeter (P) of triangle ABC is formulated as P = AB + AC + BC

Example

Determine the perimeter of the following triangle!

[Gambar segitiga]

P = AB + AC + BC

= 10 + 4 + 8

= 22

Hence, the perimeter of ABC is 22 cm

The perimeter of a triangle-like garden is 60 m. The length of the two sides of the garden are 15 m and 28 m. Determine the length of its other side!

Solution:

Let, the length of the unknown side is b, then

P = b + 15 + 28

60 = b + 15 + 28

= b + 43

b = 60-43 = 17

Hence, the length of the other side of the garden is 17 m.

Exercise

Find out the perimeter of the following triangles!

16 cm, 10 cm, and 20 cm

30 cm, 45 cm, and 35 cm

75 cm, 100 cm, and 120 cm

The perimeter of a triangle is 55 cm. The length of the two sides are 14 cm and 16 cm. Determine the length of the third side!

Determining the Area of Triangles

At any triangle there are base and height. Each side of triangle can be its base. The height of triangle is the straight line perpendicular to the base, passing through the opposite apex.

Followings are the examples of triangles with their base and height.

[Gambar tiga segitiga dengan alas dan garis tinggi]

You can determine the area of triangle using the following formula

Area of triangle = ½ x base x height

Example

Determine the area of triangle in which its base is 8 cm and the height is 6 cm!

Solution:

[Gambar segitiga yang dimaksud]

Area of triangle = ½ x base x height = ½ x 8 x 6 = 24

Hence, the area of triangle is 24 cm2

Exercise

Determine the area of the following triangles!

[Gambar segitiga 1 sd 5]

Angles of Triangles

The Sum of the Angles in a Triangle

The sum of the three angles of a triangle is 180°. Consider the following figures:

[Gambar guntingan segitiga]

Tear off each angle, and then put the three angles together like the following figure.

You noticed that the sum of the three angles is 180°.

Example

Determine the measure of angles in the following triangle!

[Gambar segitiga siku-siku]

Solution:

Triangle ABC is the right triangle at A. Therefore, ∠ CAB=90°

On the other hand we have

∠ CAB+∠ABC+ ∠BCA=180° then

90° + 2 x° + x° = 180°

90° + 3 x° = 180°

3 x° = 90°

x° = 30°

Hence the measure of each angle of triangle ABC are:

∠ CAB=90° , ∠ BCA=x°=30° , and ∠ ABC=2x°=2(30°)=60°

The comparison among angles of a triangles is 2 : 3 : 4

Determine the measure of each angle!

Solution:

Let, the measure of the angles are 2x°, 3x° and 4x°.

On the other hand we have the sum of the three angles of a triangle is 180°, then

2x°+ 3x°+ 4x° = 180°

9x° = 180°

x° = (180°)/9=20°

Hence, 2x°=2(20°)=40°; 3x°=3 (20°)=60°; 4x°=4(20°)= 80°

Thus, the measure of the angles are 40°, 60° and 80°.

Exercise

Determine the value of y in the following triangles!

[Gambar Segitiga a, b c]

Then, determine also the measure of each angle!

Proportion of angles is 5: 2: 3

Determine the angles.

Interior and Exterior Angles of Triangles

In the following triangle, ∠ BAC,∠ BCA,and ∠ ABC are called interior angles of triangle ABC, while ∠ ACD is called exterior angle of triangle ABC.

[Gambar segitiga]

An exterior angle is equal to the sum of the two interior angles that do not share a side

with the exterior angle.

You know that the sum of the three angles of a triangle is 180°.

or , ∠ CAB+ ∠ ACB+ ∠ ABC= 180°

Due to ∠ ACB shares a side with ∠ ACD, then ∠ ACB+ ∠ ACD= 180°

Hence, ∠ ACD= ∠ BAC+ ∠ ABC

Example

Consider the following figure

[Gambar segitiga]

Determine the measure of ∠ ACD and the measure of ∠ ACB,

when ∠ ABC= 80° and ∠ CAB= 35°!

Solution:

The sum of the three angles of a triangle is 180°, therefore,

∠ CAB+ ∠ ACB+ ∠ ABC= 180°

= 35°+80°=115°

Due to ∠ ACD= ∠ CAB+∠ ACB, hence

∠ ACB= 180°- ∠ ACD

= 180°-115°=65°

Exercise

Triangle ABC, ∠ ABC= 75° and ∠ ACB= 55°. Determine:

∠ BAC;

Exterior angle of A;.

Exterior angle of B.

Determine the exterior angles of P in the following triangles!

[Gambar segitiga]

Quadrilateral

Rectangle

Rectangle is a quadrilateral with two pairs of parallel sides, four right angles, opposite sides equal in length, equal diagonals bisecting one another. Consider the following rectangle ABCD.

[Gambar persegi panjang]

The elements of rectangle consist of length, width, and diagonal.

AB and CD of the rectangle ABCD are called the length

AD and BC of the rectangle ABD are called the width

AC and BD of the rectangle ABD are called diagonal

Followings are the properties of rectangle.

Opposite sides of rectangle are equal and parallel

Each interior angle of rectangle is a right angle

Diagonals of rectangle are equal

Diagonals of rectangle have the same midpoint

Let, a rectangle has its length p and width l then:

Perimeter (K) of the rectangle is K = 2 (p + l)

Area (L) of the rectangle is L = p x l

Example

Draw a rectangle ABCD in which AD = 4 cm and CD = 6 cm!

Is the length of AB equal to CD?

Is the length of AD equal to BC?

Determine the perimeter and the area of the rectangle ABCD!

Solution:

The following is a drawing of rectangle ABCD

[Gambar persegi panjang ABCD]

The length of AB is equal to the length of CD because the opposite sides of rectangle are equal

The length of AD is equal to the length of BC because the opposite sides of rectangle are equal

The length of rectangle ABCD is 6 cm. The width of rectangle ABCD is 4 cm. Therefore,

we can determine the perimeter and the area as:

K = 2 (p + l)

= 2 (6 + 4)

= 2 (10)

= 20

L = p x l

= 6 x 4

= 24

Hence, the perimeter and the area of rectangle are 20 cm and 24 cm2

Exercise

Complete the following table of rectangle!

[Tabel ]

Mr Karto has a field in the form of rectangle. The length of the field is twice as the width.

If the perimeter of Mr. Karto’s fiels is 48 m, determine:

the length and the width of the field;

the area of the field;

if the field of Mr. Karto produces 5 kg casava for each m2

, how many kg casava will produced by Mr. Karto’s field totally?

Square

Square is two-dimensional figure with four straight sides, whose four interior angles are right angles (90°), and whose four sides are of equal length. Consider the following square ABCD.

[Gambar persegi]

The elements of square are:

AB, BC, CD, and AD are the sides of square ABCD.

AC and BD are diagonals of square ABCD.

The properties of square:

all four sides of a square are of equal length

the diagonals of a square are the bisectors

the diagonals of a square are perpendicular bisector of each other.

Let, a square has its side of s cm length:

The perimeter of the square K = 4 s

The area of the square L = s2

Example

The side of a square is 16 cm. Then, each angle is torn off 2 cm. Determine the perimeter and the area of the torn off square!

Solution

The following figure is resulted from the torn off square of 16 cm length of side.

[Gambar bangun persegi yang telah terpotong]

You can indicate the area of the last figure by subtracting the area of the square before it is torn off by the total area of torn off parts.

The area of square before it is torn off is 16 cm x 16 cm = 256 cm2

The area of each torn off part is 2 cm x 2 cm = 4 cm2

Therefore, the total area of torn off parts = 4 x 4 cm2 = 16 cm2

The area of figure after it is torn off = 256 – 16 = 240 cm2

The perimeter of the figure after it is torn off is:

K = 4 x (2 + 12 + 2)

= 4 x 16

= 64 cm

Exercise

Complete the following table!

[ Tabel soal}

A bricklayer is to set the square marbles of the size 20 cm x 20 cm each, to fit together without leaving any space of a rectangle-flour with the size of 4 m x 3 m.

Calculate the number of marbles to tessellate the flour.

Determine the area of the flour to be tessellated.

[Gambar lantai]

Parallelogram

We can form the parallelogram by combining two congruence triangles.

The followings are the elements of parallelogram.

AB, BC, CD and AD are the sides of parallelogram

AC and BD are the diagonals of parallelogram

AB is the base of parallelogram

t is the height of parallelogram

Note:

Triangles which have the exact same size and shape are congruence

[Gambar jajaran genjang]

The properties of parallelogram are:

The opposite sides are equal and parallel to each other

The opposite angles are equal

The sum of the consecutive angles is 1800

The diagonals are perpendicular bisector to each other

We can determine the perimeter of parallelogram by adding up all sides.

The perimeter of parallelogram, K = 2 (AB + BC)

[Gambar jajaran genjang]

The area of parallelogram can be found by the following formula.

The area of parallelogram, L = base x height

= a x t

Example

The length of AB and AD of a parallelogram ABCD are consecutively 12 cm and 7 cm. The height of parallelogram is 5 cm. Determine the perimeter and the area of parallelogram ABCD!

Solution:

You can draw the parallelogram ABCD as follow.

[Gambar jajaran genjang]

Hence, the perimeter of the parallelogram is 38 cm and the area is 60 cm2

Exercise

Complete the following table of parallelogram

[Tabel jajarangenjang]

Let, PQRS is a parallelogram in which its diagonals intersect at point O.

Indicate two couples of parallel sides of the parallelogram PQRS!

Indicate two couples of obtuse triangle in the parallelogram PQRS!

Indicate two couples of equal angles of the parallelogram PQRS!

Consider the following parallelogram

[Gambar jajarangenjang]

Calculate the value of x.

And then, calculate the length of AB.

Calculate the area of the parallelogram.

Trapezoid

Trapezoid is quadrilateral figure with two parallel sides. or bases, of unequal length.

The elements of trapezoid are lower base, upper base, height and legs. Consider the following trapezoid figure.

[Gambar Trapesium]

AB, BC, CD, and AD are the sides of trapezoid

AB has a special name, i.e. the lower base of trapezoid

CD has a special name, i.e. the upper base of trapezoid

AD and BC have a special name, i.e. the legs of trapezoid

Line t is called the height of trapezoid

Type of Trapezoid Can be formed from The Figure

Right Trapezoid A rectangle and a right triangle

gambar

A square and a right triangle

gambar

Isosceles Trapezoid A rectangle and two congruence right triangles

gambar

A square and two congruence right triangles

gambar

Scalene Trapezoid Some plane figures

gambar

The properties of Trapezoid

[Gambar trapesium]

Right Trapezoid

It has exactly two right angles, i.e. ∠ BAD and ∠ ADC

∠ BAD+∠ ADC=1800

∠ ABC+∠ BCD= 1800

[Gambar trapesium]

Isosceles Trapezoid

∠ BAD= ∠ ABC

∠ ADC= ∠ BCD

∠ BAD+ ∠ ADC=1800

∠ ABC+ ∠ BCD=1800

The two diagonals are equal in length (AC = BD)

[Gambar trapesium]

Scalene Trapezoid

∠ BAD+∠ ADC=1800

∠ ABC+ ∠ BCD=1800

The perimeter and area of trapezoid are:

K = AB + BC + CD + AD

L = ½ x (AB + DC) x t

[Gambar trapasium]

Example

Consider the following figure

[Gambar trapesium]

Determine:

The perimeter of trapezoid ABCD

The area of trapezoid ABCD

Solution:

The perimeter of trapezoid can be determined by adding up all sides

K = AB + BC + CD + AD

= 8 + 5 + 5 + 4

= 22

Hence, the perimeter of trapezoid ABCD is 22 cm

Trapezoid ABCD is the right trapezoid. Therefore, the height of trapezoid is (t) equals AD.

L = ½ x (AB + CD) x t

= ½ x ( 8 + 5)) x 4

= ½ x 13 x 4

= 26

Hence, the area of trapezoid ABCD is 26 cm2

BAGIAN KE TIGA (Hal 160 sd 166)

Bagian Terakhir (Hal 167 -172: Soal-soal) menyusul

Exercise

Complete the angles of the following trapezoid

[Gambar 3 bua trapesium]

Trapezoid PQRS is isosceles trapezoid in which PQ // RS. The length of PQ is 15 cm, PR = 1/3 PQ, and the height of trapezoid PQRS is 12 cm. Determine the area of trapezoid PQRS!

The field of Mr. Arman has its square form with 30 m length of side. Mr. Arman then buys new field to extend the field on the left and on the right hand such that the ultimate form of Mr. Aman’s field is isosceles trapezoid as it is shown in the figure. Mr. Arman wish to fence the field, calculate how long the fence that Mr. Aman needs?

Kite

A kite can be formed from two congruence triangles by joining them through their bases.

[Gambar layang-layang]

The elements of a kite:

PQ, PS, QR and SR are called the sides of the kite

PR and QS are called diagonals of the kite

The properties of a kite are:

Two pairs of adjacent sides are equal

One of the diagonals is a perpendicular bisector of the other.

Two pairs of opposite angles are equal

The perimeter and the area of a kite are:

K = 2 (PQ + QR)

L = ½ x QS x PR

Note:

Axis of symmetry is a line producing a figure identical to the original or a mirror image of the original figure.

Example

Consider the following figure.

[Gambar layang-layang]

Determine:

the perimeter of the kite PQRS

the area of the kite PQRS

Solution:

K = 2(PQ + QR)

= 2(13 + 20)

= 2 (33)

= 66

Hence, the perimeter of the kite PQRS is 66 cm

L = ½ x QS x PR

= ½ x (12 + 12) x (5 + 16)

= ½ x 24 x 21

= 252

Hence, the area of the kite PQRS is 252 cm2

Exercise

Consider the following figure

[Gambar layang-layang]

Determine the unknown angles of the kite ABCD

Consider the following figure

[Gambar layang-layang]

Determine:

the perimeter of the kite PQRS

the area of the kite PQRS

Rhombus

Rhombus can be formed by combining two congruence isosceles triangles through its bases.

[Gambar belah ketupat]

The elements of rhombus:

AB, BC, CD, and AD are the sides of rhombus ABCD

AC and BD are the diagonals of rhombus ABCD

The properties of rhombus:

The four sides of rhombus are equal

The opposite angles of rhombus are equal

The diagonals are perpendicular bisector to each other

The perimeter and the area of rhombus are:

K = 4 s

L = ½ x AC x BD

Example

Consider the following figure

[Gambar belah ketuat]

Determine:

the perimeter of rhombus ABCD

the area of rhombus ABCD

Solution:

The side of rhombus ABCD is 15 cm

K = 4 s = 4 x 15 = 60

Hence, the perimeter of rhombus ABCD is 60 cm

We know that ½ AC = 12 cm and ½ BD = 9 cm, hence AC = 2 x 12 cm = 24 cm

and BD = 2 x 9 = 18 cm.

L = ½ x BD x AC

= ½ x 18 x 24

= 216

Hence, the area of rhombus ABCD is 216 cm2

Exercise

Complete the following table of rhombus

[Tabel belah ketupat]

A cassette-shelf is in the form of rhombus as it is shown by the following figure.

[Gambar rak kaset]

Determine the perimeter and the area of that cassette-shelf!

(Hint: to determine the length of OD, use the formula OD = ...)

EXERCISE 9

Chose the correct answer

The measure of the angles of a triangle are consecutively 2xo, (x + 40)o and (4x + 35)o.

The value of x is ...

55

40

35

15

Of the isosceles triangle ABC, the length AB = BC and ∠ ABC=30o. The measure of ∠ BAC is…

150o

120o

75o

60o

Of the isosceles triangle PQR, the length PR = QR and ∠ ABC=37,5o. The measure of

∠ PRQ is…

a.

b.

c.

d.

(lihat dok)

The perimeter of triangle ABC is 120 cm. If the proportion of the sides AB : BC : AC = 3: 4 : 5, the length of AB is...

a.

b.

c.

d.

(lihat dok)

The following ∆ABC can be constructed, except..

a.

b.

c.

d.

(lihat dok)

Consider the following figure

[Gambar segitiga]

The area of ∆KLM is ...

a.

b.

c.

d.

(lihat dok)

The area of a triangle is 135 cm2. The base of the triangle is 18 cm2. The height of the triangle is...

a.

b.

c.

d.

(lihat dok)

The area of ∆ABC in the following figure is..

a.

b.

c.

d.

(lihat dok)

The area of ∆ABC in the following figure is..

a.

b.

c.

d.

(lihat dok)

The length and the width of a rectangle are consecutively (3x + 4) cm and (2x + 4) cm. The perimeter of the rectangle is 36 cm. The value of x is ...

a.

b.

c.

d.

(lihat dok)

A yard has its form of rectangle with the size of 10 m x 8 m. In the middle of the yard is built a pond with the size of 4m x 6m. Therefore, the area of the yard out of the pond is..

a.

b.

c.

d.

(lihat dok)

The proportion of the length and the width of a rectangle is 2:5. If the area of the rectangle is 40 cm2, hence the width of the rectangle is..

a.

b.

c.

d.

(lihat dok)

A square yard has it sides to be planting with akasias tree. The distance between every two akasias is 3 m. If the side of the yard is 15 m, then the number of akasia tree is..

a.

b.

c.

d.

(lihat dok)

The length and the width of a rectangle are 30 cm and 18 cm. If you have a square PQRS in which its perimeter is ½ of the perimeter of the rectangle ABCD, then the side of the square PQRS is..

a.

b.

c.

d.

(lihat dok)

The perimeter of shaded-area in the following figure is...

[Gambar]

a.

b.

c.

d.

(lihat dok)

Consider the following figure.

[Gambar trapesium]

The area of trapezoid showing in the figure is...

a.

b.

c.

d.

(lihat dok)

The area of the following shaded-area is ...

a.

b.

c.

d.

(lihat dok)

Consider the following figure of a kite

[Gambar layang-layang]

If the perimeter of the kite is 240 cm and AD = ½ BC then the value of ½ x + 23 is...

a.

b.

c.

d.

(lihat dok)

[Gambar trapesium]

The area of shaded-area in the above figure is...

a.

b.

c.

d.

(lihat dok)

The perimeter of a rhombus is 52 cm. If the length of its diagonal AC = 10 cm then the area of rhombus ABCD is..

a.

b.

c.

d.

(lihat dok)

Solve the following problems correctly

The perimeter of ∆ABC is 35 cm. The length AB = BC. Determine the length of BC if the length of AC is 10 cm!

The perimeter of a rectangle is 62 cm. The width of the rectangle is 12 cm. Determine:

the length of the rectangle

the area of the rectangle

The length of one of the diagonals of rhombus is 10 cm. While the proportion of the length of both diagonals is 2 : 5. Determine the area of the rhombus!

Consider the following figure of a kite!

[Gambar layang-layang]

Determine the area of shaded-area of the kite if the length of Ac is 10 cm, the length of BD is 4 cm, and the proportion of AO and OC is 2:3!

The size of the flour if 12 m x 9 m. The flour will be tessellated using the marbles with the size of 30 cm x 30 cm each. Determine the number of marbles to be used to fit together without leaving any space of the flour.

## Wednesday, February 25, 2009

### Mathematics for Junior High School

By Marsigit

Three-dimensional Curved Surfaces Shape

Basic Competency:

To identify the elements of cylinder, cone, and sphere.

To estimate the lateral area and volume of cylinder, cone, and sphere.

To solve the problem that is related with cylinder, cone, and sphere.

What will be learned with this chapter?

Cylinder

Cone

Sphere

Cylinder

Definition of Cylinder

Cylinder or tube is a three-dimensional shape consists of two circular bases of equal area that are in parallel planes, and are connected by a lateral surface that intersects the boundaries of the bases.

The Elements of Cylinder

The elements of the cylinder are as follows:

Cylinder has an upper side (cap) and under side (base) in form of congruent circle (equal in its form and its size).

AB is called the diameter of cylinder base.

PE, PA, and PB are called the radius of cylinder base.

BC and AD is called the height of cylinder

The surface which is perpendicular to the bases is called the lateral of cylinder.

Two circular bases

The cap, the base and the lateral surface is called the surface of cylinder.

The Nets of Cylinder

Pay attention to the Figure 2.2 (a) is a cylinder that the radius is r and the height is t. If the cylinder as in the Figure 2.2 (a) is sliced as long as its height line (along AD or BC) and along its curved sides (along its perimeter of circle base and cap) as in the Figure 2.2 (b) then you will get the net of the cylinder as it seemed in the Figure 2.2 (c).

Figure 2.2

The cylinder which has the radius r and the height t.

The cylinder that is sliced along its curved sides of the base and the cap and along its height of cylinder.

The nets of cylinder.

In the nets of cylinder, it shows that the upper side (cap) and under side (base) are circles which has a radius r, while the net of the curved side (cylinder lateral) is rectangle ABCD.

Surface Area of Cylinder

The surface of the cylinder consists of cylinder lateral, upper side (cap), and under side (base). The cylinder lateral side is formed by a rectangle with length 2Ï€r and width t.

The following formulas of areas are usually used in the cylinder:

Example:

The radius of the base of the cylinder is 7 cm and the height is 10 cm. Determine:

the length of the the lateral net.

the area of lateral surface, and

the area of cylinder surface.

Solution:

The height (t) is 10 cm and the cylinder base radius (r) is 7 cm.

The length of the the lateral net = 2Ï€r

= 2 x x 7 = 44

So, the length of the the lateral net is 44 cm.

The area of lateral surface = 2Ï€r x t

= 44 x 10

= 440

So, the area of lateral surface is 440 cm².

The area of cylinder surface = 2Ï€r(t + r )

= 44 x (10 + 7)

= 44 x 17

= 748

So, area of cylinder surface is 749 cm².

Exercise:

An un-caped cylinder of glass has a diameter of 7 cm and a height of 20 cm. Determine:

the area of lateral surface, and

the area of cylinder surface.

A water pipe in the form of cylinder has the radius of 2.1 cm and the length of 28 cm. If the water pipe is hollow out at both its ends, determine the surface area of the pipe!

A clay-made flowerpot is in the form of cylinder. The base radius of the pot is 10 cm and the height is 20 cm. If the un-caped flowerpot will be painted on its lateral side and its base, determine the surface area should be painted!

A tart-cake for celebrating a birthday is in the form of cylinder with diameter 28 cm and height 8 cm. If it coated with chocolate, determine the surface area of the chocolate!

A can of milk is in the form of cylinder with diameter 7 cm and height 8 cm. The lateral side will be coated with paper containing information about the milk product. Determine the area of the paper!

5. The Volume of Cylinder

The way to identify the volume of cylinder is identical with that of the volume of a right prism. Consider the Figure 2.3!

Figure 2.3 A cylinder is a right prism with circular bases

If the base and the cap of regular prism as it shown in Figure 2.3 has a lot of facets so the form of the base and cap of the prism will close to the form of circle. The right prism with circular bases is called cylinder. Therefore, you will get the volume of the cylinder as the following:

The volume of the cylinder = the area of the base x the height of the cylinder

= the area of the circle x the height of the cylinder

= (Ï€r²) x t

= Ï€r²t.

So, the volume of the cylinder is Ï€r²t, with r is radius of cylinder and t is height of cylinder.

Example:

Calculate the volume of a cylinder of radius 7 cm and height 20 cm!

Solution:

The radius of cylinder base (r) is 7 cm and the height of cylinder (t) is 20 cm. Therefore it goes in effect,

The volume of cylinder = Ï€r²t

= x 7 x 7 x 20

= 22 x 7 x 20

= 3,080

So the volume of the cylinder is 3,080 cm³.

A cylinder is fully filled by 5,024 cm³ of water. The cylinder base radius is 10 cm.

Calculate the height of the water!

Solution:

The volume of cylinder is 5,024 cm³ and the radius (r) of the base is 10 cm. Let, the height of water is t cm, therefore it goes in effect,

Cylinder volume = Ï€r²t

5,024 = 3.14 x 10² x t

5,024 = 3.14 x 100 x t

5,024 = 314 x t

t =

t = 16

So, the height of cylinder height is 16 cm.

Exercise:

A can of food that in form of cylinder has a height of 10 cm and a diameter of 7 cm. Determine the volume of the can!

A can of food that in form of cylinder has a radius of 10 cm. The can is loaded full by 11,000 cm³ of water. Determine the height of the cylinder!

A cylindrical drum of radius 30 cm and height 100 cm is fully filled with kerosene. Determine the volume of kerosene in the drum!

The lateral surface of a cylindrical can of biscuit is covered by a piece of gift paper. Ani wants to know the volume of the biscuit in the can. Ani then spreads out the cover of the can and finds out that the length is 88 cm and the width is 30 cm.

Determine the radius of the can?

Determine the volume of biscuit in the can!

A cylindrical can of paint has a height of 25 cm and a volume of 7,850 cm³. Determine the radius of the paint can!

Cone

Definition of Cone

A Cone is three-dimensional curved surfaces which has circular base of radius r and one side surface of revolution in the form of a sector of a circle.

The Elements of Cone

The elements of cone are as following:

A Cone consists of curved side surface called the lateral and the base in the form of circle.

PA and PC are called the radius of cone base

BP is called the height of a cone.

BA and BC are called the slant height.

Slant height (s) is a line that is connecting the peak of the cone to the point on base perimeter.

Figure 2.4 The component of a cone.

he Nets of Cone

Consider the Figure 2.5! Figure 2.5 (a) is a cone that has a radius r and a slant height s. If the cone as shown in Figure 2.5 (a) is sliced along its slant height s and along the curved base side (along the perimeter of base circle) then we will get the net of the cone as shown in Figure 2.5 (b).

Figure 2.5

A cone that has a radius r and a slant height s.

Radius of cone.

In the nets of a cone we see that the side of the base is a circle of radius r and a curved side (lateral of cone) is a sector of circle ABC of radius s.

The Area of the Surface of Cone

The surface of the cone consists of cone lateral and cone base. The cone lateral area (the circle sector area ABC with radius s) can be determined by the following comparison:

(The area of Circle sector ABC)/(The area of big circle )=( The lenght of Small arch AC)/(The perimeter of big circle )

(The area of cone lateral )/Ï€s²= 2Ï€r/2Ï€s

The area of cone lateral = (Ï€^2 s^2 r)/Ï€s= Ï€sr.

The area of cone surface = The area of cone lateral + the area of cone base

= Ï€rs + Ï€r² = Ï€r(s + r)

Example:

The radius of a cone base is 6 cm. If the cone height is 8 cm, determine:

the area of cone lateral, and

the area of cone surface.

Solution:

The height (s) of the cone slant can be determined as follows:

s=√(r²+t²)= √(6²+8²)= √(36+64)= √100=10

Hence,

The area of cone lateral = Ï€rs

= 3.14 x 6 x (10 + 6)

=188.4

So, The area of cone lateral is 188.4 cm².

The area of cone surface = Ï€r (s + r)

= 3.14 x 6 x (10 + 6)

= 18.84 x 16

= 301.44

So, the area of cone surface is 301.44 cm².

Exercise:

A cone has a base radius of 7 cm and slant height of 20 cm. Determine the cone surface area!

A farmer has a wizard’s hat with the slant height 28 cm. The hat is made of bamboo matting for the width of 1,232 cm². Determine diameter of the farmer hat!

A trumpet in the form of cone is made of carton. If the area of the carton to make the trumpet is 550 cm² and yield the trumpet slant height 25 cm, determine the long of the trumpet!

A building has a shape of cone with the diameter of 12 m and height of 8 m. Determine the lateral area of the building!

Mother will make a hat for younger brother. The hat is in the formed of cone that has a base diameter of 21 cm and height of 14 cm. Determine the area of the component to make the hat!

5. The Volume of Cone

Consider the following Figure 2.6!

Figure 2.6 A cone is the regular pyramid with a lot of numbers of facets.

If the base of regular pyramid as in Figure 2.6 has a lot of facets, hence the pallet form of the regular pyramid is almost seems like circle. Pyramid that its base is in the form of circle is called a cone.

The volume of the cone = 1/3 x cone base area x cone height.

= 1/3 x Ï€r² x t

Example:

Calculate the volume of the cone that has a radius of 3 cm and slant height of 5 cm!

Solution:

The base radius (r) of the cone is 3 cm and the slant height (s) of the cone is 5 cm. The height of the cone is determined as following:

s² = r² + t²

t² = s² - r² = 5² - 3² = 25 – 9 = 16

t = √16

t = 4

Therefore,

Volume of cylinder = 1/3 Ï€r²t

= 1/3 x 3.14 x 3² x 4

= 37. 68

So, the volume of the cone is 37. 68 cm³.

Exercise:

A cone with radius 9 cm and slant height 15 cm. Determine the volume of the cone!

Mrs. Tuti will make a ceremonial dish of yellow rice served in a cone shape. It has a height of 56 cm and base radius of 42 cm. Determine the volume of the dish made by Mrs. Tuti!

Let, the perimeter of a base circle of cone is 132 cm and a slant height is 35 cm. Determine the volume of the cone!

A truncated-cone has a height of 21 cm and a cap radius of 20 cm. Determine the volume of the cone!

A glass is in the form of a cone. The perimeter of the upper glass is 22 cm. If the height is 10 cm, determine the volume of the glass!

Sphere

The Definition of Sphere

Sphere is a three-dimensional shape described by the rotation of a semicircle about its diameter.

The Area of Sphere Surface

It is not like a cylinder and a cone that have curved edge, not also like a cone that has an angle point, sphere hasn’t curved edge and angle point. Sphere only has one curved surface called the surface of sphere.

The area of the surface of the sphere can be found by use the formula as follow:

The surface area of the sphere = 2 x the surface area of hemisphere

= 2 x (2 x the area of circle)

= 2 x (2 x Ï€r²)

= 4Ï€r².

Example:

The radius of the sphere is 10 cm. Determine the surface area of the sphere!

Solution:

The surface area of sphere = 4Ï€r²

= 4 x 3.14 x 10²

= 12.56 x 100

= 1,256

So, the surface area of sphere is 1,256 cm². Figure of sphere.

Exercise:

A sphere with radius 7 cm. Determine the surface area of the sphere!

To cover the surface of a sphere it needs material of 1,386 cm². Determine the diameter of the sphere!

A bowl in the form of hemisphere. The perimeter of the upper edge of the bowl is 31.4 cm. Determine surface area of the bowl!

A building has a roof in the form of hemisphere with diameter 14 m. The roof is made of glass. If the price of the glass is Rp 500,000.00 per m², determine the cost of the entire roof surface!

The perimeter of the bisector of sphere is 50.24 cm. Determine the surface area of the sphere!

The Volume of a Sphere

Volume of sphere can be determined by the following formula:

The volume of sphere = 2 x thevolume of hemisphere.

= 2 x (2 x volume of cone)

= 4 x 1/3Ï€r²t = 4/3Ï€r²t

= 4/3Ï€r³ (remember: height of cone (t) = radius of cone (r))

Example:

Calculate the volume of the sphere with radius 10 cm!

Solution:

The volume of sphere = 4/3Ï€r³ = 4/3 x 3,140 = 4,186.67

So, the volume of the sphere is 4,186.67 cm³.

Exercise:

A sphere has a radius of 14 cm. Calculate the volume of the sphere!

Let, the surface area of the sphere is 616 cm². Determine:

the radius of sphere, and

the volume of sphere.

Let, the volume of the sphere is 288Ï€ cm³. Determine the diameter of the sphere!

The inside part of an unripe coconut contains full of coconut milk. After the coconut milk is poured, its volume is 1,437 1/3 cm³ (the unripe coconut is assumed to have a spherical shape). Determine the diameter of the coconut if the thickness of the coconut and its shell is 0.5 cm! (Use Ï€ = 22/( 7))

An orange is sliced athwartly in equal size. It found that the orange diameter is 7 cm (the orange is assumed as spherical shape). Determine the volume of the half of orange!

Exercise 2

Choose the Correct Answer of the following problems

A sardine can is in the form of cylinder. The can has a radius of 7 cm and a height of 10 cm. The volume of the can is….

1,550 cm³ c. 1,504 cm³

1,540 cm³ d. 1450 cm³

The lateral area of the cylinder with radius 10 cm and height 20 cm is…

2,356 cm² c. 1,265 cm²

1,356 cm² d. 1,256 cm²

A can of drink in the form of cylinder on which its lateral side will be covered by paper. After the paper is unfolded, the size of the paper has a length of 62.8 cm and a width of 12 cm. The volume of the can is…

6,378 cm³ c. 3,678 cm³

3, 768 cm³ d. 3,578 cm³

A cylindrical shape of a can is filled up by wall paint. The can has a diameter of 20 cm and a height of 19 cm. The can volume is…

5,696 cm³ c. 5,969 cm³

5,966 cm³ d. 5,996 cm³

The younger brother buys milk as much as 2,009.6 cm³. He looks for the can to save the milk. He found the can in form of cylinder with the height of 10 cm. He found that the can is exactly filled by the milk. The diameter of the can is…

20 cm c. 16 cm

18 cm d. 14 cm

A can of drink is in formed of cylinder. The drink of the drink has a diameter of 2.8 cm and a high of 10 cm. The volume of the drink is….

63.6 cm³ c. 61.6 cm³

62.6 cm³ d. 60.6 cm³

A drum of kerosene has a height of 100 cm. The drum can load a full of kerosene as much as 138,600 cm³. The diameter of the drum is…

21 cm c. 35 cm

28 cm d. 42 cm

A cone has a height of 28 cm and a circle base radius of 21 cm. The lateral area is….

38,808 cm² c. 3,210 cm²

12,936 cm² d. 2,310 cm²

My brother buys boiled peanut in a vendor. The seller wraps it with the conical shape of paper that has a cap radius of 5 cm and a height of 15 cm. The volume of the cone is…

235.5 cm³ c. 392.5 cm³

382.5 cm³ d. 1,177.5 cm³

Tono was born in 21st. Therefore, on his birthday for 14th, a ceremonial dish of yellow rice in the shape of cone has its diameter of 14 cm and height of 21 cm. The volume of the shape is….

3,324 cm³ c. 1,780 cm³

3,234 cm³ d. 1,078 cm³

Ani will make a hat in form of cone that has a circle base perimeter of 44 cm. If the slant height is 10 cm, so, the lateral area of Ani’s hat is…

1,540 cm² c. 440 cm²

513.33 cm² d. 220 cm²

The area of paper that as the lateral of the cone is 753.6 cm². The slant height is 20 cm. The radius of the cone base is…

8 cm c. 12 cm

10 cm d. 16 cm

A cone is fully filled with 2.198 dm³ of fried nut. If the diameter of the cone’s cap is 20 cm, so the height of the cone is…

10 cm c. 21 cm

20 cm d. 22 cm

Brother sliced the orange athwartly into two similar parts. The diameter of the orange is 7 cm (the orange is assumed to have a spherical form). The area of the orange peel is…

616 cm² c. 154 cm²

166 cm² d. 145 cm²

Mother buys a watermelon. The perimeter of the surrounding watermelon is 62.8 cm. (the watermelon is assumed as sphere). The volume of the watermelon is…

628 cm³ c. 2,093.33 cm³

1.256 cm³ d.4,186.67 cm³

A sphere has a radius of 9 cm. The volume of the sphere is…

339.12 cm³ c. 1,017.36 cm³

678.24 cm³ d. 3,052.08 cm³

The surface area of the sphere is 1,808.64 cm². The volume of the sphere is…

2,411.52 cm³ c. 7,234.56 cm³

4,823.04 cm³ d. 7,236.56 cm³

A ball is fully filled up with 1,4371/3 cm³ of sands. The diameter of the ball is…

7 cm c. 14 cm

12 cm d. 21 cm

A ball has a volume of 904.32 cm³. The surface area of the ball is…

2,1712.96 cm² c. 425.16 cm²

452.16 cm² d. 254.16 cm²

A ball has diameter 24 cm. The surface of the ball will be covered with decorative paper. The area of the paper is…

1,880.64 cm² c. 150.72 cm²

1,808.64 cm² d. 105.72 cm²

Solve the following problems

A gasholder is in the form of cylinder. It is fully filled up by 2,355 dm³ of kerosene. If the height of the gasholder is 300 cm, determine:

the diameter of the gasholder, and

the surface area of the gasholder.

Budi wants to make a ceremonial dish of yellow rice in the shape of cone (tumpeng) that has a height of 30 cm. If he wants the base area of the tumpeng is 616 cm²,

How long the radius of the tumpeng?

How big the volume of the tumpeng?

Brother buys pop corn in conic paper bag. If the volume of the pop corn is 314 cm³ and the cap diameter is 10 cm, calculate:

The height of the bag, and

The paper area that is the pop corn packer.

Andi has two globes made of glass. One of the globes has a diameter of 15 cm and the other has a diameter of 0.5 cm.

What the volume of the globe?

What the area of the glass to make the globe?

Three-dimensional Curved Surfaces Shape

Basic Competency:

To identify the elements of cylinder, cone, and sphere.

To estimate the lateral area and volume of cylinder, cone, and sphere.

To solve the problem that is related with cylinder, cone, and sphere.

What will be learned with this chapter?

Cylinder

Cone

Sphere

Cylinder

Definition of Cylinder

Cylinder or tube is a three-dimensional shape consists of two circular bases of equal area that are in parallel planes, and are connected by a lateral surface that intersects the boundaries of the bases.

The Elements of Cylinder

The elements of the cylinder are as follows:

Cylinder has an upper side (cap) and under side (base) in form of congruent circle (equal in its form and its size).

AB is called the diameter of cylinder base.

PE, PA, and PB are called the radius of cylinder base.

BC and AD is called the height of cylinder

The surface which is perpendicular to the bases is called the lateral of cylinder.

Two circular bases

The cap, the base and the lateral surface is called the surface of cylinder.

The Nets of Cylinder

Pay attention to the Figure 2.2 (a) is a cylinder that the radius is r and the height is t. If the cylinder as in the Figure 2.2 (a) is sliced as long as its height line (along AD or BC) and along its curved sides (along its perimeter of circle base and cap) as in the Figure 2.2 (b) then you will get the net of the cylinder as it seemed in the Figure 2.2 (c).

Figure 2.2

The cylinder which has the radius r and the height t.

The cylinder that is sliced along its curved sides of the base and the cap and along its height of cylinder.

The nets of cylinder.

In the nets of cylinder, it shows that the upper side (cap) and under side (base) are circles which has a radius r, while the net of the curved side (cylinder lateral) is rectangle ABCD.

Surface Area of Cylinder

The surface of the cylinder consists of cylinder lateral, upper side (cap), and under side (base). The cylinder lateral side is formed by a rectangle with length 2Ï€r and width t.

The following formulas of areas are usually used in the cylinder:

Example:

The radius of the base of the cylinder is 7 cm and the height is 10 cm. Determine:

the length of the the lateral net.

the area of lateral surface, and

the area of cylinder surface.

Solution:

The height (t) is 10 cm and the cylinder base radius (r) is 7 cm.

The length of the the lateral net = 2Ï€r

= 2 x x 7 = 44

So, the length of the the lateral net is 44 cm.

The area of lateral surface = 2Ï€r x t

= 44 x 10

= 440

So, the area of lateral surface is 440 cm².

The area of cylinder surface = 2Ï€r(t + r )

= 44 x (10 + 7)

= 44 x 17

= 748

So, area of cylinder surface is 749 cm².

Exercise:

An un-caped cylinder of glass has a diameter of 7 cm and a height of 20 cm. Determine:

the area of lateral surface, and

the area of cylinder surface.

A water pipe in the form of cylinder has the radius of 2.1 cm and the length of 28 cm. If the water pipe is hollow out at both its ends, determine the surface area of the pipe!

A clay-made flowerpot is in the form of cylinder. The base radius of the pot is 10 cm and the height is 20 cm. If the un-caped flowerpot will be painted on its lateral side and its base, determine the surface area should be painted!

A tart-cake for celebrating a birthday is in the form of cylinder with diameter 28 cm and height 8 cm. If it coated with chocolate, determine the surface area of the chocolate!

A can of milk is in the form of cylinder with diameter 7 cm and height 8 cm. The lateral side will be coated with paper containing information about the milk product. Determine the area of the paper!

5. The Volume of Cylinder

The way to identify the volume of cylinder is identical with that of the volume of a right prism. Consider the Figure 2.3!

Figure 2.3 A cylinder is a right prism with circular bases

If the base and the cap of regular prism as it shown in Figure 2.3 has a lot of facets so the form of the base and cap of the prism will close to the form of circle. The right prism with circular bases is called cylinder. Therefore, you will get the volume of the cylinder as the following:

The volume of the cylinder = the area of the base x the height of the cylinder

= the area of the circle x the height of the cylinder

= (Ï€r²) x t

= Ï€r²t.

So, the volume of the cylinder is Ï€r²t, with r is radius of cylinder and t is height of cylinder.

Example:

Calculate the volume of a cylinder of radius 7 cm and height 20 cm!

Solution:

The radius of cylinder base (r) is 7 cm and the height of cylinder (t) is 20 cm. Therefore it goes in effect,

The volume of cylinder = Ï€r²t

= x 7 x 7 x 20

= 22 x 7 x 20

= 3,080

So the volume of the cylinder is 3,080 cm³.

A cylinder is fully filled by 5,024 cm³ of water. The cylinder base radius is 10 cm.

Calculate the height of the water!

Solution:

The volume of cylinder is 5,024 cm³ and the radius (r) of the base is 10 cm. Let, the height of water is t cm, therefore it goes in effect,

Cylinder volume = Ï€r²t

5,024 = 3.14 x 10² x t

5,024 = 3.14 x 100 x t

5,024 = 314 x t

t =

t = 16

So, the height of cylinder height is 16 cm.

Exercise:

A can of food that in form of cylinder has a height of 10 cm and a diameter of 7 cm. Determine the volume of the can!

A can of food that in form of cylinder has a radius of 10 cm. The can is loaded full by 11,000 cm³ of water. Determine the height of the cylinder!

A cylindrical drum of radius 30 cm and height 100 cm is fully filled with kerosene. Determine the volume of kerosene in the drum!

The lateral surface of a cylindrical can of biscuit is covered by a piece of gift paper. Ani wants to know the volume of the biscuit in the can. Ani then spreads out the cover of the can and finds out that the length is 88 cm and the width is 30 cm.

Determine the radius of the can?

Determine the volume of biscuit in the can!

A cylindrical can of paint has a height of 25 cm and a volume of 7,850 cm³. Determine the radius of the paint can!

Cone

Definition of Cone

A Cone is three-dimensional curved surfaces which has circular base of radius r and one side surface of revolution in the form of a sector of a circle.

The Elements of Cone

The elements of cone are as following:

A Cone consists of curved side surface called the lateral and the base in the form of circle.

PA and PC are called the radius of cone base

BP is called the height of a cone.

BA and BC are called the slant height.

Slant height (s) is a line that is connecting the peak of the cone to the point on base perimeter.

Figure 2.4 The component of a cone.

he Nets of Cone

Consider the Figure 2.5! Figure 2.5 (a) is a cone that has a radius r and a slant height s. If the cone as shown in Figure 2.5 (a) is sliced along its slant height s and along the curved base side (along the perimeter of base circle) then we will get the net of the cone as shown in Figure 2.5 (b).

Figure 2.5

A cone that has a radius r and a slant height s.

Radius of cone.

In the nets of a cone we see that the side of the base is a circle of radius r and a curved side (lateral of cone) is a sector of circle ABC of radius s.

The Area of the Surface of Cone

The surface of the cone consists of cone lateral and cone base. The cone lateral area (the circle sector area ABC with radius s) can be determined by the following comparison:

(The area of Circle sector ABC)/(The area of big circle )=( The lenght of Small arch AC)/(The perimeter of big circle )

(The area of cone lateral )/Ï€s²= 2Ï€r/2Ï€s

The area of cone lateral = (Ï€^2 s^2 r)/Ï€s= Ï€sr.

The area of cone surface = The area of cone lateral + the area of cone base

= Ï€rs + Ï€r² = Ï€r(s + r)

Example:

The radius of a cone base is 6 cm. If the cone height is 8 cm, determine:

the area of cone lateral, and

the area of cone surface.

Solution:

The height (s) of the cone slant can be determined as follows:

s=√(r²+t²)= √(6²+8²)= √(36+64)= √100=10

Hence,

The area of cone lateral = Ï€rs

= 3.14 x 6 x (10 + 6)

=188.4

So, The area of cone lateral is 188.4 cm².

The area of cone surface = Ï€r (s + r)

= 3.14 x 6 x (10 + 6)

= 18.84 x 16

= 301.44

So, the area of cone surface is 301.44 cm².

Exercise:

A cone has a base radius of 7 cm and slant height of 20 cm. Determine the cone surface area!

A farmer has a wizard’s hat with the slant height 28 cm. The hat is made of bamboo matting for the width of 1,232 cm². Determine diameter of the farmer hat!

A trumpet in the form of cone is made of carton. If the area of the carton to make the trumpet is 550 cm² and yield the trumpet slant height 25 cm, determine the long of the trumpet!

A building has a shape of cone with the diameter of 12 m and height of 8 m. Determine the lateral area of the building!

Mother will make a hat for younger brother. The hat is in the formed of cone that has a base diameter of 21 cm and height of 14 cm. Determine the area of the component to make the hat!

5. The Volume of Cone

Consider the following Figure 2.6!

Figure 2.6 A cone is the regular pyramid with a lot of numbers of facets.

If the base of regular pyramid as in Figure 2.6 has a lot of facets, hence the pallet form of the regular pyramid is almost seems like circle. Pyramid that its base is in the form of circle is called a cone.

The volume of the cone = 1/3 x cone base area x cone height.

= 1/3 x Ï€r² x t

Example:

Calculate the volume of the cone that has a radius of 3 cm and slant height of 5 cm!

Solution:

The base radius (r) of the cone is 3 cm and the slant height (s) of the cone is 5 cm. The height of the cone is determined as following:

s² = r² + t²

t² = s² - r² = 5² - 3² = 25 – 9 = 16

t = √16

t = 4

Therefore,

Volume of cylinder = 1/3 Ï€r²t

= 1/3 x 3.14 x 3² x 4

= 37. 68

So, the volume of the cone is 37. 68 cm³.

Exercise:

A cone with radius 9 cm and slant height 15 cm. Determine the volume of the cone!

Mrs. Tuti will make a ceremonial dish of yellow rice served in a cone shape. It has a height of 56 cm and base radius of 42 cm. Determine the volume of the dish made by Mrs. Tuti!

Let, the perimeter of a base circle of cone is 132 cm and a slant height is 35 cm. Determine the volume of the cone!

A truncated-cone has a height of 21 cm and a cap radius of 20 cm. Determine the volume of the cone!

A glass is in the form of a cone. The perimeter of the upper glass is 22 cm. If the height is 10 cm, determine the volume of the glass!

Sphere

The Definition of Sphere

Sphere is a three-dimensional shape described by the rotation of a semicircle about its diameter.

The Area of Sphere Surface

It is not like a cylinder and a cone that have curved edge, not also like a cone that has an angle point, sphere hasn’t curved edge and angle point. Sphere only has one curved surface called the surface of sphere.

The area of the surface of the sphere can be found by use the formula as follow:

The surface area of the sphere = 2 x the surface area of hemisphere

= 2 x (2 x the area of circle)

= 2 x (2 x Ï€r²)

= 4Ï€r².

Example:

The radius of the sphere is 10 cm. Determine the surface area of the sphere!

Solution:

The surface area of sphere = 4Ï€r²

= 4 x 3.14 x 10²

= 12.56 x 100

= 1,256

So, the surface area of sphere is 1,256 cm². Figure of sphere.

Exercise:

A sphere with radius 7 cm. Determine the surface area of the sphere!

To cover the surface of a sphere it needs material of 1,386 cm². Determine the diameter of the sphere!

A bowl in the form of hemisphere. The perimeter of the upper edge of the bowl is 31.4 cm. Determine surface area of the bowl!

A building has a roof in the form of hemisphere with diameter 14 m. The roof is made of glass. If the price of the glass is Rp 500,000.00 per m², determine the cost of the entire roof surface!

The perimeter of the bisector of sphere is 50.24 cm. Determine the surface area of the sphere!

The Volume of a Sphere

Volume of sphere can be determined by the following formula:

The volume of sphere = 2 x thevolume of hemisphere.

= 2 x (2 x volume of cone)

= 4 x 1/3Ï€r²t = 4/3Ï€r²t

= 4/3Ï€r³ (remember: height of cone (t) = radius of cone (r))

Example:

Calculate the volume of the sphere with radius 10 cm!

Solution:

The volume of sphere = 4/3Ï€r³ = 4/3 x 3,140 = 4,186.67

So, the volume of the sphere is 4,186.67 cm³.

Exercise:

A sphere has a radius of 14 cm. Calculate the volume of the sphere!

Let, the surface area of the sphere is 616 cm². Determine:

the radius of sphere, and

the volume of sphere.

Let, the volume of the sphere is 288Ï€ cm³. Determine the diameter of the sphere!

The inside part of an unripe coconut contains full of coconut milk. After the coconut milk is poured, its volume is 1,437 1/3 cm³ (the unripe coconut is assumed to have a spherical shape). Determine the diameter of the coconut if the thickness of the coconut and its shell is 0.5 cm! (Use Ï€ = 22/( 7))

An orange is sliced athwartly in equal size. It found that the orange diameter is 7 cm (the orange is assumed as spherical shape). Determine the volume of the half of orange!

Exercise 2

Choose the Correct Answer of the following problems

A sardine can is in the form of cylinder. The can has a radius of 7 cm and a height of 10 cm. The volume of the can is….

1,550 cm³ c. 1,504 cm³

1,540 cm³ d. 1450 cm³

The lateral area of the cylinder with radius 10 cm and height 20 cm is…

2,356 cm² c. 1,265 cm²

1,356 cm² d. 1,256 cm²

A can of drink in the form of cylinder on which its lateral side will be covered by paper. After the paper is unfolded, the size of the paper has a length of 62.8 cm and a width of 12 cm. The volume of the can is…

6,378 cm³ c. 3,678 cm³

3, 768 cm³ d. 3,578 cm³

A cylindrical shape of a can is filled up by wall paint. The can has a diameter of 20 cm and a height of 19 cm. The can volume is…

5,696 cm³ c. 5,969 cm³

5,966 cm³ d. 5,996 cm³

The younger brother buys milk as much as 2,009.6 cm³. He looks for the can to save the milk. He found the can in form of cylinder with the height of 10 cm. He found that the can is exactly filled by the milk. The diameter of the can is…

20 cm c. 16 cm

18 cm d. 14 cm

A can of drink is in formed of cylinder. The drink of the drink has a diameter of 2.8 cm and a high of 10 cm. The volume of the drink is….

63.6 cm³ c. 61.6 cm³

62.6 cm³ d. 60.6 cm³

A drum of kerosene has a height of 100 cm. The drum can load a full of kerosene as much as 138,600 cm³. The diameter of the drum is…

21 cm c. 35 cm

28 cm d. 42 cm

A cone has a height of 28 cm and a circle base radius of 21 cm. The lateral area is….

38,808 cm² c. 3,210 cm²

12,936 cm² d. 2,310 cm²

My brother buys boiled peanut in a vendor. The seller wraps it with the conical shape of paper that has a cap radius of 5 cm and a height of 15 cm. The volume of the cone is…

235.5 cm³ c. 392.5 cm³

382.5 cm³ d. 1,177.5 cm³

Tono was born in 21st. Therefore, on his birthday for 14th, a ceremonial dish of yellow rice in the shape of cone has its diameter of 14 cm and height of 21 cm. The volume of the shape is….

3,324 cm³ c. 1,780 cm³

3,234 cm³ d. 1,078 cm³

Ani will make a hat in form of cone that has a circle base perimeter of 44 cm. If the slant height is 10 cm, so, the lateral area of Ani’s hat is…

1,540 cm² c. 440 cm²

513.33 cm² d. 220 cm²

The area of paper that as the lateral of the cone is 753.6 cm². The slant height is 20 cm. The radius of the cone base is…

8 cm c. 12 cm

10 cm d. 16 cm

A cone is fully filled with 2.198 dm³ of fried nut. If the diameter of the cone’s cap is 20 cm, so the height of the cone is…

10 cm c. 21 cm

20 cm d. 22 cm

Brother sliced the orange athwartly into two similar parts. The diameter of the orange is 7 cm (the orange is assumed to have a spherical form). The area of the orange peel is…

616 cm² c. 154 cm²

166 cm² d. 145 cm²

Mother buys a watermelon. The perimeter of the surrounding watermelon is 62.8 cm. (the watermelon is assumed as sphere). The volume of the watermelon is…

628 cm³ c. 2,093.33 cm³

1.256 cm³ d.4,186.67 cm³

A sphere has a radius of 9 cm. The volume of the sphere is…

339.12 cm³ c. 1,017.36 cm³

678.24 cm³ d. 3,052.08 cm³

The surface area of the sphere is 1,808.64 cm². The volume of the sphere is…

2,411.52 cm³ c. 7,234.56 cm³

4,823.04 cm³ d. 7,236.56 cm³

A ball is fully filled up with 1,4371/3 cm³ of sands. The diameter of the ball is…

7 cm c. 14 cm

12 cm d. 21 cm

A ball has a volume of 904.32 cm³. The surface area of the ball is…

2,1712.96 cm² c. 425.16 cm²

452.16 cm² d. 254.16 cm²

A ball has diameter 24 cm. The surface of the ball will be covered with decorative paper. The area of the paper is…

1,880.64 cm² c. 150.72 cm²

1,808.64 cm² d. 105.72 cm²

Solve the following problems

A gasholder is in the form of cylinder. It is fully filled up by 2,355 dm³ of kerosene. If the height of the gasholder is 300 cm, determine:

the diameter of the gasholder, and

the surface area of the gasholder.

Budi wants to make a ceremonial dish of yellow rice in the shape of cone (tumpeng) that has a height of 30 cm. If he wants the base area of the tumpeng is 616 cm²,

How long the radius of the tumpeng?

How big the volume of the tumpeng?

Brother buys pop corn in conic paper bag. If the volume of the pop corn is 314 cm³ and the cap diameter is 10 cm, calculate:

The height of the bag, and

The paper area that is the pop corn packer.

Andi has two globes made of glass. One of the globes has a diameter of 15 cm and the other has a diameter of 0.5 cm.

What the volume of the globe?

What the area of the glass to make the globe?

### Mathematics for Junior High School

By Marsigit

Proportion

Basic Competency :

To Use Proportion to solve the problems.

Have you ever seen a house, building, monument, or sport stadium?. Of course you have. How to build them? are they just built anyway?. Of course, they did not. Before to build them the worker develop the plan first. For example, the needed materials, the cost, and design a scale model. So what is the definition of a scale model? A scale model is a decried building with a certain proportion scale. By making the scale model, you can see the detailed of the buildings anyway.

What will you learn in this chapter?

• Scaled Picture

• Direct Proportion

• Inverse Proportions

A. Scaled Pictures

Scaled picture is a picture with a certain proportion or scale. Scaled picture

is made to represent the reality with a ratio (scale). Some example of scaled pictures are maps, design of a building, and cars model.

1. Definition of Scale

Scale is the ratio between the size of the picture and the actual size.

Scale is usually written as a ratio. For example, the map of Indonesia has a scale of 1 : 1.000.000. It means that 1 cm on the map represents 1.000.000 cm = 10 km on the actual size.

Example

The distance between Yogyakarta and Cilacap is 150 km. Find the distance between the cities on the map with the scale of 1 : 5.000.000.

Solution:

The real distance between Yogyakarta and Cilacap is 150 km = 15.000.000 cm.

The scale of the map is 1 : 5.000.000 =

As we know that,

Thus,

Distance on the map =

Hence, the distance between Yogyakarta and Cilacap on the map is 3 cm.

Exercise

1. Explain the meaning of the scale of 1 : 1.000.000 ?

2. Find the scale on a map if 5 cm on the map represent 30 km.

3. Let, the distance between Denpasar and Singaraja on the map is 3 cm. Find the real distance if the scale of the map is 1 : 6.000.000.

4. The length of a hedge is 16 m. Find the length of the hedge on the sketch with the scale of 1 : 400.

5. The distance between Makassar and Pare-pare is 155 km. If the distance between the cities on the map is 10 cm, find the scale of the map.

2. Scale Factor

Scale factor is the ratio between the size of the model and the actual size. Scale factor is used to determine the reduction or the enlargement of a scaled model or design.

If you enlarge a scaled model, you will get the scale factor k > 1. If you reduce a scaled model, you will get the scale factor 0 < k < 1.

Example

The length and the width of the handicraft model is 10 cm and 7,5 cm. The real length of the handicraft is 40 cm. Find the scale factor of the handicraft model, and then find the real width of the handicraft.

Solution:

The length of the handicraft model is 10 cm, and the width is 7.5 cm.

The real length is 40 cm.

• Scale factor,

Therefore, the scale factor of the handicraft model is .

•

Therefore, the real width of the handicraft is 30 cm.

Exercise

1. Complete the following ratio with the right number.

a. b.

2. The length and the width of the scaled picture of a football yard are 11 cm and 7 cm. Find the scale factor and the real width of the football yard if the real length is 110 m.

[Gambar lapangan sepak bola]

3. The size of photo is 4 x 6 cm. It will be enlarged by the scale factor of 1.5. Find the height of the photo after the enlargement.

[Gambar foto sawah]

4. The ratio of the length and the width of a photograph is 2 : 1. Find the length of the photograph if it’s width is reduced to be 4 cm.

5. The length and the width of a field are 100 m and 30 m. Rita wants to sketch the field. Determine the length of the sketch if the width of the sketch is 3 cm.

B. Direct Proportion

1. The Concept of Direct Proportion

The illustration below shows the example of a direct proportion. At the unit measurement of a gas station it shows the amount of the gas (liter) and the money should be paid by the buyer.

Amount of the gas (litre) Price (Rp.)

1 6.000

2 12.000

10 60.000

23 138.000

It shows that the more the gas, the more the price should be paid. This proportion is called a direct proportion. Direct proportion is indicated by a series of number. See the table below.

Number Direct Proportion

2, 7, 8, 28

18, 20, 27, 30

8, 4, 2, 1

Example

Determine the direct proportions of the following numbers.

1. 3, 7, 9, and 21 2. 52, 26, 12, and 6 3. 24, 16, 15, and 10

Solution:

1. 2. 3.

Exercise

Determine the direct proportions of the following numbers.

1. 450, 210, 60 and 28

2. 24, 42, 248 and 434

3. 320, 240, 48 and 36

4. 9, 26, 63, and 182

5. 6, 7, 24 and 28

2. Calculating Direct Proportion

There are two ways to calculate a direct proportion i.e. by unit method and by proportion.

a. Unit Method

To calculate a direct proportion by unit method, first the letter you choose must represent the unit of measure common to each part.

Example

Mom needs 10 eggs to make two cakes. How many eggs are needed for 7 cakes ?

Solution

First, find out how many eggs needed for one cake.

Two cakes need 10 eggs, therefore one cake needs,

eggs.

So, mom needs 7 x 5 eggs = 35 eggs to make 7 cakes.

Exercise

1. A motorcycle needs 4 liter of gas for 144 km. How much gas is needed for 54 km?

2. If the price of one dozen of spoons is Rp 18.000,00, how much it costs for 27 spoons?

b. Proportion

You can determine a direct proportion by proportion. See the example below.

Example

The ratio between the number of girls and boys in a class is 5 : 9. The total number of students in the class is 28 students. Determine the number of the girls and the boys in the class ?

Solution

The ratio between the girls and the boys in the class is 5 : 9.

The total number of students in the class is 28 students.

So the total boys in the class is,

students

By the same way you’ll find the total girls in the class, that is,

students

Check the answer by adding the total girls and the total boys in the class.

You find,

10 + 18 = 28 students.

Exercise

1. The perimeter of a garden is 30 meter. The ratio between the length and the width of the garden is 3 : 2. Determine the length and the width of the garden.

2. In the shelve there are Mathematic books, Indonesian Language books, and Physics books with the ratio of 4 : 2 : 3. Determine the numbers of Physics books and Mathematics books if the Indonesian Language books is 6 books.

3. The sum of two numbers is 72. The ratio between both two numbers is 4 : 5. Determine each of the numbers.

4. The total amount of Mr. Tanto’s money in two Banks are Rp 20.000.000,00. The ratio between the sum of money in each Bank is 3 : 2. How much Mr. Tanto’s has the money in each Bank?

5. Mother has spent Rp 60.000,00 to buy meat, rice, and vegetables by the ratio of 10 : 9 : 5. Determine the price of each of the goods.

2. The Graph of Direct Proportion

Consider the gasoline prices below. Can you draw a graph of the data from the table? To draw a graph, you need to make two axis, that is horizontal axis and vertical axis.

Indicate the horizontal line as the amount of gas and the vertical as the prices of the gasoline.

Amount of the gas (litre) Price (Rp.)

1 6.000

2 12.000

10 60.000

23 138.000

The graph is shows as follow.

Exercise

The table below shows the relation between the time and the distance of the journey by a car.

Time (hrs) 1 2 3 4 5

Distance (km) 60 120 … … …

1. Copy the table on your work-sheet and fill in the blank with appropriate numbers.

2. Sketch the graph representing the data of the table.

3. How long the distance traveling by the car for 11 hrs ?

4. How long time the car needs to travel for 780 km ?

5. How long time the car needs to travel for 20 km ?

C. Inverse Proportions

1. The Concept of Inverse Proportion

You can find some inverse proportions in daily life. For example, the ratio between the numbers of days needed to construct a house and the number of workers is the inverse proportion. The more the workers the sooner the house will be finished.

2. Calculating the Inverse Proportions

You can calculate the inverse proportion in two ways, based on the result of multiplication and proportion

a. Based on multiplication

An inverse proportion can be calculated in term of the result of multiplication. Look at the following example.

Example

A house can be built in two months by 10 workers. How long time it can be done by six workers?

Solution:

Make a table to help you to solve the problem.

Number of Workers

Time

10 2 months (60 days)

6

The multiplication results of each raw should be the similar, so you have an inverse proportion as follows.

It is concluded that if there are six workers, it can be done in 100 days.

Exercise

1. Complete the following proportion with the correct number!

a. b. c.

2. A bridge can be done by 18 workers in 20 days. If there are 12 workers, how long time it can be done?

b. Based on Proportion

An inverse proportion can be calculated by proportion method. Look at the

following example.

Example

Let, a box of candy has been distributed to 20 children, in such a way that

each child gets 12 candies. If there are 30 children, how many candies each child will

get?

Solution

Look at the following table!

Number of children

candies for each child

20 12

30

Based on the proportion, you can determine the amount of candies for each child.

. Hence, if there are 30 children. every child will get 8 candies.

Exercise

1. A basket of carrots can be finished by 10 rabbits in 6 days. How long time for 15 rabbits to eat them?

2. Father needs 20 minutes to go to the office by car in 60 km/hrs. Today he just takes 15 minutes to go to his office. Determine the speed of the car used to go to his office today!

3. The Graph of the Inverse Proportion

Do you know how to measure the speed? The speed can be measured by the following formula , with = velocity, distance, and = time. For example, you will go by bus to the city which the distance of 180 km from your house. The time you take for the trip depends on the speed of the bus you ride, as it shown in the following table.

[Tabel t –v]

Based on the table above, you can make a velocity graph towards time as follows.

[Grafik kecepatan-waktu]

Exercise

Anton is going to school by bike. He measures the speed of his bike and the time he needs to arrived at school on time. Look at the following result of Anton’s measurement!

(minute)

20 40 50

(minute)

15 7,5 6

1. Sketch a graph based on the table above!

2. How far the distance of Anton’s house to school? (use the formula )

3. If Anton want to arrive at school in 30 minutes, what does the speed of Anton’s bike?

Proportion

Basic Competency :

To Use Proportion to solve the problems.

Have you ever seen a house, building, monument, or sport stadium?. Of course you have. How to build them? are they just built anyway?. Of course, they did not. Before to build them the worker develop the plan first. For example, the needed materials, the cost, and design a scale model. So what is the definition of a scale model? A scale model is a decried building with a certain proportion scale. By making the scale model, you can see the detailed of the buildings anyway.

What will you learn in this chapter?

• Scaled Picture

• Direct Proportion

• Inverse Proportions

A. Scaled Pictures

Scaled picture is a picture with a certain proportion or scale. Scaled picture

is made to represent the reality with a ratio (scale). Some example of scaled pictures are maps, design of a building, and cars model.

1. Definition of Scale

Scale is the ratio between the size of the picture and the actual size.

Scale is usually written as a ratio. For example, the map of Indonesia has a scale of 1 : 1.000.000. It means that 1 cm on the map represents 1.000.000 cm = 10 km on the actual size.

Example

The distance between Yogyakarta and Cilacap is 150 km. Find the distance between the cities on the map with the scale of 1 : 5.000.000.

Solution:

The real distance between Yogyakarta and Cilacap is 150 km = 15.000.000 cm.

The scale of the map is 1 : 5.000.000 =

As we know that,

Thus,

Distance on the map =

Hence, the distance between Yogyakarta and Cilacap on the map is 3 cm.

Exercise

1. Explain the meaning of the scale of 1 : 1.000.000 ?

2. Find the scale on a map if 5 cm on the map represent 30 km.

3. Let, the distance between Denpasar and Singaraja on the map is 3 cm. Find the real distance if the scale of the map is 1 : 6.000.000.

4. The length of a hedge is 16 m. Find the length of the hedge on the sketch with the scale of 1 : 400.

5. The distance between Makassar and Pare-pare is 155 km. If the distance between the cities on the map is 10 cm, find the scale of the map.

2. Scale Factor

Scale factor is the ratio between the size of the model and the actual size. Scale factor is used to determine the reduction or the enlargement of a scaled model or design.

If you enlarge a scaled model, you will get the scale factor k > 1. If you reduce a scaled model, you will get the scale factor 0 < k < 1.

Example

The length and the width of the handicraft model is 10 cm and 7,5 cm. The real length of the handicraft is 40 cm. Find the scale factor of the handicraft model, and then find the real width of the handicraft.

Solution:

The length of the handicraft model is 10 cm, and the width is 7.5 cm.

The real length is 40 cm.

• Scale factor,

Therefore, the scale factor of the handicraft model is .

•

Therefore, the real width of the handicraft is 30 cm.

Exercise

1. Complete the following ratio with the right number.

a. b.

2. The length and the width of the scaled picture of a football yard are 11 cm and 7 cm. Find the scale factor and the real width of the football yard if the real length is 110 m.

[Gambar lapangan sepak bola]

3. The size of photo is 4 x 6 cm. It will be enlarged by the scale factor of 1.5. Find the height of the photo after the enlargement.

[Gambar foto sawah]

4. The ratio of the length and the width of a photograph is 2 : 1. Find the length of the photograph if it’s width is reduced to be 4 cm.

5. The length and the width of a field are 100 m and 30 m. Rita wants to sketch the field. Determine the length of the sketch if the width of the sketch is 3 cm.

B. Direct Proportion

1. The Concept of Direct Proportion

The illustration below shows the example of a direct proportion. At the unit measurement of a gas station it shows the amount of the gas (liter) and the money should be paid by the buyer.

Amount of the gas (litre) Price (Rp.)

1 6.000

2 12.000

10 60.000

23 138.000

It shows that the more the gas, the more the price should be paid. This proportion is called a direct proportion. Direct proportion is indicated by a series of number. See the table below.

Number Direct Proportion

2, 7, 8, 28

18, 20, 27, 30

8, 4, 2, 1

Example

Determine the direct proportions of the following numbers.

1. 3, 7, 9, and 21 2. 52, 26, 12, and 6 3. 24, 16, 15, and 10

Solution:

1. 2. 3.

Exercise

Determine the direct proportions of the following numbers.

1. 450, 210, 60 and 28

2. 24, 42, 248 and 434

3. 320, 240, 48 and 36

4. 9, 26, 63, and 182

5. 6, 7, 24 and 28

2. Calculating Direct Proportion

There are two ways to calculate a direct proportion i.e. by unit method and by proportion.

a. Unit Method

To calculate a direct proportion by unit method, first the letter you choose must represent the unit of measure common to each part.

Example

Mom needs 10 eggs to make two cakes. How many eggs are needed for 7 cakes ?

Solution

First, find out how many eggs needed for one cake.

Two cakes need 10 eggs, therefore one cake needs,

eggs.

So, mom needs 7 x 5 eggs = 35 eggs to make 7 cakes.

Exercise

1. A motorcycle needs 4 liter of gas for 144 km. How much gas is needed for 54 km?

2. If the price of one dozen of spoons is Rp 18.000,00, how much it costs for 27 spoons?

b. Proportion

You can determine a direct proportion by proportion. See the example below.

Example

The ratio between the number of girls and boys in a class is 5 : 9. The total number of students in the class is 28 students. Determine the number of the girls and the boys in the class ?

Solution

The ratio between the girls and the boys in the class is 5 : 9.

The total number of students in the class is 28 students.

So the total boys in the class is,

students

By the same way you’ll find the total girls in the class, that is,

students

Check the answer by adding the total girls and the total boys in the class.

You find,

10 + 18 = 28 students.

Exercise

1. The perimeter of a garden is 30 meter. The ratio between the length and the width of the garden is 3 : 2. Determine the length and the width of the garden.

2. In the shelve there are Mathematic books, Indonesian Language books, and Physics books with the ratio of 4 : 2 : 3. Determine the numbers of Physics books and Mathematics books if the Indonesian Language books is 6 books.

3. The sum of two numbers is 72. The ratio between both two numbers is 4 : 5. Determine each of the numbers.

4. The total amount of Mr. Tanto’s money in two Banks are Rp 20.000.000,00. The ratio between the sum of money in each Bank is 3 : 2. How much Mr. Tanto’s has the money in each Bank?

5. Mother has spent Rp 60.000,00 to buy meat, rice, and vegetables by the ratio of 10 : 9 : 5. Determine the price of each of the goods.

2. The Graph of Direct Proportion

Consider the gasoline prices below. Can you draw a graph of the data from the table? To draw a graph, you need to make two axis, that is horizontal axis and vertical axis.

Indicate the horizontal line as the amount of gas and the vertical as the prices of the gasoline.

Amount of the gas (litre) Price (Rp.)

1 6.000

2 12.000

10 60.000

23 138.000

The graph is shows as follow.

Exercise

The table below shows the relation between the time and the distance of the journey by a car.

Time (hrs) 1 2 3 4 5

Distance (km) 60 120 … … …

1. Copy the table on your work-sheet and fill in the blank with appropriate numbers.

2. Sketch the graph representing the data of the table.

3. How long the distance traveling by the car for 11 hrs ?

4. How long time the car needs to travel for 780 km ?

5. How long time the car needs to travel for 20 km ?

C. Inverse Proportions

1. The Concept of Inverse Proportion

You can find some inverse proportions in daily life. For example, the ratio between the numbers of days needed to construct a house and the number of workers is the inverse proportion. The more the workers the sooner the house will be finished.

2. Calculating the Inverse Proportions

You can calculate the inverse proportion in two ways, based on the result of multiplication and proportion

a. Based on multiplication

An inverse proportion can be calculated in term of the result of multiplication. Look at the following example.

Example

A house can be built in two months by 10 workers. How long time it can be done by six workers?

Solution:

Make a table to help you to solve the problem.

Number of Workers

Time

10 2 months (60 days)

6

The multiplication results of each raw should be the similar, so you have an inverse proportion as follows.

It is concluded that if there are six workers, it can be done in 100 days.

Exercise

1. Complete the following proportion with the correct number!

a. b. c.

2. A bridge can be done by 18 workers in 20 days. If there are 12 workers, how long time it can be done?

b. Based on Proportion

An inverse proportion can be calculated by proportion method. Look at the

following example.

Example

Let, a box of candy has been distributed to 20 children, in such a way that

each child gets 12 candies. If there are 30 children, how many candies each child will

get?

Solution

Look at the following table!

Number of children

candies for each child

20 12

30

Based on the proportion, you can determine the amount of candies for each child.

. Hence, if there are 30 children. every child will get 8 candies.

Exercise

1. A basket of carrots can be finished by 10 rabbits in 6 days. How long time for 15 rabbits to eat them?

2. Father needs 20 minutes to go to the office by car in 60 km/hrs. Today he just takes 15 minutes to go to his office. Determine the speed of the car used to go to his office today!

3. The Graph of the Inverse Proportion

Do you know how to measure the speed? The speed can be measured by the following formula , with = velocity, distance, and = time. For example, you will go by bus to the city which the distance of 180 km from your house. The time you take for the trip depends on the speed of the bus you ride, as it shown in the following table.

[Tabel t –v]

Based on the table above, you can make a velocity graph towards time as follows.

[Grafik kecepatan-waktu]

Exercise

Anton is going to school by bike. He measures the speed of his bike and the time he needs to arrived at school on time. Look at the following result of Anton’s measurement!

(minute)

20 40 50

(minute)

15 7,5 6

1. Sketch a graph based on the table above!

2. How far the distance of Anton’s house to school? (use the formula )

3. If Anton want to arrive at school in 30 minutes, what does the speed of Anton’s bike?

### Mathematics for Junior High School

By Marsigit

C. Probability with relative Frequency

Probability can be explained from the definition of relative frequency. See the explanation below!

• In the experiment of tossing a coin, the ratio between the occurrence of the head and the total number of the experiment is called the relative frequency of obtaining the head.

• If in the experiment, the coin balances enough then the more the experiment is taken out, the more the relative frequency of getting a head is close to .

• In this case, we can tell that in the experiment of throwing a balance coin, the probability of obtaining a head (P(G)) is .

If n is large enough, then the probability of event A is

Where f represents the frequency of event A and n represents the total number of experiment.

Example:

From the experiment of 40 times tossing of 1 coin , the occurrence of head is 22 times. Find the relative frequency of obtaining a head !

Solution :

The total number of experiment (n) is 40 times.

The frequency of the occurrences of the head, f=22

So, the relative frequency of the occurrences of the head is=

Exercises

1. In the experiment of rolling once a six-sided dice , find :

a. the probability of occurrences of the 3-spot side

b. the probability of occurrences of the odd-spot side

2. From tossing a coin 30 times, there are 16 times tail occurrences. Find :

a. the relative frequency of obtaining of the tail

b. the relative frequency of obtaining of the head

3. Expectation Frequency

In an experiment, if A is an event and the probability of event A is P(A) then the expectation frequency of event A for n times experiments is determined by the formula stated below

Example :

A coin was tossed 30 times (assume the coin balances ). What is the expectation frequency of obtaining a head ?

Solution :

The total number of tossing is 30 times. If H is an event of obtaining a head, the probability of obtaining a head is . Hence the frequency of expectation of obtaining a head is

So, the expectation frequency of obtaining the head is 15 times.

Exercises

1. Two coins were tossed together 10 times. What is the expectation frequency of obtaining a head from the first coin and the tail from the other coin?

2. A six-sided dice was thrown 30 times. What is the frequency of expectation of occurrences of the even-spot side?

3. A coin and a dice were rolled together 4 times. What is the expectation frequency of obtaining a tail from the coin and even-spot side from the dice?

4. Two dice were rolled together. How many experiments that have been taken out if the expectation frequency of occurrences of the same side from both of the dice is 10 times?

4. Compound Events

Compound event is an event that contains 2 or more events. In the next explanation, we will discuss about the probability of some compound events contained 2 or more events.

A. Probability Of Non Mutually Exclusive Event

Event A and event B are non mutually exclusive if . Consider the Venn diagram of event A and B that are non mutually exclusive events below

….

The probability of 2 events ( event A and event B) that are non mutually exclusive is as follows

Remember:

is an intersection operation in sets algebra

is an empty sets that is a set which has no elements.

is an union operation in sets algebra

Note:

In mathematics, the union operation is presented by term 'or'. While the intersect operation is presented by term 'and'. For example, P(AUB) is stated as the probability of event A or event B. While P(A∩B) is stated as the probability of event A and event B..

Example

A six-sided dice is rolled once. Find the probability of occurrences of the primary number spot side or the sides with spots less than 5.

Solution:

The sample space of the results of rolling a six sided dice once is S={1,2,3,4,5,6}. So, the total member of sample space S is n(S)=6

If A is an event of obtaining primary number spot side, then A={2,33,5}. So, the total member of event A is n(A)=3.

If B is an event of obtaining the sides with spot less than 5. So, the total member of event B is n(B)=4.

Hence, we can conclude that

So, the total member of event is . So, A and B are non mutually exclusive events.

As a results

The probability of obtaining primary number spot side or sides with spot less that 5 is 5/6

B. Probability of Mutually Exclusive Events

Event A and event B is called mutually exclusive event if . See the Venn diagram of event A and B that are mutually exclusive event below

….

Example

A six sided dice is thrown once. Find the probability of occurrences of the sides with spots less than 3 or the sides with spots greater than or equal to 5!

Solution :

The sample space of the results of throwing a six sided dice once is S={1,2,3,4,5,6}. So the total member of sample space S is n(S)=6

If A is an event of occurrences of the side wit spots less than3, then A={1,2}. The total member of event A is n(A)=2

If B is an event of occurrences of the side with spots greater than or equal with 5 then B={5,6}. The total member of event B is n(B)=2

We can conclude that . So A and B are mutually exclusive events.

as a results

So the probability of occurrences of sides with spots less than 3 or the sides with spots greater then or equal with 5 is 4/6

c. The Probability of Independent Events

Event A and event B are called independent events if the occurrence or non-occurrence of event A is not in any way influenced by the occurrence or non-occurrence of another event. The probability of independent events is as follows

Examples

A basket contains 4 oranges and 6 apples. The other basket contains 5 oranges and 15 apples. What is the probability of obtaining an orange from the first basket and an orange too from the other basket?

Answer :

If,

S1 is sample space of fruits in the first basket, then the number of fruits in the first basket is n(S1)=10

S2 is a sample space of fruits in the second basket, then the number of fruits in the second basket is n(S2)=20

A is an event of obtaining an orange from the first basket n(A)=4

B is an event of obtaining an orange from the second basket, then n(B)=5

So the probability of obtaining an orange from the first basket and an orange from the second basket is 1/10

Exercises

1. A six sided dice was rolled once. Find:

a. the probability of obtaining the even-spot side or the odd-spot side

b. the probability of obtaining the even spot side or the sides with spot less than 3

2. From 52 cards, a card is taken randomly. Determine:

a. the probability that the red card or black card is selected

b. the probability that the red ace card or black ace card is selected

c. the probability that the black card or the 9th card is selected

d. the probability that the red card or the black ace card is selected

3. There are 80 students from a certain school joining some extracurricular programs. It was found that 25 students were joining karate, 15 students were joining swimming and 10 students were joining both karate and swimming program. While the rests were joining the other programs. If we select randomly 1 student from the students that were joining the extracurricular programs, what is the probability that the students which were joining karate or swimming program is selected?

4. Class IX A consists of 40 students. It is found that 10 from 40 students like Biology subject, 20 like English and 5 students like both Biology and English. The other students like the other subject. If we select randomly 1 student from class IX A, what is the probability that the students that like Biology or English is selected?

5. Two six-sided dice were rolled once together. Determine:

a. the probability of occurrences of 1 spot side of the first dice and the primary number spot side of the second dice.

b. the probability of occurrences of 2 spot side of the first dice and 3 spot side of the other dice.

Exercise Chapter 3

A. Chose the correct answer

1. Mrs. Tuti tastes 1 spoon of soup from a bowl of soup. The population from the illustration above is

a. 1 bowl of soup

b. 1 spoon of soup tasted by mrs Tuti

c. 1 pan of soup

d. 1 plate of soup

2. Given the data presented as below

….

The average of the data is....

a. 18

b. 18.25

c. 18.50

d. 18.75

3. Given the data of the student’s age from a certain organization is presented below

The median of the data is ....

a. 13

b. 13.50

c. 15.50

d. 16

4. The data of the weight of a group of athletes is presented below...

The modus of data above is ....

a. 42

b. 44.5

c. 47

d. 42 and 47

5. The average of the art test score from a certain student group consist of 5 students is 75. After a new member join the group, the average turns into 73. The art test score of the new member is....

a. 73

b. 70

c. 63

d. 60

6. See the data of English test score from 50 students presented in the bar diagram below.

The modus of the data of the English test score is ....

a. 5

b. 7

c. 9

d. 5 and 7

7. The money ( in rupiahs ) owned by a student in a week is presented in the diagram below.

The average of the student's money in a week is ....

a. Rp 6000,00

b. Rp 6200,00

c. Rp 6500,00

d. Rp 6800,00

8. If 2 coins were tossed once then the probability of obtaining a tail from one of the coins is ...

a. 1/4

b. 1/2

c. 3/4

d. 1

9. Tono checks his 2 game CD that has not being used for a long time. He wants to know whether his CDs are still working or not. The number of possible outcome that might occur from the checking is....

a. 1

b. 2

c. 4

d. 6

10. The total member of a certain Arisan group is 40 people. Every round, there are 4 people that get the Arisan money. The probability of a member getting the money in the first round is ....

a. 1/10

b. 1/10

c. 1/4

d. 1

11. In the experiment of throwing a coin three times, B is an event of obtaining a tail once. The B event can be expressed as….

a. {T}

b. {THH}

c. {THH, HTH, HHT}

d. {THH, HTH, HHT, HHH}

12. If from 1 pack of cards is taken 1 card randomly, then the probability of obtaining an Ace is ….

a.

b.

c.

d.

13. A box contains 4 red marble, 5 white and 6 green ones. If 1 marble is taken randomly, the probability of occurrences of obtaining a red marble is ….

a.

b.

c.

d.

14. A manager from a certain company gets information that 3 of 100 products from that company are damaged. If 1 product from that company is taken randomly, the probability that the product is good is ….

a.

b.

c.

d.

15. A coin and a six-sided dice are rolled together. The probability of obtaining the 5 spot side is ….

a.

b.

c.

d.

16. Two dices are thrown together. The probability that both of the sides are same is ….

a.

b.

c.

d.

17. A six-sided dice is rolled 18 times. The relative frequency ( the expectation frequency) of obtaining a side with spots less that 4 is ….

a. 2

b. 3

c. 6

d. 9

19. A six sided dice and a coin were rolled together once. The probability of occurrences of obtaining a primary number spot side and the tail from the coin is ….

a. 0.25

b. 0.5

c. 0.75

d. 0.85

20. From 52 cards we took 1 card randomly. The probability of getting a black Ace or a red card is ….

a.

b.

c.

d.

B. Do the exercises below

1. The data of favorite sports of 120 students is presented as a pie diagram below.

a. How many students that like badminton ?

b. What is the modus of student favourite sports? Explain your answer

2. The try out score of Science National Examination from a certain school is presented in table below

a. How many students whose score is more than 5?

b. Calculate the average of the try out score !

c. Determine the median and modus of the data above ?

3. A student is observing whether 3 laboratory equipments are still working or not.

a. Determine the sample space of the observation

b. If A is an event of getting 2 equipment damaged, write the member of event A !

c. Determine the probability of event that 1 equipment is damaged !

4. Given 2 box, each contains 5 balls. The balls in every box were labeled 1 to 5. Then, from each box was taken 1 ball randomly all at once.

a. What is the probability of obtaining that the balls of each box have the same label.

b. What is the probability of obtaining an odd numbered ball from the first box?

5. The probability of a student to be chosen as representatives in the scientific paper competition for teenagers in a certain school is 0.025. How many students that will be the representatives in the competition if the total number of students joining is 720?

Evaluation 1-3

A. Chose the correct answers

1. Two triangles below are congruent in the case of the relation….

a. side, side, side

b. side, side, angle

c. side, angle, angle

d. angle, angle, angle

….

2. Two triangles below are congruent in the case of the relation ….

a. side, side, side

b. side, side, angle

c. side, angle, angle

d. angle, angle, angle

….

3. Two triangles below are congruent in the case of the relation….

a. side, side, side

b. side, side, angle

c. side, angle, angle

d. angle, angle, angle

….

4. Given ABC and CBD. If AB=10 cm and BC=7 cm then the length of BD is ....

a. 10 cm

b. cm

c. cm

d. 4.9 cm

….

5. Given ABC and PQR are similar. CAB=RPQ, ABC= PQR, and BCA=QRP then the length of PR is ….

a. 7.14 cm

b. 3.5 cm

c. 0.56 cm

d. 0.14 cm

….

6. Given ABC and CBD are similar. CAB=DCB=90o and CBA=DBC. If AC=6 cm and BC=10 cm then the length of AB and the length of CD are ….

a. AB=7.5 cm and CD=8 cm

b. AB=8 cm and CD=4.8 cm

c. AB=8 cm and CD=7.5 cm

d. AB=8 cm and CD=13.3 cm

….

7. Given AOB and XOY are similar. If AO=2 cm, XA=3cm, AB=4 cm, and BY=5.1 cm then the length of XY and the length of OB are ….

a. XY=3.4 cm and OB=1.6 cm

b. XY=3.4 cm and OB=10 cm

c. XY=1.6 cm and OB=3.4 cm

d. XY=10 cm and OB=3.4 cm

….

8. Given ABC and DEC are similar. If CD=5 cm and AD=3 cm then the value of is ….

a.

b.

c.

d.

….

9. Given ABX and CDX are similar. If AX= 8 cm XC=10 cm, and BD=27 cm then the length of DX is ….

a. 12 cm

b. 13 cm

c. 14 cm

d. 15 cm

….

10. Given ABC and DEC are similar. If BCA:CAB:ABC = 1:2:1 then the value of DEC is ….

a. 25o

b. 30o

c. 45o

d. 50o

….

11. A cylinder has a radius 6 cm and the side area is 37.86 cm2. The height of the cylinder is ....

a. 12 cm

b. 10 cm

c. 2 cm

d. 1 cm

12. Oil 4.71 liter was poured into a can formed cylinder so the cylinder is filled with the oil until 15 cm high. The diameter of the cylinder is ....

a. 10 cm

b. 15 cm

c. 20 cm

d. 30 cm

13. The side area of a cylinder is 440 cm2. If the height of the cylinder is 10 cm then the top and bottom area of the cylinder are ....

a. 308 cm2

b. 154 cm2

c. 88 cm2

d. 44 cm2

14. A biscuit can has diameter 20 cm and height 10 cm. The surface area of the biscuit can is ....

a. 12,560 cm2

b. 1,256 cm2

c. 628 cm2

d. 125.6 cm2

15. A trumpet made from cardboard and formed a cone has a diameter of 14 cm. If the slant height is 30 cm then the area of the cardboard needed to make the trumpet is ....

a. 1,320 cm2

b. 660 cm2

c. 132 cm2

d. 66 cm2

16. A cone has a slant height of 7 cm and radius 3.5 cm. The surface area of the cone is ....

a. 1,155 cm2

b. 770 cm2

c. 115.5 cm2

d. 77 cm2

17. If the volume of a cone having perpendicular height 9 cm is 462 cm3 then the base area of the cone is ....

a. 44 cm2

b. 154 cm2

c. 1,386 cm2

d. 7,546 cm2

18. Mom has a mold to make jelly formed a half sphere and having a diameter of 21 cm. The volume of the jelly if it was made using the mold is ....

a. 38,808 cm3

b. 19,404 cm3

c. 4,851 cm3

d. 2,425.5 cm3

19. A ball is made from woven rattan. If the total area of woven rattan used to make the sphere is 616 cm2 then the diameter of the ball is ....

a. 4.9 cm

b. 7 cm

c. 9.8 cm

d. 14 cm

20. An aquarium formed a sphere has diameter of 35 cm. The volume of the aquarium is ....

a. 67,375 liter

b. 22,458.33 liter

c. 67.375 liter

d. 22.458 liter

Do number 21-22 based on the table below. We were given a data of weight of an apple from some different trees below.

….

21. The total number of apples that are being measured is….

a. 43 apples

b. 44 apples

c. 48 apples

d. 50 apples

22. If the apples that have weight less than 80 gram are not sold then the total number of apples that are sold is ….

a. 33 apples

b. 26 apples

c. 24 apples

d. 17 apples

Do number 23-25 based on the circle graph below. The data of student favorite activities in their spare time is presented below.

….

23. The value of x is …..

a. 100o

b. 110o

c. 120o

d. 130o

24. The percentage of students that like reading is ….

a. 80 %

b. 40 %

c. 33.33%

d. 22.22%

25. If the number of students who like reading is 40 people, then the number of students who like watching TV is ….

a. 50 students

b. 60 students

c. 80 students

d. 120 students

26. We were given a data of the total number of goal occurrences of 15 soccer games in a certain season

….

The median of the data above is ….

a. 1

b. 2

c. 3

d. 4

27. A box contains 4 red marble, 3 white, and 5 green ones. If we took 1 marble randomly then the probability of obtaining a white marble is ….

a. 0.25

b. 0.33

c. 0.42

d. 0.60

28. If the probability of a student getting scholarship is 0.125 then the probability of a student not getting that scholarship is …

a. 0.875

b. 0.785

c.

d.

29. A six sided dice and a coin were rolled together. The probability of occurrences of odd-spot side of the dice and a tail of the coin is ….

a. 0.75

b. 0.5

c. 0.25

d. 0.125

30. A card is taken randomly from 52 cards. The probability of getting an Ace card is ….

a.

b.

c.

d.

B. Solve the following problems

1. In the afternoon, a boy who has height 1.5 m has a shadow with length of 4.5 m. Tina’s height is 1.3 m. What is the length of Tina’s shadow if she stands in the same time and the same place as the boy?

2. See the picture below. A fisherman wants to know about the distance of his friend boat that sailing in the ocean from the coast line. Based on the position of the boat, the fisherman is located in point D. Then he make another points ABC along the coast according to the sketch below. Determine the length of AA'!

….

3. Prove that ABC and ADC are congruent!

4. A cylinder has maximum capacity of 2.156 liter. If the diameter of the cylinder is 14 cm, then calculate the height of the cylinder!

5. A traditional cap formed a cone has a slant height of 28 cm and radius 21 cm. It is made from woven rattan. Find the area of the woven rattan used to make the traditional cap!

6. A bowl is formed half-sphere with diameter 14 cm. Calculate the volume of the bowl!

7. A ball has a circumference of the mid circle of 66 cm. Determine the total area of material used to cover the leather!

8. A student has a collection of 50 books, consists of 20 school text books, 10 fiction books, 15 motivation books, and 5 biography books of world famous person. Draw a circle graph to illustrate the information!

9. A trader mixes 10 kg oranges that priced Rp 15.000,00 per kg and 5 kg oranges priced Rp 10.000,00. What is the average price of that mixed oranges now?

10. In a certain school, there are 100 students who propose for scholarship. Each student has the probability of 0.25 to get the scholarship. How many students who will get the schoolarship?

C. Probability with relative Frequency

Probability can be explained from the definition of relative frequency. See the explanation below!

• In the experiment of tossing a coin, the ratio between the occurrence of the head and the total number of the experiment is called the relative frequency of obtaining the head.

• If in the experiment, the coin balances enough then the more the experiment is taken out, the more the relative frequency of getting a head is close to .

• In this case, we can tell that in the experiment of throwing a balance coin, the probability of obtaining a head (P(G)) is .

If n is large enough, then the probability of event A is

Where f represents the frequency of event A and n represents the total number of experiment.

Example:

From the experiment of 40 times tossing of 1 coin , the occurrence of head is 22 times. Find the relative frequency of obtaining a head !

Solution :

The total number of experiment (n) is 40 times.

The frequency of the occurrences of the head, f=22

So, the relative frequency of the occurrences of the head is=

Exercises

1. In the experiment of rolling once a six-sided dice , find :

a. the probability of occurrences of the 3-spot side

b. the probability of occurrences of the odd-spot side

2. From tossing a coin 30 times, there are 16 times tail occurrences. Find :

a. the relative frequency of obtaining of the tail

b. the relative frequency of obtaining of the head

3. Expectation Frequency

In an experiment, if A is an event and the probability of event A is P(A) then the expectation frequency of event A for n times experiments is determined by the formula stated below

Example :

A coin was tossed 30 times (assume the coin balances ). What is the expectation frequency of obtaining a head ?

Solution :

The total number of tossing is 30 times. If H is an event of obtaining a head, the probability of obtaining a head is . Hence the frequency of expectation of obtaining a head is

So, the expectation frequency of obtaining the head is 15 times.

Exercises

1. Two coins were tossed together 10 times. What is the expectation frequency of obtaining a head from the first coin and the tail from the other coin?

2. A six-sided dice was thrown 30 times. What is the frequency of expectation of occurrences of the even-spot side?

3. A coin and a dice were rolled together 4 times. What is the expectation frequency of obtaining a tail from the coin and even-spot side from the dice?

4. Two dice were rolled together. How many experiments that have been taken out if the expectation frequency of occurrences of the same side from both of the dice is 10 times?

4. Compound Events

Compound event is an event that contains 2 or more events. In the next explanation, we will discuss about the probability of some compound events contained 2 or more events.

A. Probability Of Non Mutually Exclusive Event

Event A and event B are non mutually exclusive if . Consider the Venn diagram of event A and B that are non mutually exclusive events below

….

The probability of 2 events ( event A and event B) that are non mutually exclusive is as follows

Remember:

is an intersection operation in sets algebra

is an empty sets that is a set which has no elements.

is an union operation in sets algebra

Note:

In mathematics, the union operation is presented by term 'or'. While the intersect operation is presented by term 'and'. For example, P(AUB) is stated as the probability of event A or event B. While P(A∩B) is stated as the probability of event A and event B..

Example

A six-sided dice is rolled once. Find the probability of occurrences of the primary number spot side or the sides with spots less than 5.

Solution:

The sample space of the results of rolling a six sided dice once is S={1,2,3,4,5,6}. So, the total member of sample space S is n(S)=6

If A is an event of obtaining primary number spot side, then A={2,33,5}. So, the total member of event A is n(A)=3.

If B is an event of obtaining the sides with spot less than 5. So, the total member of event B is n(B)=4.

Hence, we can conclude that

So, the total member of event is . So, A and B are non mutually exclusive events.

As a results

The probability of obtaining primary number spot side or sides with spot less that 5 is 5/6

B. Probability of Mutually Exclusive Events

Event A and event B is called mutually exclusive event if . See the Venn diagram of event A and B that are mutually exclusive event below

….

Example

A six sided dice is thrown once. Find the probability of occurrences of the sides with spots less than 3 or the sides with spots greater than or equal to 5!

Solution :

The sample space of the results of throwing a six sided dice once is S={1,2,3,4,5,6}. So the total member of sample space S is n(S)=6

If A is an event of occurrences of the side wit spots less than3, then A={1,2}. The total member of event A is n(A)=2

If B is an event of occurrences of the side with spots greater than or equal with 5 then B={5,6}. The total member of event B is n(B)=2

We can conclude that . So A and B are mutually exclusive events.

as a results

So the probability of occurrences of sides with spots less than 3 or the sides with spots greater then or equal with 5 is 4/6

c. The Probability of Independent Events

Event A and event B are called independent events if the occurrence or non-occurrence of event A is not in any way influenced by the occurrence or non-occurrence of another event. The probability of independent events is as follows

Examples

A basket contains 4 oranges and 6 apples. The other basket contains 5 oranges and 15 apples. What is the probability of obtaining an orange from the first basket and an orange too from the other basket?

Answer :

If,

S1 is sample space of fruits in the first basket, then the number of fruits in the first basket is n(S1)=10

S2 is a sample space of fruits in the second basket, then the number of fruits in the second basket is n(S2)=20

A is an event of obtaining an orange from the first basket n(A)=4

B is an event of obtaining an orange from the second basket, then n(B)=5

So the probability of obtaining an orange from the first basket and an orange from the second basket is 1/10

Exercises

1. A six sided dice was rolled once. Find:

a. the probability of obtaining the even-spot side or the odd-spot side

b. the probability of obtaining the even spot side or the sides with spot less than 3

2. From 52 cards, a card is taken randomly. Determine:

a. the probability that the red card or black card is selected

b. the probability that the red ace card or black ace card is selected

c. the probability that the black card or the 9th card is selected

d. the probability that the red card or the black ace card is selected

3. There are 80 students from a certain school joining some extracurricular programs. It was found that 25 students were joining karate, 15 students were joining swimming and 10 students were joining both karate and swimming program. While the rests were joining the other programs. If we select randomly 1 student from the students that were joining the extracurricular programs, what is the probability that the students which were joining karate or swimming program is selected?

4. Class IX A consists of 40 students. It is found that 10 from 40 students like Biology subject, 20 like English and 5 students like both Biology and English. The other students like the other subject. If we select randomly 1 student from class IX A, what is the probability that the students that like Biology or English is selected?

5. Two six-sided dice were rolled once together. Determine:

a. the probability of occurrences of 1 spot side of the first dice and the primary number spot side of the second dice.

b. the probability of occurrences of 2 spot side of the first dice and 3 spot side of the other dice.

Exercise Chapter 3

A. Chose the correct answer

1. Mrs. Tuti tastes 1 spoon of soup from a bowl of soup. The population from the illustration above is

a. 1 bowl of soup

b. 1 spoon of soup tasted by mrs Tuti

c. 1 pan of soup

d. 1 plate of soup

2. Given the data presented as below

….

The average of the data is....

a. 18

b. 18.25

c. 18.50

d. 18.75

3. Given the data of the student’s age from a certain organization is presented below

The median of the data is ....

a. 13

b. 13.50

c. 15.50

d. 16

4. The data of the weight of a group of athletes is presented below...

The modus of data above is ....

a. 42

b. 44.5

c. 47

d. 42 and 47

5. The average of the art test score from a certain student group consist of 5 students is 75. After a new member join the group, the average turns into 73. The art test score of the new member is....

a. 73

b. 70

c. 63

d. 60

6. See the data of English test score from 50 students presented in the bar diagram below.

The modus of the data of the English test score is ....

a. 5

b. 7

c. 9

d. 5 and 7

7. The money ( in rupiahs ) owned by a student in a week is presented in the diagram below.

The average of the student's money in a week is ....

a. Rp 6000,00

b. Rp 6200,00

c. Rp 6500,00

d. Rp 6800,00

8. If 2 coins were tossed once then the probability of obtaining a tail from one of the coins is ...

a. 1/4

b. 1/2

c. 3/4

d. 1

9. Tono checks his 2 game CD that has not being used for a long time. He wants to know whether his CDs are still working or not. The number of possible outcome that might occur from the checking is....

a. 1

b. 2

c. 4

d. 6

10. The total member of a certain Arisan group is 40 people. Every round, there are 4 people that get the Arisan money. The probability of a member getting the money in the first round is ....

a. 1/10

b. 1/10

c. 1/4

d. 1

11. In the experiment of throwing a coin three times, B is an event of obtaining a tail once. The B event can be expressed as….

a. {T}

b. {THH}

c. {THH, HTH, HHT}

d. {THH, HTH, HHT, HHH}

12. If from 1 pack of cards is taken 1 card randomly, then the probability of obtaining an Ace is ….

a.

b.

c.

d.

13. A box contains 4 red marble, 5 white and 6 green ones. If 1 marble is taken randomly, the probability of occurrences of obtaining a red marble is ….

a.

b.

c.

d.

14. A manager from a certain company gets information that 3 of 100 products from that company are damaged. If 1 product from that company is taken randomly, the probability that the product is good is ….

a.

b.

c.

d.

15. A coin and a six-sided dice are rolled together. The probability of obtaining the 5 spot side is ….

a.

b.

c.

d.

16. Two dices are thrown together. The probability that both of the sides are same is ….

a.

b.

c.

d.

17. A six-sided dice is rolled 18 times. The relative frequency ( the expectation frequency) of obtaining a side with spots less that 4 is ….

a. 2

b. 3

c. 6

d. 9

19. A six sided dice and a coin were rolled together once. The probability of occurrences of obtaining a primary number spot side and the tail from the coin is ….

a. 0.25

b. 0.5

c. 0.75

d. 0.85

20. From 52 cards we took 1 card randomly. The probability of getting a black Ace or a red card is ….

a.

b.

c.

d.

B. Do the exercises below

1. The data of favorite sports of 120 students is presented as a pie diagram below.

a. How many students that like badminton ?

b. What is the modus of student favourite sports? Explain your answer

2. The try out score of Science National Examination from a certain school is presented in table below

a. How many students whose score is more than 5?

b. Calculate the average of the try out score !

c. Determine the median and modus of the data above ?

3. A student is observing whether 3 laboratory equipments are still working or not.

a. Determine the sample space of the observation

b. If A is an event of getting 2 equipment damaged, write the member of event A !

c. Determine the probability of event that 1 equipment is damaged !

4. Given 2 box, each contains 5 balls. The balls in every box were labeled 1 to 5. Then, from each box was taken 1 ball randomly all at once.

a. What is the probability of obtaining that the balls of each box have the same label.

b. What is the probability of obtaining an odd numbered ball from the first box?

5. The probability of a student to be chosen as representatives in the scientific paper competition for teenagers in a certain school is 0.025. How many students that will be the representatives in the competition if the total number of students joining is 720?

Evaluation 1-3

A. Chose the correct answers

1. Two triangles below are congruent in the case of the relation….

a. side, side, side

b. side, side, angle

c. side, angle, angle

d. angle, angle, angle

….

2. Two triangles below are congruent in the case of the relation ….

a. side, side, side

b. side, side, angle

c. side, angle, angle

d. angle, angle, angle

….

3. Two triangles below are congruent in the case of the relation….

a. side, side, side

b. side, side, angle

c. side, angle, angle

d. angle, angle, angle

….

4. Given ABC and CBD. If AB=10 cm and BC=7 cm then the length of BD is ....

a. 10 cm

b. cm

c. cm

d. 4.9 cm

….

5. Given ABC and PQR are similar. CAB=RPQ, ABC= PQR, and BCA=QRP then the length of PR is ….

a. 7.14 cm

b. 3.5 cm

c. 0.56 cm

d. 0.14 cm

….

6. Given ABC and CBD are similar. CAB=DCB=90o and CBA=DBC. If AC=6 cm and BC=10 cm then the length of AB and the length of CD are ….

a. AB=7.5 cm and CD=8 cm

b. AB=8 cm and CD=4.8 cm

c. AB=8 cm and CD=7.5 cm

d. AB=8 cm and CD=13.3 cm

….

7. Given AOB and XOY are similar. If AO=2 cm, XA=3cm, AB=4 cm, and BY=5.1 cm then the length of XY and the length of OB are ….

a. XY=3.4 cm and OB=1.6 cm

b. XY=3.4 cm and OB=10 cm

c. XY=1.6 cm and OB=3.4 cm

d. XY=10 cm and OB=3.4 cm

….

8. Given ABC and DEC are similar. If CD=5 cm and AD=3 cm then the value of is ….

a.

b.

c.

d.

….

9. Given ABX and CDX are similar. If AX= 8 cm XC=10 cm, and BD=27 cm then the length of DX is ….

a. 12 cm

b. 13 cm

c. 14 cm

d. 15 cm

….

10. Given ABC and DEC are similar. If BCA:CAB:ABC = 1:2:1 then the value of DEC is ….

a. 25o

b. 30o

c. 45o

d. 50o

….

11. A cylinder has a radius 6 cm and the side area is 37.86 cm2. The height of the cylinder is ....

a. 12 cm

b. 10 cm

c. 2 cm

d. 1 cm

12. Oil 4.71 liter was poured into a can formed cylinder so the cylinder is filled with the oil until 15 cm high. The diameter of the cylinder is ....

a. 10 cm

b. 15 cm

c. 20 cm

d. 30 cm

13. The side area of a cylinder is 440 cm2. If the height of the cylinder is 10 cm then the top and bottom area of the cylinder are ....

a. 308 cm2

b. 154 cm2

c. 88 cm2

d. 44 cm2

14. A biscuit can has diameter 20 cm and height 10 cm. The surface area of the biscuit can is ....

a. 12,560 cm2

b. 1,256 cm2

c. 628 cm2

d. 125.6 cm2

15. A trumpet made from cardboard and formed a cone has a diameter of 14 cm. If the slant height is 30 cm then the area of the cardboard needed to make the trumpet is ....

a. 1,320 cm2

b. 660 cm2

c. 132 cm2

d. 66 cm2

16. A cone has a slant height of 7 cm and radius 3.5 cm. The surface area of the cone is ....

a. 1,155 cm2

b. 770 cm2

c. 115.5 cm2

d. 77 cm2

17. If the volume of a cone having perpendicular height 9 cm is 462 cm3 then the base area of the cone is ....

a. 44 cm2

b. 154 cm2

c. 1,386 cm2

d. 7,546 cm2

18. Mom has a mold to make jelly formed a half sphere and having a diameter of 21 cm. The volume of the jelly if it was made using the mold is ....

a. 38,808 cm3

b. 19,404 cm3

c. 4,851 cm3

d. 2,425.5 cm3

19. A ball is made from woven rattan. If the total area of woven rattan used to make the sphere is 616 cm2 then the diameter of the ball is ....

a. 4.9 cm

b. 7 cm

c. 9.8 cm

d. 14 cm

20. An aquarium formed a sphere has diameter of 35 cm. The volume of the aquarium is ....

a. 67,375 liter

b. 22,458.33 liter

c. 67.375 liter

d. 22.458 liter

Do number 21-22 based on the table below. We were given a data of weight of an apple from some different trees below.

….

21. The total number of apples that are being measured is….

a. 43 apples

b. 44 apples

c. 48 apples

d. 50 apples

22. If the apples that have weight less than 80 gram are not sold then the total number of apples that are sold is ….

a. 33 apples

b. 26 apples

c. 24 apples

d. 17 apples

Do number 23-25 based on the circle graph below. The data of student favorite activities in their spare time is presented below.

….

23. The value of x is …..

a. 100o

b. 110o

c. 120o

d. 130o

24. The percentage of students that like reading is ….

a. 80 %

b. 40 %

c. 33.33%

d. 22.22%

25. If the number of students who like reading is 40 people, then the number of students who like watching TV is ….

a. 50 students

b. 60 students

c. 80 students

d. 120 students

26. We were given a data of the total number of goal occurrences of 15 soccer games in a certain season

….

The median of the data above is ….

a. 1

b. 2

c. 3

d. 4

27. A box contains 4 red marble, 3 white, and 5 green ones. If we took 1 marble randomly then the probability of obtaining a white marble is ….

a. 0.25

b. 0.33

c. 0.42

d. 0.60

28. If the probability of a student getting scholarship is 0.125 then the probability of a student not getting that scholarship is …

a. 0.875

b. 0.785

c.

d.

29. A six sided dice and a coin were rolled together. The probability of occurrences of odd-spot side of the dice and a tail of the coin is ….

a. 0.75

b. 0.5

c. 0.25

d. 0.125

30. A card is taken randomly from 52 cards. The probability of getting an Ace card is ….

a.

b.

c.

d.

B. Solve the following problems

1. In the afternoon, a boy who has height 1.5 m has a shadow with length of 4.5 m. Tina’s height is 1.3 m. What is the length of Tina’s shadow if she stands in the same time and the same place as the boy?

2. See the picture below. A fisherman wants to know about the distance of his friend boat that sailing in the ocean from the coast line. Based on the position of the boat, the fisherman is located in point D. Then he make another points ABC along the coast according to the sketch below. Determine the length of AA'!

….

3. Prove that ABC and ADC are congruent!

4. A cylinder has maximum capacity of 2.156 liter. If the diameter of the cylinder is 14 cm, then calculate the height of the cylinder!

5. A traditional cap formed a cone has a slant height of 28 cm and radius 21 cm. It is made from woven rattan. Find the area of the woven rattan used to make the traditional cap!

6. A bowl is formed half-sphere with diameter 14 cm. Calculate the volume of the bowl!

7. A ball has a circumference of the mid circle of 66 cm. Determine the total area of material used to cover the leather!

8. A student has a collection of 50 books, consists of 20 school text books, 10 fiction books, 15 motivation books, and 5 biography books of world famous person. Draw a circle graph to illustrate the information!

9. A trader mixes 10 kg oranges that priced Rp 15.000,00 per kg and 5 kg oranges priced Rp 10.000,00. What is the average price of that mixed oranges now?

10. In a certain school, there are 100 students who propose for scholarship. Each student has the probability of 0.25 to get the scholarship. How many students who will get the schoolarship?

## Monday, February 23, 2009

### Elegi Pengakuan Orang Tua Berambut Putih

Oleh Marsigit

Kini aku merasa sudah saatnya aku berterus terang. Aku ingin berterus terang kepada semuanya. Tetapi ketahuilah tidak berterus terang bukanlah berarti bohong. Mengapa aku ingin berterus terang? Karena aku tidak tahu kapan aku berakhir, sedemikian juga seperti ketidaktahuanku kapan aku mulai. Supaya pengakuanku mempunyai bobot yang cukup baik, maka aku akan mencari saksi. Tetapi sebelum aku bertanya kepada saksi-saksi terlebih aku ingin sampaikan bahwa sebenar-benar sebenar-benar saksi tidak lain tidak bukan adalah aku sendiri. Aku ingin mengaku bahwa aku mengetahui segala sesuatu, tetapi jika mereka bertanya kepadaku maka dengan serta merta aku tidak mengetahuinya. Mengapa? Karena sebenar-benar diriku adalah pertanyaan mereka. Bagaimana mungkin aku mengerti tentang pertanyaanku sendiri? Maka sebodoh-bodoh orang adalah diriku ini, karena aku selalu bertanya tentang segala hal, tetapi aku selalu tidak dapat menjelaskannya. Tetapi bukankah mengerti bahwa aku tidak dapat menjawab itu pertanda bahwa tidak tidak mengerti. Maka dapat aku katakan bahwa sebenar-benar diriku adalah tidak tidak mengerti. Hendaknya jangan terkejut bahwa tidak tidak mengerti itu sebenarnya adalah mengerti pula. Maka bukanlah aku yang ingin mengatakan, tetapi mereka yang boleh mengatakan bahwa aku selalu mengerti tentang pertanyaanku itu. Bukankah di sini mereka tahu bahwa sebenar-benar diriku adalah kontradiksi, karena aku sekaligus mengerti dan tidak mengerti. Itulah sebenar-benar ilmu, yaitu kontradiktif. Tetapi janganlah salah paham, kontradiksi itu adalah ilmu, tempat tinggalnya ada dalam pikiranmu.

Orang tua berambut putih berjumpa Socrates, Plato, Aristoteles, George Berkely, Rene Descartes, dan Immanuel Kant.

Wahai Socrates, Plato, Aristoteles, Rene Descartes, David Hume dan Immanuel Kant, ..., bolehka aku bertanya kepadamu. Menurut kesaksianmu, siapakah sebenar-benar dirimu dan sebenar-benar diriku?

Socrates:

Wahai orang tua berambut putih. Sebenar-benar diriku adalah pertanyaanku. Maka aku akan bertanya kepada siapapun tentang segala hal yang aku sukai. Sedangkan sebenar-benar dirimu adalah diriku juga. Maka sebenar-benar dirimu tidak lain adalah pertanyaanku juga pertanyaanmu. Jawaban para pakar dan para ahli itulah sebenar-benar tempat tinggalmu. Itulah sebenar-benar ilmu yaitu pertanyaanmu.

Plato:

Sebenar-benar dirimu adalah pikiranku. Sebenar-benar dirimu adalah imajinasiku. Tidak aku panggil, engkau sudah ada dalam pikiranku. Maka dirimu yang absolut itulah yang selalu aku pikirkan. Semuanya tentang dirimu sudah ada dalam pikiranku. Hanya terkadang aku sulit mengenalimu. Tetapi aku sadar, bahwa banyak orang mencoba mirip-mirip dengan mu. Itulah mereka yang dapat aku lihat dan dapat aku raba. Mereka jumlahnya sangat banyak. Tetapi mereka semuanya bersifat sementara. Maka sebenar-benar dirimu adalah ide-ide ku. Tempat tinggalmu ada dalam pikiranku. Tetapi aku selalu mengkhawatirkanmu, karena dalam pikiranku selalu ada lubang gelap seperti gua. Gua-gua seperti itulah yang menyebabkan aku sulit mengenalmu. Aku juga mengkhawatirkan akan badanku, karena badanku inilah penyebab munculnya gua-gua itu. Maka sebenar-benar dirimu adalah hamba yang terlepas dari penjara badanku. Maka sebenar-benar dirimu adalah pikiranku yang berada di luar gua kegelapanku dan terbebas dari badanku. Itulah sebenar-benar ilmu, tidak lain tidak bukan adalah pikiranku. Tetapi aku mengalami kesulitan untuk menjelaskan kepada orang-orang tentang tempat tinggalmu itu.

Aristoteles:

Aku agak berbeda dengan guruku Plato. Menurutku, sebenar-benar dirimu adalah adalah pengalamanku. Maka tempat tinggalmu adalah pada pengalamanku. Begitu aku menggapai pengalamanku maka dengan serta merta muncullah dirimu itu. Maka menurutku, sebenar-benar dirimu adalah yang dapat aku lihat, aku raba, dan aku indera. Maka sebenar-benar dirimu adalah diluar diriku. Padahal aku tahu bahwa dirimu adalah diriku. Maka aku tidak lain tidak bukan adalah sekaligus bukan aku. Mengapa? Karena aku mengalami kesulitan memahami engkau yang berada diluar diriku. Dan juga mengalami kesulitan bagaimana aku dapat menjelaskan kepada orang-orang bahwa engkau yang berada di luar diriku itu sebenar-benarnya adalah diriku.

George Berkely:

Menurutku, engkau adalah yang aku lihat atau aku persepsi. Jikalau engkau tidak adalah di situ, maka tidak adalah sebenar-benarnya engkau itu. Esse est percipi itulah kata-kataku. Maka sebenar-benar engkau adalah tipuanmu belaka. Engkau adalah fatamorgana. Maka sebenar-benar engkau yang adapat aku lihat adalah tipuan juga. Jadi apalah artnya sesuatu yang dapat engkau lihat itu, kecuali hanya tipuan belaka. Maka dunia ini tidak lain tidak bukan adalah tipuan belaka. Itulah sebenar-benar dirimu itu. Padahal aku tahu bahwa dirimu adalah diriku juga, maka sebenar-benar ilmu itu tidak lain tidak bukan adalah tipuan butamu.

Rene Descartes:

Menurutku, sebenar-benar dirimu adalah mimpiku. Tiadalah aku dapat menemukanmu tanpa mimpi-mimpi itu. Bagiku mimpi adalah nyata. Dan yang nyata bisa juga menjadi mimpiku. Maka sebenar-benar diriku adalah tidak dapat membedakan apakah kenyataan atau mimpiku. Namun aku selalu risau karena jika aku ingin berjumpa denganmu aku selalu dihadang makhluk hitam yang akan menyesatkanku, sehingga aku selalu salah mengertimu. Maka selalulah terjadi bahwa aku selalu meragukan keberadaanmu. Maka sebenar-benar dirimu adalah keraguanku itu sendiri. Itulah sebenar-benar ilmu, yaitu keraguanmu. Aku meragukan semuanya tanpa kecuali. Hanyalah ada satu yang tidak dapat aku ragukan yaitu diriku sendiri yang meragukan itu. Itulah satu-satunya kepastian bagiku. Cogito ergosum itulah kata-kataku, yaitu bahwa diriku itu ada karena aku tahu sedang meragukannya. Bukankah itu sebenar-benar dirimu. Maka sebenar-benar keberadaanku adalah diriku yang berpikir ini.

Immanuel Kant:

Menurutku engkau adalah pikiranku sekaligus pengalamanku. Tiadalah dirimu itu ada di situ tanpa pikiranku atau tanpa pengalamanku. Sedangkan tempat tinggalmu adalah di dalam intuisiku. Ketahuilah bahwa aku mempunyai dua intuisi yaitu intuisi ruang dan intuisi waktu. Itulah sebenar-benar tempat tinggalmu. Tetapi untuk mengerti tentang dirimu aku harus bersifat kritis. Maka sebanar-benar dirimu adalah pikiranku yang kritis. Itulah sebenar-benar ilmu, tidak lain tidak bukan adalah pikiran kritismu. Namun ketahui pula bahwa tidaklah mudah menggapai dirimu. Ketika engkau kukejar dengan pikiran kritisku, maka engkau lari menuju keputusanmu. Itulah setinggi-tinggi tempat tinggalmu, yaitu pada keputusanku. Maka sebenar-benar ilmu tidak lain tidak bukan adalah keputusanmu.

Orang tua berambut putih tidak puas dengan jawaban Socrates, Plato, Aristoteles, George Berkely, Rene Descartes, dan Immanuel Kant.

Ah, ngacau semua mereka itu. Jawaban mereka sesuai selera masing-masing. Maka belum puaslah diriku alan jawaban mereka semua. Tetapi aku sadar bahwa aku sedang berada pada jaman jauh sebelum diriku sekarang. Padahal aku belum sempat bertanya kepada orang sekarang yang berseliweran didepanku. Maka aku berjanji akan meneruskan perjalananku untuk bertanya siapa sebenarnya diriku itu?

Kini aku merasa sudah saatnya aku berterus terang. Aku ingin berterus terang kepada semuanya. Tetapi ketahuilah tidak berterus terang bukanlah berarti bohong. Mengapa aku ingin berterus terang? Karena aku tidak tahu kapan aku berakhir, sedemikian juga seperti ketidaktahuanku kapan aku mulai. Supaya pengakuanku mempunyai bobot yang cukup baik, maka aku akan mencari saksi. Tetapi sebelum aku bertanya kepada saksi-saksi terlebih aku ingin sampaikan bahwa sebenar-benar sebenar-benar saksi tidak lain tidak bukan adalah aku sendiri. Aku ingin mengaku bahwa aku mengetahui segala sesuatu, tetapi jika mereka bertanya kepadaku maka dengan serta merta aku tidak mengetahuinya. Mengapa? Karena sebenar-benar diriku adalah pertanyaan mereka. Bagaimana mungkin aku mengerti tentang pertanyaanku sendiri? Maka sebodoh-bodoh orang adalah diriku ini, karena aku selalu bertanya tentang segala hal, tetapi aku selalu tidak dapat menjelaskannya. Tetapi bukankah mengerti bahwa aku tidak dapat menjawab itu pertanda bahwa tidak tidak mengerti. Maka dapat aku katakan bahwa sebenar-benar diriku adalah tidak tidak mengerti. Hendaknya jangan terkejut bahwa tidak tidak mengerti itu sebenarnya adalah mengerti pula. Maka bukanlah aku yang ingin mengatakan, tetapi mereka yang boleh mengatakan bahwa aku selalu mengerti tentang pertanyaanku itu. Bukankah di sini mereka tahu bahwa sebenar-benar diriku adalah kontradiksi, karena aku sekaligus mengerti dan tidak mengerti. Itulah sebenar-benar ilmu, yaitu kontradiktif. Tetapi janganlah salah paham, kontradiksi itu adalah ilmu, tempat tinggalnya ada dalam pikiranmu.

Orang tua berambut putih berjumpa Socrates, Plato, Aristoteles, George Berkely, Rene Descartes, dan Immanuel Kant.

Wahai Socrates, Plato, Aristoteles, Rene Descartes, David Hume dan Immanuel Kant, ..., bolehka aku bertanya kepadamu. Menurut kesaksianmu, siapakah sebenar-benar dirimu dan sebenar-benar diriku?

Socrates:

Wahai orang tua berambut putih. Sebenar-benar diriku adalah pertanyaanku. Maka aku akan bertanya kepada siapapun tentang segala hal yang aku sukai. Sedangkan sebenar-benar dirimu adalah diriku juga. Maka sebenar-benar dirimu tidak lain adalah pertanyaanku juga pertanyaanmu. Jawaban para pakar dan para ahli itulah sebenar-benar tempat tinggalmu. Itulah sebenar-benar ilmu yaitu pertanyaanmu.

Plato:

Sebenar-benar dirimu adalah pikiranku. Sebenar-benar dirimu adalah imajinasiku. Tidak aku panggil, engkau sudah ada dalam pikiranku. Maka dirimu yang absolut itulah yang selalu aku pikirkan. Semuanya tentang dirimu sudah ada dalam pikiranku. Hanya terkadang aku sulit mengenalimu. Tetapi aku sadar, bahwa banyak orang mencoba mirip-mirip dengan mu. Itulah mereka yang dapat aku lihat dan dapat aku raba. Mereka jumlahnya sangat banyak. Tetapi mereka semuanya bersifat sementara. Maka sebenar-benar dirimu adalah ide-ide ku. Tempat tinggalmu ada dalam pikiranku. Tetapi aku selalu mengkhawatirkanmu, karena dalam pikiranku selalu ada lubang gelap seperti gua. Gua-gua seperti itulah yang menyebabkan aku sulit mengenalmu. Aku juga mengkhawatirkan akan badanku, karena badanku inilah penyebab munculnya gua-gua itu. Maka sebenar-benar dirimu adalah hamba yang terlepas dari penjara badanku. Maka sebenar-benar dirimu adalah pikiranku yang berada di luar gua kegelapanku dan terbebas dari badanku. Itulah sebenar-benar ilmu, tidak lain tidak bukan adalah pikiranku. Tetapi aku mengalami kesulitan untuk menjelaskan kepada orang-orang tentang tempat tinggalmu itu.

Aristoteles:

Aku agak berbeda dengan guruku Plato. Menurutku, sebenar-benar dirimu adalah adalah pengalamanku. Maka tempat tinggalmu adalah pada pengalamanku. Begitu aku menggapai pengalamanku maka dengan serta merta muncullah dirimu itu. Maka menurutku, sebenar-benar dirimu adalah yang dapat aku lihat, aku raba, dan aku indera. Maka sebenar-benar dirimu adalah diluar diriku. Padahal aku tahu bahwa dirimu adalah diriku. Maka aku tidak lain tidak bukan adalah sekaligus bukan aku. Mengapa? Karena aku mengalami kesulitan memahami engkau yang berada diluar diriku. Dan juga mengalami kesulitan bagaimana aku dapat menjelaskan kepada orang-orang bahwa engkau yang berada di luar diriku itu sebenar-benarnya adalah diriku.

George Berkely:

Menurutku, engkau adalah yang aku lihat atau aku persepsi. Jikalau engkau tidak adalah di situ, maka tidak adalah sebenar-benarnya engkau itu. Esse est percipi itulah kata-kataku. Maka sebenar-benar engkau adalah tipuanmu belaka. Engkau adalah fatamorgana. Maka sebenar-benar engkau yang adapat aku lihat adalah tipuan juga. Jadi apalah artnya sesuatu yang dapat engkau lihat itu, kecuali hanya tipuan belaka. Maka dunia ini tidak lain tidak bukan adalah tipuan belaka. Itulah sebenar-benar dirimu itu. Padahal aku tahu bahwa dirimu adalah diriku juga, maka sebenar-benar ilmu itu tidak lain tidak bukan adalah tipuan butamu.

Rene Descartes:

Menurutku, sebenar-benar dirimu adalah mimpiku. Tiadalah aku dapat menemukanmu tanpa mimpi-mimpi itu. Bagiku mimpi adalah nyata. Dan yang nyata bisa juga menjadi mimpiku. Maka sebenar-benar diriku adalah tidak dapat membedakan apakah kenyataan atau mimpiku. Namun aku selalu risau karena jika aku ingin berjumpa denganmu aku selalu dihadang makhluk hitam yang akan menyesatkanku, sehingga aku selalu salah mengertimu. Maka selalulah terjadi bahwa aku selalu meragukan keberadaanmu. Maka sebenar-benar dirimu adalah keraguanku itu sendiri. Itulah sebenar-benar ilmu, yaitu keraguanmu. Aku meragukan semuanya tanpa kecuali. Hanyalah ada satu yang tidak dapat aku ragukan yaitu diriku sendiri yang meragukan itu. Itulah satu-satunya kepastian bagiku. Cogito ergosum itulah kata-kataku, yaitu bahwa diriku itu ada karena aku tahu sedang meragukannya. Bukankah itu sebenar-benar dirimu. Maka sebenar-benar keberadaanku adalah diriku yang berpikir ini.

Immanuel Kant:

Menurutku engkau adalah pikiranku sekaligus pengalamanku. Tiadalah dirimu itu ada di situ tanpa pikiranku atau tanpa pengalamanku. Sedangkan tempat tinggalmu adalah di dalam intuisiku. Ketahuilah bahwa aku mempunyai dua intuisi yaitu intuisi ruang dan intuisi waktu. Itulah sebenar-benar tempat tinggalmu. Tetapi untuk mengerti tentang dirimu aku harus bersifat kritis. Maka sebanar-benar dirimu adalah pikiranku yang kritis. Itulah sebenar-benar ilmu, tidak lain tidak bukan adalah pikiran kritismu. Namun ketahui pula bahwa tidaklah mudah menggapai dirimu. Ketika engkau kukejar dengan pikiran kritisku, maka engkau lari menuju keputusanmu. Itulah setinggi-tinggi tempat tinggalmu, yaitu pada keputusanku. Maka sebenar-benar ilmu tidak lain tidak bukan adalah keputusanmu.

Orang tua berambut putih tidak puas dengan jawaban Socrates, Plato, Aristoteles, George Berkely, Rene Descartes, dan Immanuel Kant.

Ah, ngacau semua mereka itu. Jawaban mereka sesuai selera masing-masing. Maka belum puaslah diriku alan jawaban mereka semua. Tetapi aku sadar bahwa aku sedang berada pada jaman jauh sebelum diriku sekarang. Padahal aku belum sempat bertanya kepada orang sekarang yang berseliweran didepanku. Maka aku berjanji akan meneruskan perjalananku untuk bertanya siapa sebenarnya diriku itu?

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