Math 7

We know that, each interior angle of a regular hexagon

= (n – 2) × 180°
n

= (6 – 2) × 180° (where n = 6)
6

= 64 × 180° = 120°

And each exterior angle of a regular hexagon = 360°
n

= 360° = 60°
6

Hence, each interior angle and exterior angle of regular hexagon are 120° and 60° respectively.

Example: 2
Find the number of sides of a polygon, the sum of whose interior angles is 1260°.

Solution:

Sum of interior angles of polygon = 1260°
Number of sides of polygon (n) = ?

We know that,

Sum of interior angles of a polygon = (n – 2) × 180°

or, 1260° = (n – 2) × 180°

or, 1260° = n – 2
or 180°

7 = n–2

∴ n = 7 + 2 = 9.

Hence, number of sides of polygon (n) = 9.

Example: 3 110º 145º
In the adjoining figure, find the value of x.
Solution: 95º 160º xº
125º
Here, number of sides of given polygon (n) = 6

We know that,

Sum of interior angles of polygon = (n – 2) × 180°

or 95° + 125° + 160° + x + 145° + 110° = (6 – 2) × 180°

or x + 635 = 4 × 180°

or x = 720° – 635°

∴ x = 85°

Oasis School Mathematics – 7 45

Exercise 2.6

1. Find the sum of interior angles of following polygons.

(i) Pentagon (ii) Hexagon (iii) Octagon

2. Using the formula, find the size of each interior angle and exterior angle in the

following regular polygons.

(i) Pentagon (ii) Octagon (iii) Decagon

(iv) Square (v) Do-decagon (vi) Heptagon

3. Find the values of 'a' and 'b' in the given figure.

(a) (b) (c)

a a
b b

a b

4. (a) Find the number of sides of a polygon, the sum of whose interior angles are;

(i) 1800° (ii) 720°

(b) If an exterior angle of a regular polygon is 60°, find the number of sides of the

polygon.

(c) If an interior angle of a regular polygon is 108°, find the number of sides of the

polygon.

(d) If an interior angle of a regular polygon is 150°, find the number of sides of the

polygon.

5. (a) If the interior angles of a polygon are in the ratio 1 : 2 : 3 : 4 : 5, find the value of

each angles.

(b) The interior angles of a hexagon, 1 : 2 : 3 : 5 : 4 : 3, find the values of each angles.
6. F i nd the value of x or y in the following polygons.

(a) 150º (b) 70º 80º (c) 60º

70º 70º 4x
xº

150º 60º

160º xy 80º 5x 40º
yº
50º

Answer
1. (a) 5400 (b) 7200 (c) 10800 (d) 12400
2. (a) (i) 1080, 720 (ii) 1350, 450 (iii) 1440, 360 (iv) 900, 900 (v) 1500, 300 (vi) 128.570, 51.430
3. (i) a = 1200, b = 600 (ii) a = 1080, b = 720 (iii) a = 600, b = 1200 4. (a) 12 (b) 6 (c) 5
(d) 12 5. (a) 360, 720, 1080, 1440, 1800 (b) 400, 800, 1200, 2000, 1600, 1200,
6. (a) x = 2100, y = 600 (b) x = 800, y = 1000 (c) x = 200

46 Oasis School Mathematics – 7

2.7 Circle and Its Different Parts radius O

Put your compass needle at a plane paper and trace out the
path described by pencil needle as shown in figure. You will get
a curved path bounded in a plane and that path is called circle.
The point at which the needle of the compass is situated is called
centre of the circle and the outer path is called circumference of
the circle.

Radius

In the above activity, the distance between pencil needle and
compass needle is called radius. In other words, radius is the
distance between centre and circumference. The plural form of
radius is radii. There are many radii of a circle. They all are
equal in length.

Here, O is the centre of circle, so:

a = b = c = d.

Chord of a circle A
When we take any two points on the circumference of a circle B
and join them we will get a straight line segment that is called
chord. In figure, AB is a chord. P
O
Diameter r
A straight line passing through the centre and terminated by Q
circumference, as shown in figure, is called diameter of the
circle. C
PQ is a diameter of the circle. arc
• Diameter is the longest chord of a circle. Diameter divides
A
the circle into two equal parts. O
• Diameter is two times the radius of a circle.
A Sector C
Arc of a circle
In the adjoining figure, AC is a certain part of
circumference of a circle so AC is called arc of the circle.

Sector of a circle
In the adjoining figure, AC is an arc and AO and OC are radii so
OAC is the sector of a circle.

Oasis School Mathematics – 7 47

Semi–circle
The half part of circle as shown in the diagram
is called semi–circle.

Segment of a circle X

It is the region between a chord and arc of a circle as shown in Segment
the figure.

Central angle O

In the figure, OA and OB are radii of the circle Central
and AXB is an arc so ∠AOB is an angle at
center or central angle. A angle B
X

Angle at circumference (Inscribed angle) A
In the adjoining figure, AB and AC are two chords and BC BC
is an arc so the ∠BAC is called angle at circumference or
inscribed angle.

Exercise 2.7

1. Identify whether the given statements are true or false.
a) Diameter is the longest chord of the circle.
b) Diameter of a circle is equal to its radius.
c) The region between the two chords is called a sector.

d) Half part of the circle is called semi-circle.

e) Angle made by two chords of a circle at the circumference of a circle is called
inscribed angle.

f) Diameter of a circle is two times its radius.

2. What does the shaded part of given figures represent?

(a) (b) (c) (d) Q

OA B BO

OA A A P POQ
AB AB

3. What does the shaded part of given figures represent?

48 Oasis School Mathematics – 7

(a) A X (b) A (c)

B B

O C

4. Identify whether the given angles are inscribed angle or the central angle.

(a) (b) P (c) B

O SO AC
R D
A B
C

(d) A (e) S (f) P

OU O

BC T QT

D

5. Name the centre, a diameter and a chord in each of the following circle. Make

each of the given figure in your copy and shade the sector and segment in each of

given figure.

(a) A M (b) Y
P
O Z
O

C N XQ
B

6. Find the diameter of the circle whose radii are given below.

(a) 4cm (b) 5cm (c) 6cm (d) 18 cm.

7. Find the radius of a circle having following diameters.

(a) 8 cm (b) 6 cm (c) 9 cm (d) 7 cm.

8. Draw the circles having given radius.

(a) 5 cm (b) 3.6 cm (c) 5.2 cm (d) 4.1 cm.

Answer
Consult your teacher.

Oasis School Mathematics – 7 49

Unit

3 Congruency and Similarity

Introduction

Let's compare the following pair of figures.


Similar in shape and equal in size
Similar in shape but different in size

Similar in shape and equal in size Similar in shape but different in size

Similar in shape and equal in size Similar in shape but different in size

The pair of figures in the first column are congruent figures.

The pair of figures in the second column are similar figures.

Hence, two figures are said to be congruent figures if they are similar in shape and
equal in size.

Two figures are said to be similar figure if they are similar in shape even though
different in size.

Let's investigate !

• Are all circles similar?

• Are all rectangles similar?

• Are all triangles similar?

Congruency and Similarity in different shapes

3cm B 3cm
A Q 3cm

3cm 60° 60°
P 4cm 4cm

Congruent lines Congruent angles

50 Oasis School Mathematics – 7

Similar lines Similar angles

Cube Cube

Similar pictures Similar cubes

Exercise 3.1

1. Identify whether the given pair of figures are congruent or not. Justify your answer.

(a) 4cm B (b) A P
Q
A

4cm BC R
P Q

(c) A B P Q (d) A X
R CY
D CS B
Z
(e) (f)

2. Identify whether the following pair of figures are similar or not.

(a) A P (b) BP Q
R
A

B CQ R D CS

(c) (d)

Oasis School Mathematics – 7 51

(e) A P Q (f)
R
B EU
S
C DT

3. Draw the following figures and identify whether they are congruent or similar.
Justify your answer.

(a) A circle of radius 3cm and a circle of radius 4cm.

(b) A square having a side 4cm and a square having a side 5cm.

(c) A rectangle ABCD where AB = 5cm, BC = 4cm and a rectangle PQRS where
PQ = 5cm, QR = 4cm.

4. (a) Draw four pairs of congruent figures in your copy.

(b) Draw four pairs of similar figures in your copy.

Answer
Consult your teacher.




52 Oasis School Mathematics – 7

Unit Vertices, Edges and Faces

4 of Solid Object

4.1 Vertices, Edges and Faces of Solid Object

Vertices, Edges and Faces of Solid Object

Lets study the given figures and try understand the concept of vertices, edges and
faces of solid objects.

A A BO O

B CD C

A BP
D HG

Q

E F E Cube F D Pyramid C R Tetrahedron
Triangular Prism
Edges: AB, AD, Edges: OA, OB Edges: OP, OQ,
Edges: AB, AC, DC, BC, DE, AH, , OC, OD, AB, BC, OR, PQ, QR and
BC, DE, EF, DF, CF, BG, EH, EF, CD and AD PR
BE, AD and CF. FG and HG
Vertices: O, A, B, Vertices: O, P, Q
Vertices: A, B, C, Vertices: A, B, C, C, and D and R
D, E and F. D, E, F, G, and H
Faces: OAB, OAD, Faces: OPQ, OPR,
Faces: ABED, Faces: ABCD, OBC, OCD and OQR and PQR
ABCD
ADFC, BCFE, ADEH, EFGH,

ABC and DEF. BCFG, DCFE and

ABGH.

Lets count the edges, vertices and faces of above figures.

Name of solid Number of faces Number of Number of edges
object (F) vertices (V) (E)

Cube 68 12
8
Pyramid 55 6
9
Tetrahedron 4 4

Prism 56

In each solid object we can see that F + V = E + 2.

This formula is called Eulers formula.

Oasis School Mathematics – 7 53

4.2 Construction of Some Models of Solids

Name of Solids Figures Net Skeleton
Cube

Tetrahedron

Octahedron

Net of some solid objects

Solid objects Net Solid Objects Net

Cube Prism
Pyramid
Cuboid
Cone
Cylinder
54 Oasis School Mathematics – 7

Exercise 4.1

1. Name the following solid figures.

(a) (b) (c) (d)


(g)

(e) (f) (h)

(i) (j)

BF
AE

2. In the adjoining cube, count the number of : C G
D H
(a) Faces (b) Vertices (c) Edges

Hence, show that Number of faces + Numbers of vertices = Number of edges + 2

3. In the adjoining prism, count the number of D

(a) Faces (F) : (b) Vertices (V) : C E
(c) Edges (E) : A

Hence, show that F + V = E + 2 BF

4. Draw the following solid figures. Write down the number of faces (F), vertices (V)

and number of edges (E) and show that F + V = E + 2

(a) Cuboid (b) Pyramid (c) Tetrahedron.

5. Draw the net of following solid figures:

(a) Cube (b) Cuboid (c) Cylinder
6. Draw the net of : (c) Cone
(a) Prism (b) Pyramid

7. Name the solid objects having the following nets:

(a) (b) (c)

Oasis School Mathematics – 7 55

8. (a) A geometrical object has 6 faces and 8 vertices then find its number of edges.
(b) A geometrical body has 9 edges and 6 vertices, find its number of faces.

Answer
Consult your teacher.

Activities

1. Draw the following nets on a paper and cut the outlines of the net. Fold along the
dotted lines and paste the edges of the folded faces. Name the solid so formed.

(a) (b) (c) (d)

Objective Questions

1. If all the sides of a parallelogram are equal, then it is a

(i) square (ii) rhombus (iii) rectangle

2. An exterior angle of a regular pentagon is

(i) 108° (ii) 72° (iii) 120°

3. The shaded part of the given circle
represents

(i) segment (ii) sector (iii) chord

4. The given figure is the net of
( i) cube (ii) pyrami d (iii) prism.

5. Which of the following statements is not true?

(i) All angles of a rhombus are equal.

(ii) Diagonals of a rhombus bisect each other

(iii) All sides of a rhombus are equal.

6. Formula to calculate the interior angle of regular polygon is

(i) 3600 (ii) (n-2) × 1800 (iii) (n-2)×1800
n n

56 Oasis School Mathematics – 7

7. Which of the following is a regular polygon?

(i) rectangle (ii) square (iii) rhombus

8. Which of the following statement is not true?
(i) All the rectangles are similar
(ii) An equilateral triangle and a right angled triangle are similar figures.
(iii) All the squares are similar
9. What is the sum of exterior angles of a regular hexagon?

(i) 1800 (ii) 3600 (iii) 3600
n

10. What is a quadrilateral called whose opposite sides and opposite angles are equal?

(i) parallelogram (ii) kite (iii) trapezium

11. Which of the following statement is not true?

(i) The longest chord of a circle is a diameter.

(ii) Half part of a circle is called a semi-circle.

(iii) Radius of a circle is equal to its diameter.

12. Which one of the following is not a property of a parallelogram

(i) all sides are equal. (ii) opposite sides are equal. (iii) opposite angles are equal.

Assessment Test Paper

Full Marks – 35

Attempt all the questions. Group 'A' [11 × 1 = 11]

A

1. (a) From the given figure, write the names

of a pair of adjacent angles. AO B

(b) From the given figure, write the names DC
of a pair of complementary angles.

2. Find the value of x. B C
S
(a) A (b) R

x 55° x
D 80° 55°
B C P Q T

3. Find the value of x. E

(a) P x R B (b) A GB
130°
A

C 20° S D Cx D
H
Q F

Oasis School Mathematics – 7 57

4. What is the name of shaded part of the given circle? O
AB
5. Find the value of x.

(a) A 2x + 5 B (b) P Q
x

C S 60° R

D x+8

6. (a) What type of figures are said to be congruent figures?

(b) If two figures are similar in shape but different in size, what type of figures
are they?

Group 'B' [6 × 2 = 12] E

7. (a) Construct an angle 120° with the help of a protractor. P y B
x D
A
z


(b) In the given figure, find the value of x, y and z. C 60° 50°

QR

F
A

(c) Find the value of 'x' from the given triangle. 600

Bx 700 C y B

8. (a) Find the value of 'x', 'y' and 'z' in the given figure. A (3x + 10)°

(b) Find the interior angles of regular octagon.

(c) Draw the net of a cylinder. D (2x – 10)° z
C

Group 'C' [3 × 4 = 12] 88° B C D

A x y
E 40° F
9. From the given figure, find the value of x and y.

10. Verify experimentally that diagonals of a parallelogram bisect each other.

11. Construct a parallelogram ABCD where AB = 6.8cm, ∠B = 60° and BC = 4cm.

58 Oasis School Mathematics – 7

Co-ordinates

5Estimated Teaching Hours

Contents

• Co-ordinates and Graph

Expected Learning Outcomes

At the end of this unit, students will be able to develop the
following competencies:
• Plot the given points on the graph paper
• Identify the co-ordinates of given points

Teaching Materials

• Graph paper, pencil, etc.

Oasis School Mathematics – 7 59

Unit

5 Coordinates

5.1 Rectangular Coordinate Axes

We are familiar with number line. There is natural number 'zero', positive numbers to
its right and negative numbers to its left as shown in the figure.

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

If we draw similar vertical number line in which upper part contains positive

numbers, lower part contains negative numbers and zero at the centre and combine

them, we get the number lines as below: Y

Here, the horizontal number line XOX' 6

is called X–axis and vertical line YOY' is 5

called Y –axis and the point of intersection 4
O is called origin. In the figure, the plane 3

2

is divided into four parts which are called 1

quadrants. Here we discuss the nature of X' -6 -5 -4 -3 -2 -1 O 123456 X
-1
each quadrant separately.
-2

(a) First quadrant: The XOY plane which -3
is determined by positive part of both -4
X–axis and Y –axis is called first quadrant.
-5
-6

Any point in this quadrant contains both Y'

positive coordinates.

(b) Second quadrant: The X'OY plane which is determined by negative part of X–axis
and positive part of Y–axis is called second quadrant. Any point in this quadrant has
X – coordinate negative and Y – coordinate positive.

(c) Third quadrant: The X'OY' plane which is determined by negative part of both
X–axis and Y –axis is called third quadrant. Any point in this quadrant contains both
x– coordinate and y – coordinate negative.

(d) Fourth quadrant: The XOY' plane which is determined by positive part of X–axis
and negative part of Y –axis is called fourth quadrant. Any point in this quadrant
contains x– coordinate positive and y – coordinate negative.

The origin is considered as central point which is at a distance of 0 units from both
X–axis and Y–axis. So, its coordinates are given by (0, 0). In general, a point in any

60 Oasis School Mathematics – 7

quadrant is represented by (x, y) where x is a perpendicular distance from the
Y–axis and y is a perpendicular distance from the X–axis.

Representation of point on a Cartesian plane (on graph paper)

Let us consider two lines intersecting each other through origin perpendicularly as
XOX' and YOY' as shown below. The coordinate of point O is represented by O (0,0),
i.e. X – coordinate of O (0,0) is zero as well as Y – coordinate is also zero, we can
also define X – coordinate as abscissa of the point while Y – coordinate as ordinate.

Now in order to represent any point on the Cartesian plane we consider the distance
of given point along X – axis and along the Y – axis from the origin. Here the dis-
tance is measured in terms of unit selected on the graph paper.

Starting from origin right direction along X – axis Y X
shows the "+ ve" coordinate while left direction
along X – axis shows "– ve" coordinate. Similarly, X'
starting from origin, upward direction shows "+ O (0,0)
ve" coordinate of Y – axis and downward direction
shows "– ve" coordinate of Y – axis.

Hence, according to the coordinate of a point it can
lie on any four quadrant of Cartesian plane.

Plotting a point on the graph paper Y'

We consider any four points A (1, 1), B (–2, 1), C(–1, –3) and D (3, – 1). Here we can
plot these points on the Cartesian plane by using the pre–knowledge of coordinate
of a point and its position.

For example, A(1, 1) is a point with X – coordinate 1 and Y – coordinate also 1.

Y Steps:
• Start from origin.
• Move one unit right along OX.

B(–2, 1) A (1, 1) • Move one unit up along OY.
X' O (0,0) D(3, – 1) X

C(–1, – 3)


Y'

Hence we get the position of the point A (1,1) as shown in the figure.

In the same way we can plot the points B ( – 2, 1), C (–1, – 3) and D (3, – 1) etc. After
we plot them we can find that A (1,1) lies on 1st quadrant. B (– 2, 1) lies on 2nd quad-
rant, C( – 1, – 3) on 3rd quadrant and D (3, – 1) on the 4th quadrant.

Oasis School Mathematics – 7 61

Positiveness and negativeness of coordinates in different quadrants

The sign of coordinate of any point defines the Y

quadrant on which it falls. The figure gives a

clear idea about the positiveness and negative- (–, +) (+, +)
ness of the coordinate of any point on different

quadrants. X' X
O (0,0) (+, –)

(–, –)

Y'

The signs of each coordinate in different quadrants are summarized in the following
table:

Quadrants x-coordinate y-coordinate
+ ve
First + ve + ve
– ve
Second – ve – ve

Third – ve

Fourth + ve

Note : • Co-ordinates of origin is (0, 0)
• X - co-ordinate is also known as abscissa.
• Y - co-ordinate is also known as ordinates.
• X- co-ordinate in y axis is 0.
• Y - co-ordinate in x- axis is 0.


Exercise 5.1

1. Write the coordinate of the points shown on the graph paper.

Y
B

A
QP

X' C O (0, 0) D EX

F

S
R

Y'

62 Oasis School Mathematics – 7

2. Plot the given points on a graph paper.

(a) (1, 2) (b) (2, 4) (c) ( – 3, 4) (d) ( – 5, 2)

(e) ( – 1, – 2) (f) ( – 4, – 3) (g) (5, – 1) (h) (2, – 3)

3. Plot the following points A, B and C in the graph paper and join AB, BC and AC.

(a) A(1, 2), B(7, 4) and C (3, – 2) (b) A (3 – 4), B (1, 2) and C(–1, 5)

(c) A (–2, 3), B(1, 3) and C(–3, – 5)

4. In which quadrant do the following points lie?

(a) (2, 6) (b) (4, – 2) (c) ( – 8, 3) (d) ( – 6, – 4)

(e) (2, 5) (f) (3, – 5) (g) ( – 9, 1) (h) ( – 4, – 1)

5. Identify the points which lie on X – axis.

(a) (5, – 4) (b) (2, 0) (c) ( – 8, 0) (d) (0, 5)

(e) ( – 6, – 4) (f) (0, – 8) (g) ( – 4, – 2) (h) (5, 0)

6. Identify the points which lie on Y – axis.

(a) (4, 0) (b) (0, – 2) (c) ( – 8, 6) (d) ( – 3, 0)

(e) (0, 5) (f) (1, – 3) (g) (4, 1) (h) (0, – 8)



Answer
Consult your teacher.

Assessment Test Paper

Attempt all the questions. Full Marks – 14

Group 'A' [6×1=6]

1. (a) What is the co-ordinates of the origin?
(b) A point lies on X-axis. It is 6 units right from the origin. What is its co-ordi-

nates?
(c) A point lies on Y-axis. It is 5 units below from the origin. What is its co-ordi-

nates?
(d) On which quadrant do the following points lie?
(i) (2, – 3) (ii) (– 5, 4)
2. (a) Plot the points (5, 4) and (– 3, – 4) on the graph paper.
(b) What is the co-ordinates of a point which is 3 units right from the origin and

5 units up from the X-axis?

Oasis School Mathematics – 7 63

Group 'B' [4×2=8] Y

3. (a) From the given graph paper, A X
write the co-ordinates of the D
points A, B, C and D. X' C O

B

Y'
(b) Plot the points A (3, 5) and B (–1, 2) on the graph sheet and join AB.

4. (a) Plot the triangle having vertices A (3, 2) B (–5, 4) and C (–2, –3) on the graph
paper.

(b) Plot the triangle having vertices P (–3, 1), Q (1, –4) and R (5, 4) on the graph
paper and join PQ, QR and PR.



Objective Questions

Choose the correct alternatives.

1. X- co-ordinate is also known as:

(i) abscissa (ii) ordinate (iii) co-ordinates

2. A point lies on Y-axis. It is 3 units above the origin. Its co-ordinates is :

(i) (3, 0) (ii) (0, 3) (iii) (3, 3)

3. A point lies on X-axis. It is 4 units left from the origin. Its co-ordinates is :

(i) (–4, 0) (ii) (4, 0) (iii) (0, – 4)

4. On which quadrant does the point (3, – 2) lie?

(i) second (ii) third (iii) fourth

5. A point moves 5 units left from the origin and 2 units down from that position. Then

its co-ordinates is :

(i) (5, 2) (ii) (– 5, 2) (iii) (–5, – 2)

6. A point moves 4 units right from the origin and 3 units down from that position. Its

co-ordinates is :

(i) (4, – 3) (ii) (4, 3) (iii) (–4, 3)

7. Which of the following point lies on X-axis?

(i) (2, 0) (ii) (– 5, 3) (iii) (0, 4)

8. Which of the following point lies on Y-axis?

(i) (3, -2) (ii) (0, -2) (iii) (–3, 0)

9. Which of the following point lies on Third quadrant?

(i) (2, 3) (ii) (3, -2) (iii) (–5, – 4)

64 Oasis School Mathematics – 7

Mensuration

10Estimated Teaching Hours

Contents

• Perimeter
• Area

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• Find the perimeter of plane figures.
• Find the area of plane figures.
• Find the surface area of solid objects like cube,

cuboid, etc.

Teaching Materials

• paper models of triangles, quadrilaterals, etc.
• models of cube, cuboid, etc.

Oasis School Mathematics – 7 65

Unit

6 Perimeter and Area

6.1 Perimeter

Lets try to understand the given examples:
Santosh wishes to enclose his home by a boundary wall. What is the length of the

wall?

Aadhya walks 3 times around the park in the morning. How much distance does
she cover?

In the first case, the length of boundary is the perimeter. Similarly, in the second
case the distance covered by Aadhya is three times the perimeter.

Again, how to calculate the perimeter of given figures?

A AB

FC

B C ED
(i) (ii)

Perimeter of figure (i) is AB + BC + AC and perimeter of figure (ii) is
AB + BC + CD + DE + EF + FA.

Hence, the length of the outer boundary of a closed figure is called its perimeter.

A

I. Perimeter of a triangle

The perimeter of a triangle is the sum of three sides of a triangle.

Perimeter = AB + BC + AC

B C

Note: Half of the perimeter of a triangle is called semi perimeter s = a + b + c
2
where a, b and c are the sides of the triangle.

If a triangle is equilateral, then its all sides are equal. So P = 3a where, 'a' is the length
of one side.

66 Oasis School Mathematics – 7

II. Perimeter of rectangles and parallelograms

In rectangle and parallelogram opposite sides are equal, so in the given figure
AB = DC = b and AD = BC = l (suppose) then the perimeter of
rectangle or parallelogram = AB + BC + CD + DA.

Al DA lD

b b

B Parallelogram C BC

Rectangle

P = b + l + b + l l
= 2l + 2b
= 2(l + b) l

If all sides of rectangles are equal then the rectangle is a square
l

and its perimeter is l + l + l + l = 4l.

l

Remember !

Perimeter of rectangle (P) = 2 (l+b)

Perimeter of parallelogram (P) = 2(l+b)

Perimeter of square (P) = 4l

Unit of perimeter is same as the unit of length and breadth since rhombus is also a
parallelogram, perimeter of rhombus = 4l.

Worked Out Examples

Example: 1 A 12cm B

Find the perimeter of quadrilateral ABCD in the given figure. 8cm 10cm
Solution:
D 18cm C
Here, AB = 12cm, AD = 8 cm, DC = 18cm, BC = 10cm
Perimeter of the given figure = AB + BC + CD + AD

= 12cm + 10cm + 18cm + 8cm

= 48cm.

Example: 2 A 18cm B

Find the perimeter of the given rectangle. 8cm
Solution:
DC
Here, Length of the rectangle (l) = 18cm

Oasis School Mathematics – 7 67

Breadth of the rectangle (b) = 8cm

We have,

Perimeter of the rectangle (p) = 2(l+b)

= 2(18+8)

= 2 × 26cm

= 52cm.

Example: 3

The perimeter of a rectangular land is 24 m. If the breadth is 4m, find its length.
Solution:

Here, Perimeter (p) = 24m

Breadth (b) = 4m,

Length (l) = ?
We have,

Now, P = 2(l + b)

or, 24 = 2(l + 4)
or,
or, 24 = l+4
2

12 – 4 = l

∴l = 8

So, the length of the field is 8 m.

Example: 4

Find the perimeter of a square the length of whose one side is 40 cm.
Solution:

Given, Length of a side (l) = 40cm

We have, P = 4l

or, P = 4 × 40 cm = 160 cm

so, the perimeter of the square is 160 cm.

Example: 5

A room is 4m longer than its breadth and the perimeter is 32m. Find its length and
breadth.

Solution:

Given, Perimeter (p) = 32m
Let, Breadth (b) = x,
then Length (l) = (x+4)

We have, P = 2(l + b)

68 Oasis School Mathematics – 7

32 = 2(x + 4 + x)

or, 32 = 2x + 4
or, 2
or,
16 – 4 = 2x
12
2 = x

∴ x = 6

So, b = 6m, l = (6 + 4)m = 10 m

Example: 6

The length and breadth of rectangular garden is 50 m and 30 m respectively. A man
runs 10 times around the garden. Find the total distance covered by him.

Solution: 50 m

Length of the garden (l) = 50 m 20 m
Breadth of the garden (b) = 20 m

We have,

Perimeter of the garden = 2 (l+b) = 2(50 + 20) m

= 140 m.

Since a man runs 10 times around the garden, distance covered by him.

= 10 × 140 m

= 1400 m

Exercise 6.1

1. Find the perimeter of the following plane figures.

(a) A (b) P (c) E

3 cm 7 cm 4cm 4cm
S
6 cm QF I
5cm
3cm
5 cm
B C 4 cm 6 cm G 2cm H
6 cm R

2. Find the perimeter of given rectangles and parallelograms.

(a) A B (b) P 5 c m Q (c) E F
G
5cm 2cm 5cm
R H
D 6cm C S
8cm

Oasis School Mathematics – 7 69

3. Find the perimeter of given square and rhombus.

(a) P Q (b) A 4.5c m B (c) E F

S 6cm R D C H 8cm G

4. Find the perimeter of given figures. 2cm
(a) 2 cm 2 cm (b)
5 c m (c)1cm 2cm 2cm
2 cm 3cm 2cm 2cm
6 cm
4 cm 2cm 3cm 2cm 2cm
2 cm 2cm
2 cm 3 cm 2cm

4 cm 2 cm 1cm

2cm

5cm 2cm

5. Find the perimeter of triangles whose sides are:

(a) 4 cm, 5 cm, 6 cm (b) 4 cm, 3 cm, 6 cm. (c) 6 cm each.

6. Find the perimeter of squares whose sides are

(a) 4 cm (b) 16 cm (c) 8.2 cm

7. Find the perimeter of rhombus whose sides are

a) 4 cm (b) 3.2 cm (c) 14 cm

8. Find the perimeter of rectangle whose sides are

(a) length = 4.5 cm, breadth = 1.5 cm

(b) length = 14 cm breadth = 6.2 cm

(c) length = 11.2 cm breadth = 7.8 cm
9. Find the perimeter of parallelogram whose adjacent sides are:

(a) 4 cm and 6 cm, (b) 14 cm and 12 cm

10. a. Find the perimeter of a triangular field whose length of edges are 14 m, 12m
and 10m. Also find the length of wire required to fence 4 times around it.

b. Find the length of rope needed to fence a square garden of length 14m. three
times.

11. a. A rectangular garden is 50m long and 40m broad. Swasti ran around the gar-
den 5 times. Find the total distance covered by Swasti.

b. Length of a square garden is 15m. Sulav ran 10 times around the garden. Find
the total distance covered by him.

c. Length and breadth of a rectangular garden are 30m and 20m respectively. If
Ramesh runs around it 10 times and Umesh runs around it 7 times, find the
total distance covered by each of them.



70 Oasis School Mathematics – 7

Answers

1. (a) 17cm (b) 20cm (c) 18cm

2. (a) 22cm (b) 14cm (c) 26cm 3. (a) 24cm (b) 18cm (c) 32cm

4. (a) 24cm (b) 25cm (c) 24cm 5. (a) 15cm (b) 13cm (c) 18cm

6. (a) 16cm (b) 64cm (c) 32.8cm 7. (a) 16cm (b) 12.8cm (c) 56cm.

8. (a) 12cm (b) 40.4cm (c) 38cm 9. (a) 20cm (b) 52cm

10. (a) 36 m, 144m (b) 168m 11. (a) 900m (b) 600m (c) 1000 m, 700m.

Project Work

Find the perimeter of (i) your room (ii) your dinning table (iii) your bed-
room (iv) your board.

6.2 Area of Plane Figure

Square having length of one edge is 1 cm
1cm, 1cm² is called unit square or one 1 cm² 4 cm
centimeter square.
7 cm
1 cm

It has an area of 1 square centimeter (cm²).

A rectangle of sides 7cm by 4 cm can be di-

vided into 28 squares of each 1 cm² and 8 squares of 0.5 cm × 0.5 cm each = 0. 25 cm².

Since the area of 1 square is 1 cm2, and the figure contains 28 squares area of this
figure = 28 cm2

= 7 × 4 cm2
= l × b
Hence, area of a rectangle = l × b.
In square, all sides are equal so its area is A = l².

l

A=l×b b l A = l² l

l l
Rectangle Square

I. Area of triangle:

Rectangle can be divided into two equal triangles, so the area of a triangle is one–

half of the area of rectangle, i.e.

Area of a triangle = 1 area of rectangle
2

= 1 × length × height
2

Oasis School Mathematics – 7 71

= 1 base × height
∴ ∆ 2

= 1 b × h
2

Activity:

To find the area of triangle.

Let us take a triangular piece of paper ABC whose base is BC = b and height
AT = h.

Let us cut the piece as shown in diagram along XYZ and AY. Rearrange the pieces
of paper overlapping AX on BX and AZ on ZC. Then, we will get a rectangle as
shown in the figure. Then, the area of rectangle so formed is BC × TY.

A

x h z xyz
B 2 C h
yh
2 B 2C
T
T

= b × h2 A
∴ Area of triangle = 12 b × h

In a right angled triangle one angle is 90° so its area is equal

to 1 of the product of two sides containing right angle h
2 B
i.e. ∆ ABC = 21 AB.BC.
b C

II. Area of parallelogram A D
Bx
Parallelogram can be divided into two equal triangles. So, C
D
area of parallelogram ABCD = area of ∆ABC + ∆ADC.
= 21 base × height + 12 base × height. G
C
= base × height

So, the area of parallelogram = base × height. In the given figure, A

area of parallelogram ABCD = BC × AH and area of paral-

lelogram ABCD = CD × AG. B
H

72 Oasis School Mathematics – 7

Activity:

To find the area of a parallelogram.
Let us take a piece of paper ABCD in a parallelogram shape and cut the piece
ECD and rearrange it as shown in the figure.

Then, the area of rectangle BCED = BC × EC

i.e. Area of parallelogram = Base × Height

A ED D A E

BC B C

Remember!
• Area of triangle = 21 base × height
• Area of parallelogram = base × height

• Area of rectangle = length × breadth

• Area of a square = (length)2



Worked Out Examples

Example: 1 A
BDC
Find the area of the following figures.
20cm
Solution: Here, base = 20cm 12 cm

height = 12cm
We have, area of triangle = 12 base×height

= 21 × 20 × 12

= 120 cm²

Example: 2

Find the area of parallelogram ABCD. AB

Solution: 10 cm

Here, base = 16 cm D E16 cm C

height = 10 cm

Oasis School Mathematics – 7 73

We have, area of parallelogram = base × height

= 16 cm × 10 cm

= 160 cm²

Example: 3

In the given figure, ABCD is a rectangle. Find the area of shaded region from the given
figures.

E
AD

7 cm

B C
8 cm
Solution:

Here, ABCD is a rectangle.

Length (l) = 8 cm

Beadth (b) = 7cm

Area of the rectangle ABCD = l × b

= 8cm × 7cm

= 56cm2

Again, area of ∆EBC = 1 base × height
2
1
= 2 BC × AB

= 1 8 × 7
2

= 28 cm²

Hence, area of a shaded part

= Area of rectangle ABCD – Area of ∆EBC
= 56 cm2 – 28 cm2
= 28 cm2

Example: 4 A D
SR
Find the area of shaded region from the given figure. 2 cm 4 cm
Solution: P 3 cmT
6 cm
Area of parallelogram ABCD = Base × height B Q
C
= BC × height

= 6cm × 4cm

= 24 cm2

74 Oasis School Mathematics – 7

Again,

Area of parallelogram PQRS = Base × height

= PQ × ST

= 3 cm × 2cm

= 6 cm2

Hence,

Area of shaded part = Area of parallelogram ABCD – Area of parallelogram PQRS

= 24cm2 – 6 cm2

= 18 cm2

Exercise 6.2

1. Find the area of the following rectangle and square:

(a) (b) (c) 10 cm

5 cm

6 cm 4 cm 15 cm
X
2. Find the area of the following triangles:

A (b) P (c)

(a)

4 cm 4 cm
8 cm
B 6 cm C
Q 5 cm R Y

3. Find the area of the following parallelograms: 12 cm Z
4 cm 4 cm
(a) (b) 12 cm (c)

3 cm 8 cm

4 cm

4. Find the area of shaded regions of the following:

(a) (b) (c)

6 cm 4 cm 2 cm 6 cm
6 cm
8 cm

8 cm 15 cm

8 cm

5. Find the area of rectangle having the following dimensions:

(a) length = 18cm and breadth = 10 cm.

(b) length = 12 m and breadth = 7 m.

Oasis School Mathematics – 7 75

6. Find the area of a triangular field having the dimensions:

(a) Base 14m and height 16m

(b) Base 10 m and height 14 m

7. (a) Find the area of a square field whose length of one side is 11 m.
(b) Find the area of a square whose perimeter is 420 cm.

(c) Find the perimeter of a square field whose area is 441 m².

8. Find the area of the following combined plane figures:

(a) B E (b) A D

A 4 cm 6 cm E 6 cm E
D

C 6 cm F B 16 cm C

Answers

1. (a) 30cm2 (b) 16 cm2 (c) 150 cm2 2. (a) 12cm2 (b) 10cm2 (c) 48 cm2
(c) 74cm2
3. (a) 12cm2 (b) 96cm2 (c) 80 cm2 4. (a) 24cm2 (b) 32cm2

5. (a) 180m2 (b) 84dm2 6. (a) 112m2 (b) 71.4m2

7. (a) 121m2 (b) 11025cm2 (c) 84m 8. (a) 48cm2 (b) 304cm2

6.3 Total Surface Area of Cube and Cuboid

In previous class we have studied model of cube and cuboid and their nets. Let's
study the calculation of total surface area of cube and cuboid with the help of given
nets.

Let's compare the net of the cube and the model of the cube.

l

ll l l
l ll l
l
l l ll l l l
l l
l Model of cube

ll

l

Net of cube

How many square faces are there in a cube?
What is the area of each face of the cube?
What is the area of 6 faces of the cube?

76 Oasis School Mathematics – 7

Then, what is the total surface area of the cube?
Since, each side of the cube is l,
Area of a face of the cube = l2
Area of 6 faces of the cube = 6l2
∴ Total surface area of the cube = 6l2

Again,
Let's compare the net of cuboid and model of cuboid.

l

h h h l h
h l bb
l b
b b b l
b
Model of cuboid
l

h h
h h

l

Net of cuboid

How many rectangular faces are there in a cuboid?

How many rectangular faces have area l × b?

How many rectangular faces have area l × h?

How many rectangular faces have area b × h? Remember!
What is the total area of 6 faces? TSA of cube = 6l2
Hence, TSA of cuboid
Two rectangular faces have area l × b = 2(lb + lh + bh)
Two rectangular faces have area l × h

Two rectangular faces have area b × h

∴ Total surface area of cuboid = 2lb + 2lh + 2bh = 2 (lb + lh + bh)

Worked Out Examples

Example: 1

Find the total surface area of given cube.

Solution:

Here, length of each side of the cube (l) = 6cm 6cm

TSA = ? 6cm
6cm
We have,

TSA of cube = 6l2

= 6 × 62

= 216 cm2

Oasis School Mathematics – 7 77

Example: 2

Find the total surface area of given cuboid. 7cm
Solution: 6cm

Hence, Length of the cuboid (l) = 8cm

Breadth of the cuboid (b) = 6cm 8cm
Height of the cuboid (h) = 7cm

Total surface area (TSA) = ?

We have, TSA of cuboid = 2(lb + lh + bh)


= 2(8 × 6 + 8 × 7 + 6 × 7)

= 2(48 + 56 + 42)

= 2 (146) cm2

= 292 cm2

Example: 3

Total surface area of a cube is 294 cm2. Find the length of each side.
Solution: Total surface area (TSA) = 294cm2

Length of each side (l) = ?

We have,

TSA = 6l2

294 = 6 × l2

or, l2 = 294
6
or,
or, l2 = 49
l = 7cm

∴ l ength of each side of the cube = 7cm.

Example: 4

Length and breadth of a cuboid are 6cm and 5cm respectively. If its TSA is 126 cm2, find
its height.
Solution:

Hence, Length (l) = 6cm

Breadth (b) = 5cm

Height (h) = ?
We have,

TSA = 2 (lb+lh+bh)

or, 126 = 2(6×5+6×h+5×h)

78 Oasis School Mathematics – 7

or, 126 = 2 (30 + 11h)

or, 63 = 30 + 11h

or, 11h = 63–30
h = 33
or, 11
h = 3 cm.
∴

Exercise 6.3

1. Find the total surface area of given cube. (c)
(a) (b)

5cm
8cm

5cm 8cm
5cm 8cm

5.5cm

2. Find the total surface area of given cuboid.

(a) (b) (c)

6cm 4.5cm 9cm
8.5cm
5cm 9cm 8cm
8cm

3. Find the total surface area of a cube having: 10cm

(a) each side = 9cm (b) each side = 7.5 cm (c) each side = 8cm.

4. Find the total surface area of a cuboid having:
(a) length = 8 cm, breadth = 7 cm and height = 6 cm.
(b) length = 7.5cm, breadth = 6 cm and height = 5 cm.
(c) length = 10cm, breadth = 7.5 cm and height = 4 cm
5. (a) The total surface area of a cube is 600cm2. Find the length of each side.
(b) Surface area of a cube is 294cm2. Find the length of each side.
(c) Area of each face of a cube is 144cm2. Find its total surface area and length of each

side.

6. (a) Surface area of a cuboid is 214 cm2. If its length and breadth are 7cm and 6 cm,
find its height.

(b) Surface area of a cuboid is 376cm2, if its length and breadth are 10cm and 8cm
respectively, find the its height.

Oasis School Mathematics – 7 79

7. (a) A packet of ink has breadth 8cm and height 7cm. If it covers 80cm2 while
keeping on the table, find the

(i) length of the packet (ii) surface area of the packet.

(b) A box having rectangular base has length 20cm and height 15cm. It covers the
area 360cm2 while keeping it on the table. Find (i) its breadth (ii) its total sur-
face area.



Answers

1. (a) 150cm2 (b) 384 cm2 (c) 181.5cm2 2. (a) 236 cm2 (b) 297 cm2 (c) 503cm²

3. (a) 486cm2 (b) 337.5cm2 (c) 384cm2 4. (a) 292 cm2 (b) 225 cm2 (c) 290 cm2

5. (a) 10cm (b) 7cm (c) 12cm, 864cm2 6. (a) 5cm (b) 6cm

7. (a) 10cm, 412cm2 (b) 18cm, 1860cm2

Project Work
I. Take a paper box.
• Take the actual measurement of its sides.
• Open its faces.
• Draw the net.
• Calculate the area of each face.
• Add the area of each face.
• Calculate the total area.
II. Take a chart paper.
• Draw the net of cuboid
• Cut it.
• Fold it and form the cuboid.

80 Oasis School Mathematics – 7

Objective Questions

Choose the correct alternatives.

1. If 'l' represents a side of a rhombus, then the perimeter of rhombus is

(i) 2 (l + b) (ii) l2 (iii) 4l

2. Perimeter of a rectangle is 36cm. If its length is 10cm, then its breadth is

(i) 26cm (ii) 8cm (iii) 10cm

3. Total length of outer boundary of a plane figure is its

(i) perimeter (ii) area (iii) volume

4. Area of a face of a cube is 25cm2. Then the surface area of the cube is

(i) 100cm2 (ii) 150cm2 (iii) 50cm2

5. Length of each side of a cube is 6cm. Then the surface area of the cube is

(i) 180cm2 (ii) 216cm2 (iii) 36cm2

6. Surface area of a cube is 600cm2. Then the area of each face is

(i) 10cm2 (ii) 100cm2 (iii) 600cm2

7. Formula to calculate total surface area of a cuboid is (iii) l × b
(i) 6l2 (ii) 2(lb + lh + bh)

8. Area of a parallelogram is

(i) length × breadth (ii) 1/2 base × height (iii) base × height

9. If 'a' be the length of each side of a cube, then its surface area is

(i) a3 (ii) 6a2 (iii) a2

10. If 'l' be the length of each side of a cube, then its surface area is

(i) l3 (ii) 6l2 (iii) 5l2

11. Area of the shaded part of given figure is

(i) 40 cm2 (ii) 80 cm2 (iii) 120 cm2 8cm

12. If the perimeter of a square is 24 cm, then its area is 10 cm
(i) 24cm2 (ii) 16cm2
(iii) 36 cm2

Oasis School Mathematics – 7 81

Assessment Test Paper

Full Marks: 20

Attempt all the questions.

Group A [4×1 = 4]

1. (a) Write the formula which is used to calculate the perimeter of the rectangle.

(b) Calculate the perimeter of the given square.

6.5cm

2. (a) Which formula is used to calculate the surface area of a cuboid?
(b) Calculate the area of parallelogram whose base is 10cm and height is 8cm.

Group B [4×2 = 8]
3. (a) Find the perimeter of the given rectangle.

12cm A 6cm B

15cm 5.5cm
4.5cm
(b) Find the perimeter of the given figure. F3.5cm C
4.5cm
12cmA

E 4cm D

4. (a) Find the area of the given triangle. B C

20cm

(b) Find the surface area of a cuboid whose length, breadth and
height are 7cm, 5 cm and 4 cm.

Group C [2×4 = 8]
5. Length and breadth of a rectangular lawn are 45m and 25m respectively. How much

distance does a man cover if he walks 6 times around it.
6. A first aid box is 6cm long, 12cm broad and 10cm high.
(i) Draw the net of the box.
(ii) Calculate the surface area with the help of the net.
(iii) Calculate the surface area using the formula.
(iv) Compare the area in (ii) and in (iii).

82 Oasis School Mathematics – 7

Transformation

11Estimated Teaching Hours

Contents

• Reflection
• Rotation
• Translation
• Symmetry, Tessellation and Designs
• Bearing and Scale Drawing

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• Reflect the given point or figure on X-axis and Y-axis
• Rotate the given point or figure through 900 or 1800 about origin
• Translate the given point or figure along the given direction
• Identify whether the given objects are symmetrical or not
• Draw the line of symmetry of symmetrical objects
• Find the bearing of a point along the given direction
• Find the distance between two places on the map using scale

factor

Teaching Materials

• Graph paper

Oasis School Mathematics – 7 83

Unit

7 Transformation

7.1 Reflection

Look at the mirror. You will get your image on the mirror. Now investigate the following:

• distance of the image. • nature of the image.

Distance of the image from the mirror is equal to the
distance of the object from the mirror.

The image so formed is inverted. O P I
This phenomenon is the reflection. Object Image

I. Properties of reflection Mirror

The following are the properties of reflection.

• The object and its image are equidistant from P A P'
the axis of reflection.

Here in the figure, PA = P'A and QB = Q'B. Q B Q'
M
• The areas of object and its image are equal.
Here in the figure, area of ∆ABC = area of ∆ A'B'C'. A A'

B C M C' B'

• Object and image are reverse of each other.

M

• The points on the axis of reflection are invariant points C D
i.e., their images are the same points. M

Image of point C is point C itself. Image of point D is point D itself.

Note: The points which are not changed in transformation are called invariant points.

II. The method of finding the image under reflection A

In the given figure, ABC is a triangle and M is the axis of

reflection. B C
M

84 Oasis School Mathematics – 7

Now, the reflection image of ∆ABC can be obtained by the following procedures:

• Draw AX perpendicular to the axis of reflection 'M'
and produce it to A' such that AX = XA'.

• Draw BY perpendicular to the axis of reflection A
'M' and produce it to B' such that BY = YB'

• Draw CZ perpendicular to the axis of BC
reflection 'M' and produce it to C' such that
CZ = ZC'. Y X Z M
B' C'

• Join A'B', A'C' and B'C' respectively. A'

Hence, ∆A'B'C' is the image of the ∆ABC after Y
reflection in the line 'M'. P (x, y)

Reflection in using coordinates

a) Reflection about X – axis

Let us consider any point P (x, y) on the M X
Cartesian plane. Draw PM ⊥ OX and produce it X' O (0, 0)

to P' such that PM=MP' so we have OM=x, PM

= y and MP' = – y. Here, we can observe that the P' (x, – y)
coordinate of P (x, y) after reflection on X – axis Y'
appears at P' with co–ordinate (x, – y). Hence

reflection on X – axis changes the coordinate as ∴ P(x, y) Reflection on X-axis P'(x, – y)

AnypointontheX–axisafterreflectionontheX–axiswillremainunalteredi.e.anypoint
P (a, o) will have the same image point P' (a, o) when reflected about X – axis.

b) Reflection about Y – axis: Y

Let P (x, y) be any point on the Cartesian plane.

PN⊥OY is drawn and produced up to P' such P'(–x, y) P(x, y)

that PN = NP'. Here ON=y, NP = x and NP' = – N

x. i.e. P' ( – x, y) is the image point of the object X' O (0, 0) X
point P (x, y) after reflection about Y–axis.

Hence the reflection about Y – axis changes the

coordinate as

∴ P(x, y) Reflection on y-axis P'(–x, y) Y'

Remember ! P'(x, -y)
P'(–x, y)
P(x, y) Reflection on x-axis
P(x, y) Reflection on y-axis

Oasis School Mathematics – 7 85

Worked Out Examples

Example: 1

Plot a point P (2, 3) on a graph paper and then find its image after reflection.

(i) on x – axis. (ii) on y – axis.

Solution: Y (ii) Y
(i)

P(2, 3) P'(–2, 3) P(2, 3)

X' O X X' O X

P'(2, –3)

Y' Y'

Example: 2 Q' (1, 2)

Write the type of transformation in the following cases.

(i) P (3, 4) P' (3, –4) (ii) Q (-1, 2)
Solution:

(i) Here, P(3, 4) P' (3, -4)

That is P (x, y) P' (x, -y)

Which is the reflection on X-axis.

(ii) Again, Q (-1, 2) Q' (1, 2)

That is Q (x, y) Q' (-x, y)

Which is the reflection on Y-axis.

Example: 3

Write down the coordinate of the image of the following points after reflection.
(i) on X – axis
ii) on Y – axis
A(2, 4), B(–6, – 3), C( – 1, 5), D(4, – 3)
Solution:

(i) Reflection on X – axis:

We have,

P (x, y) Reflection on X – axis P' (x, – y)

86 Oasis School Mathematics – 7

Now, A (2, 4) Reflection on X – axis A' (2, – 4)
Reflection on X – axis B' (– 6, 3)
B ( – 6, – 3) Reflection on X – axis C' (– 1, – 5)
Reflection on X – axis D' (4, 3)
C ( – 1, 5)

D (4, – 3)

(ii) Reflection on Y – axis:

We have, Reflection on Y – axis
Reflection on Y – axis
P (x, y) Reflection on Y – axis P'( – x, y)
Reflection on Y – axis A' ( – 2, 4)
Now, A (2, 4) Reflection on Y – axis B' (6, – 3)
C' (1, 5)
B ( – 6, – 3) D' ( – 4, – 3)

C ( – 1, 5)

D (4, – 3)

Example: 4

Draw the image ∆A'B'C' when ∆ABC with coordinates A (1, 1), B ( – 3, 4) and C (0, 5) is

reflected on X – axis. P'(x, –y) B(–3, 4) Y
Solution: C(0, 5)
We have, P(x, y) Reflection on X – axis

∵ A (1, 1) Reflection on X – axis A'(1, –1) X' O A'(1, 1) X
B ( – 3, 4) Reflection on X – axis B'(–3, –4) A'(1, – 1)
C (0, 5) Reflection on X – axis C'(0, – 5) B'(–3, – 4)

The shaded part of the ∆A'B'C' is C'(0, – 5)
the image of given ∆ABC. Y'

Exercise 7.1

1. Draw the given figure in your copy and reflect on the given line.

(a) (b)

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2. Plot the following points on the square paper and then draw their images when
reflected on X – axis.

(a) (2, 4) (b) ( – 3, 0) (c) (4, – 2) (d) (– 8, – 1)

3. Draw the following points on the square paper and then draw their images when
reflected on Y – axis.

(a) (1, 3) (b) ( – 2, 4) (c) ( – 5, – 3) (d) (6, – 3)

4. Write down the coordinates of the following points after reflection on X – axis.

(a) (1, 2) (b) ( – 3, 5) (c) (6, – 4) (d) ( – 4, – 3)

5. Write down the coordinates of the following points after reflection on Y – axis.

(a) (4, 0) (b) (8, – 2) (c) ( – 1, – 3) (d) (0, – 3)

6. Find the type of transformation of the Object Image
following objects and images. a) (4, 2) (4, – 2)
b) (8, – 1) ( – 8, –1)
c) (2, – 6) (2, 6)
d) (– 3, – 4) (3, – 4)

7. Plot the points A (5, 0), B (0, 5) and C (1, 1). Join them to get ∆ABC. Now reflect it on

(i) X – axis (ii) Y – axis

Show the ∆ ABC and its image ∆ A'B'C' on graph paper and shade the image.

Answer
Consult your teacher

7.2 Rotation

Study the following illustrations and get the idea of rotation of a point. Now, we
take a point P, that rotates about a point O as shown in the figure.

In figure (i), ∠POP'= 90° and OP = OP' i.e. the
point P rotates about O through angle 90° in the
anti-clockwise direction.

Rotation through 900 in anticlockwise direction = +900

In figure (ii), ∠POP' = 90° and OP = OP' i.e. the
point P rotates about O through an angle 90° in the
clockwise direction.

Rotation through 900 in anticlockwise direction = –900

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In figure (iii), OP = OP' and ∠POP'= 180° i.e. the point
P rotates about O through a 180° in both direction
producing the same image.

In figure, (iv), OP = OP' and ∠POP' = 270° i.e. the
point P rotates about O through an angle 270° in
anti-clockwise direction.

In figure (v), OP = OP', ∠POP' = 360°, the point P
rotates about O through an angle of 360° in both
direction producing the same image.

Thus, a rotation is defined when its centre, the angle of rotation and the direction of
the rotation are given. The direction of rotation can be clockwise or anticlockwise
direction.

Hence rotation is a rule which shifts each point of an object in the same direction
through a certain angle about a fixed point.

I. Rotation of a Line PQ, about the Point O Through 90° in Anticlockwise
Direction

Anti-clockwise Steps:
• Join OP and OQ.

• Taking O as centre and OP as radius, draw
an arc in anticlockwise direction.

• At O, draw ∠P'OP = 90° and OP = OP'
meeting the arc through P at P'.

• Similarly, taking O as centre and OQ
as radius draw an arc in anticlockwise

P direction.

• At O, draw ∠Q'OQ = 90° and OQ = OQ',
meeting the arc drawn through Q at Q'.

• Join P'Q'. Hence, P'Q' is the image of PQ.

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II. Rotation of a ∆ABC about the Point O Through 60° in Clockwise Direction

Let ABC be a triangle and O is the point.

Clockwise Steps:
• Join AO, BO and CO with dotted lines.

• Taking O as centre and OA as radius, draw an
arc in clockwise direction.

• At O, draw ∠A'OA = 60° and OA = OA' meeting
the arc drawn through A at A'.

• Similarly, taking O as centre and OB and OC
as radius, draw arcs in clockwise direction
respectively.

• At O, draw ∠B'OB = 60° and OB = OB' and
∠C'OC = 60° and OC = OC' meeting the arcs
drawn through B and C at B' and C' respectively.

• Join A'B', B'C' and A'C'.

Hence, ∆A'B'C' is the image of ∆ABC.

Properties of Rotation

The following are the properties of rotation:

(i) Area of object and its image are equal.

(ii) The centre of rotation is the invariant point.

(iii) Each point of an object turns through an equal angular displacement in the
same direction.

(iv) The perpendicular bisector of the line segment joining a point of the object and
its corresponding image passes through the centre of rotation.

Rotation Using Co-ordinates

Just as in reflection, the images of the objects can easily be obtained by the rotation
through a given angle using co-ordinates. Look and learn the following illustrations.

(a) Rotation through 90° (or –270°) in anticlockwise direction about origin.

P(3,4) +900 P'(–4, 3)

Symbolically,
P(x, y) R(O,+900) P'(–y, x)

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(b) Rotation through 90° in clockwise direction about origin.

P(3, 4) – 900 P' (4, –3)
Symbolically,

P(x, y) R(O, –900) P' (y, –x)


(c) Rotation through 180° in anti-clockwise and clockwise direction.

Y

Anti-clockwise
Clockwise
P(3,4) P(3, 4) R(O,+1800) P'(–3, –4)
900 Symbolically,

X' X P(x, y) R(O,+1800) P'(–x, –y)

P'(-3,-4)

Y'

Remember !

• R (O, + 90°) is equivalent to R (O, – 270°).
• R (O, –90°) is equivalent to R (O, + 270°) .
• R (O, + 180°) is equivalent to R (O, – 180°).
• Positive quarter turn means rotation through + 90°.
• Negative quarter turn means rotation through –90°.
• Quarter turn in anticlockwise direction means positive quarter turn.
• Quarter turn in clocwise direction means negative quarter turn.

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Worked Out Examples

Example: 1

Find the coordinate of the image of point (5, 2) after a :
(a) positive quarter turn or rotation through + 900 about origin.
(b) half turn or 180°.
(c) three quarter turn + 270°, about the origin.
Solution:

a) We have, P(x, y) R(O,+900) P'(–y, x)
P(5, 2) R(O,+900) P'(–2, 5)

b) We have, P(x, y) R(O,+1800) P'(–x, -y)

P (5, 2) P'( -5, -2)

c) We have, P(x, y) R(O – 900) P'(y,-x)

P(5, 2) P'(2, -5)

Example: 2

Write the type of rotation in the following cases:

(a) P (2,3 ) P' (-3, 2), (b) Q(1, -2) Q' (-2, -1), (c) R (-4, 5) R'(4, -5)

Solution:

Here,

(a) P (2,3 ) P' (-3, 2)

i.e. P(x, y) P' (-y, x)

Which is rotation through +900 about origin.

(b) Q (1, -2) Q' (-2, -1)

i.e. Q(x, y) Q' (y, -x)

Which is rotation through –900 about origin.

Again,

(c) R (-4, 5) R' (4, -5)

i.e. R (x, y) R' (-y, -x)

Which is rotation through 1800 about origin.

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Example: 3

Draw a ∆ABC with vertices A (0, 0), B (5, 1) and C (2, 4) on the square paper. Now rotate
it through 2700 along clockwise direction. Find the coordinate of the vertices of image
∆A'B'C'. Plot ∆A'B'C' and show it by shaded part.

Solution: Y
B'(–1, 5)
When rotation takes place about origin
through 2700 along clockwise direction, it C(2, 4)
is equivalent to the rotation through 900
anticlockwise. Hence by formula, C'(–4, 2) A B(5, 1)
P (x, y) R, (O+90º) P' ( – y, x) X' X
A'(0,0)
∴ A (0, 0) R, (O+90º) A' (0, 0)

B (5, 1) R, (O+90º) B' ( – 1, 5)

C (2, 4) R, (O+90º) C' ( – 4, 2) Y'

The shaded part shows the image ∆ A'B'C'.

Exercise 7.2 A A
C
1. Draw the given figures in your copy and

rotate ∆ABC through + 90° about O. B CB

OO

2. Plot the following points on the square paper and then find their images when
rotated about origin through (i) 90° (ii) 180° and (iii) 270° in anticlockwise direction.

(a) (1, 4) (b) ( – 4, – 3) (c) (–5, 8) (d) (1, – 6)

3. Write down the coordinates of image after rotation about origin through + 90° for
the following points.

(a) (5, 1) (b) (2, – 4) (c) (–3, 6) (d) (– 4, – 5)

4. Write down the coordinates of image after rotation about origin through 1800 for the
following points.

(a) ( – 3, 6) (b) (2, 4) (c) (–5, – 3) (d) (1, – 2)

5. Write down the coordinates of image after rotation through 2700 about origin for the
following points.

(a) (4, 5) (b) ( – 6, 1) (c) (8, – 2) (d) ( –5, –3)

6. The image of some points and the points are given below. Write down the type of
rotation applied

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Point Image
a) (2, 4) ( – 2, – 4)
b) (1, 6) ( – 6, 1)
c) (4, – 3) ( – 3, – 4)

7. Plot the points P ( –1, 4), Q (0, 1) and R (1, 4). Join them to get ∆PQR. Now find
the image ∆P'Q'R', when rotated through (a) 90° b) 1800 and c) 270° about origin.
Also shade the image ∆P'Q'R' in each case.

8. Draw a unit square with coordinates of its vertices as (0, 0), (1, 0), (1, 1) and (0, 1).
Now find the coordinates of vertices of image when it is rotated through 1800 about
origin.

Answer 
Consult your teacher

7.3 Translation P'

Look at the given figure:

P

Q'
Q

Hence, P is shifted to P' and Q is shifted to Q'. Such that PP' = QQ'.

Again, PP' // QQ'

Q' i.e. object is shifted to the same distance and along the same direction.

Hence, each point of an object (or a geometrical figure) shifted along the same
distance and same direction is called translation or displacement.

The displacement of an object (or geometrical figure) has magnitude as well as
direction. Hence it is a vector quantity.

Example: Let AB be given line and a→ be translation vector. A a A'

Now, A and B are shifted to A' and B' such that AA' = a→ . Image

Then AB is said to be translated to A'B' by a .

Translation vector: The fixed value by which the geometrical B B'

figure will be displaced is called translation vector. Translation vector

A translation vector has two components. X-component and Y-component. If it
displaces a point right its X-component is positive. If it displaces a point left its
X-component is negative. Again, if it displaces a point up its Y-component is positive
and if it displaces a point down its Y-component is negative.

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