Prime Mathematics 7

Exercise 1.9

1. Define the following. (b) Radius (c) Centre (d) Diameter
(a) Circle

(e) Circumference (f) Chord (g) Arc

(h) Sector (i) Segment

2. Look at the figure, and name the following
O=
OC =
AB =
MN=

3. Look at the figure and name the following.
(a) The point O
(b) Shaded region OAXB.
(c) The region MYN.

4. Construct the circle of the following radii. X

(a) 3.5cm (b) 4cm (c) 5cm (d) 5.2 cm

5. Locate the centre of the following circles
a) b)

6. Locate the centre of the circle of which the following are the arc.
a) b)

44 Geometry

1.10 Geometrical Solids Estimated Period - 6

Polyhedra:

A polyhedron is a solid bounded by plane polygonal faces. Some of the simple
polyhedrons are shown below.

A prism A tetrahedron A Cuboid

A pyramid A frustum. An octahedron

If a polyhedron has all of its faces congruent regular polygons, it is called a
regular polyhedron. There are only five regular polyhedra.

A regular tetrahedron A regular hexahedron or a cube A regular Octahedron

(having 4 congruent (having 6 square faces of same (having, 8 congruent equilateral

equilateral triangular faces) size) triangular faces)

A regular dodecahedron A regular icosahedron
(having 12 congruent pentagonal faces) (having 20 equilateral triangular

Historical fact: faces)

Euler Descartes Formula

In a simple closed polyhedral surface V - E + F = 2

where V, E and F denote the number of vertices,

edges and faces respectively. This relation was first Leonhard Euler

observed by Rene descartes in 1649 and the first proof (1707-1783, Switzerland)

was given by Euler in 1752.

Prime Mathematics Book - 7 45

Lets Check Euler’s Property

Solid Name No of No of No of Remarks
vertices(V) edges(E) faces(F) V-E+F

Triangular 6 952
Prism

Tetrahedron 4 642

Cuboid/Cube 8 12 6 2

Octahedron 6 12 8 2

Pyramid 5 852

46 Geometry

Models of some of the polyhedra A regular tetrahedron

Tetrahedron

Draw the net of a regular tetrahedron
on a chart paper as shown (all the triangles
are equilateral).
Cut along the dotted lines and fold along
the bold lines.
Put glue over shaded parts
Bring the points A, B and C together with
glued part inside and join
properly.

Regular Hexahedron (cube)

Draw the net on a chart paper
(with suitable sides) as shown.
Cut the paper along dotted lines
and fold along bold lines.
Put glue over shaded parts
Bring the edge 1 and 2
together with glued part inside
and join properly to form a cube.

Note: As indicated by l, b, h, If the sides l, b, h are all equal it forms a cube. If l, b
and h are made different it forms a cuboid.

Regular Octahedron Circular base

Draw the net on a chart paper as shown
(equilateral triangle of suitable length)
Cut along the dotted lines and fold
along the bold lines
Put glue on the shaded parts .
Bring the ends A and B together and
with glued part inside join
properly to form an octahedron.

Cylinder.

A cylinder is a piece of pipe. Lateral
a solid object A tin can. Curved
bounded by a surface
regularly curved HB Circular base
lateral surface
with two equal Unshaped pencil
circular faces to
its opposite ends.

Prime Mathematics Book - 7 47

Lateral part of a cylinder is regular
curved , so called curved surface.
To the opposite ends are equal plane
circular face called bases.
It has no vertices.
It has two circular edges.
Distance between two bases is called
height or length of the cylinder.
Note : If the opposite circular bases
are not equal, it is not a cylinder, it
is a frustum of cone.

Net of a cylinder: Take a closed hollow tin cylinder and cut along the circular edges
we get two circular parts and an open cylinder.

Again cut the hollow open cylinder along length and flatten it to get a rectangle part.

M 10cm N

48 Geometry

Let’s make a model of cylinder with height 10 cm and radius 3.5cm
Draw a line XY = 17cm on a chart paper.
At X and Y draw circles of radius 3.5 cm
On remaining part AB complete a rectangle ABCD with BC = AD = 22cm.
Cut along the out lines, bring the sides AB and CD together and join the circles
along the circumference with cello tape.

Cone:

A Sharpened end Funnel Ice cream A sand heap

cone
is a solid
having
regular curved
surface emanated
from a point and a
circular base.

of pencil cone

In the given cone, O is the centre of the Vertical P Curved
circular base, OA is the radius of the height surface
circular base, OP is called vertical height h
and AP is slant height. l o Circular base

Slant height
A

Radius

Experiment:

Take a circular, piece of paper or a filter paper from science lab.
Make a fold and again double fold.
Separate a leaf and with your finger, form a cone.

Prime Mathematics Book - 7 49

Model of cone:
Take a circular piece of paper
Cut along OA and OB. where O is the
centre of the circle and get two
separate sectors.
Take one of the sectors and join along OA
and OB using a cello tape. The structure
is a cone.

Exercise 1.10

1. Write down the name of two objects having the following shapes.
(a) Cube (b) Cuboid (c) Cylinder (d) Cone

2. Draw figures of regular
(a) tetrahedron (b) hexahedron
(c) octahedron and complete the following table.

Regular No of Edges No of faces Shape of each
Polyhedron Vertices (v) (E) (F) face

Tetrahedron

Hexahedron

Octahedron

3 Count the number of vertices, edges and faces of each of the following
and check for Euler’s property.

(i) (ii) (iii) (iv)
Result
Fig No. of vertices (v) No of Edges (E) No of faces (F) V-E+F
(i)
(ii)
(iii)
(iv)

4. Draw the nets of the following.
(a) A regular tetrahedron (b) A regular hexahedron (c) A regular Octahedron.

50 Geometry

Unit Revision test I

1. Bisect the angle ABC

2. Construct the angle (a) 75o (b) 135°

3. Construct an angle PQR equivalent to ABC

4. Write the complementary and supplementary of
(a) 65o (b) 50o

5. Find the size of unknown angles in the figure.

6. Find the size of the angles represented by x , y and z in the
given figure.

7. In the figure given alongside, find the angles
represented by x and y.

8. Find the value of x and y in the following figures.

9. Verify experimentally that the opposite angles of parallelogram are equal.

Prime Mathematics Book - 7 51

Unit Revision Test II

1. Construct a triangle ABC in which AB=4.9cm, BAC = 45o and AC = 6.8 cm.
2. Construct a right angled triangle PQR in which PQ = 5.2 cm and hypotenuse

PR = 6.9cm.
3. Construct a parallelogram ABCD in which AB = 5cm, AE = 7cm and BAC = 30o
4. Construct a rhombus ABCD in which AC = 6cm and BD = 4.8cm
5. Find the size of an angle of an octagon.
6. Find the number of sides of the regular polygon the sum of whose interior angles

is 900o.
7. Find the size of the side represented by x in the given figure.

8. Prove that DABD @ DCDB in the given parallelogram ABCD.

9. Draw the net of a regular octahedron.
10 Count the number of vertices, edges and faces of the given polyhedron and verify

Euler’s property

52 Geometry

Answers

Unit:1 Geometry

Exercise: 1.1

Show to your teacher.

1. a) 52° b) 16° Exercise: 1.2 d) 0° e) 67.5°
c) 2°

2. a) 75° b) 130° c) 60° d) 115° e) 19.5°

3. a) 37.5°, 52.5° b) 120°, 60° c) 54°, 126° d) 72°

4. a) 20° b) 60° c) 75°
g) 45° h) 305°
d) x=45°, y=135°, z= 45° e) 30° b) Alternate angles c) Co-interior angles
e) Corresponding f) Co-interior angles
f) x = 80°, y=100°, z = 80°

6. a) Corresponding angles
d) Exterior alternate angles

7. a) ∠QRT b) ∠PQR c) ∠YQU b) 40°
8. 180° 9) a) 85°, 95° b) 63°, 117° d) a = 50°, b = 50°, c= 130°
10. a) 80°, 100° b) 36°, 54° c) 50° d) 105° f) a= 68°
11. a) 105° h)a=50°, b=130°, c=50°
j) a = 30°, b = 40°
c) a = 110°, b = 70°, c=110° l) a = 60°, b = 40°, c = 80°
e) a = 58°, b = 122°, c = 58°
g) b=110°
i) a = 100°, b = 80°, c = 100°, d=80°
k) a = 50°, b = 70°, c = 60°

1. Show to your teacher. Exercise: 1.3
2. Show to your teacher.

3. a) 35°, 70° b) x = y = 50° c) 46° d) 10° e) 20°

f) x = 55°, y = 70° g) x = 35°, y = 70° h) x = 65°, y = 65°

i)x = 55°, y = 55° j)x = 140°

4. a) x = 75°, y = 105° b) x = 4cm, y = 4cm c) x = 4cm, y = 7cm d) x = 11cm
e) x = 3cm f) x = 90°, y = 40°

5. a) x = 80° b) 36° c) 42° d) x = 60°, y = 60°, z = 120° e) x = 70°, y = 30°,
z = 80° f) x = 60°, y = 120°, z = 60°

6. a) 80° b) 46°, 69°, 115° c) 80°, 120°, 90°, 70°
7. Show to your teacher.

Prime Mathematics Book - 7 53

Show to your teacher. Exercise: 1.4

Show to your teacher. Exercise: 1.5

1. a) 5 b) 4 Exercise: 1.6 d) 9
2. a) 540° b) 1440° c) 7 d) 720°
f) 900° c) 360°
e) 1260° b) 144°
3. a) 108° b) 90° g) 1080° h) 180°
4. a) 120° e) 45°
b) 120° c) 135° d) 120°
d) 60° b) 7
5. a) 95° e) 11 c) 72°
6. a) 6
f) 36°
d) 14
c) 36°, 72°, 108°, 144°, 180° d) 60°

c) 10

Exercise: 1.7

1. a) AA b) SSS c) AA
d) AA
e) SAS f) AA
2. a) x=12cm
b) x= 5.4cm

1. Show to your teacher. Exercise: 1.8
2. a) x=4cm, y=4cm, z=4cm b) x= 2.5cm, y=7cm, z=4cm

c) x=10cm, y=7cm

Exercise: 1.9

Show to your teacher.

Exercise: 1.10

Show to your teacher.

54 Geometry

UNIT CO-ORDINATE Objectives:

2 GEOMETRY

Estimated periods 5

At the end of this unit the students will be able to
● give the introduction of axes, quadrants and co-ordinates.
● find the co-ordinates of the given points in the graph.
● plot the given points in the graph.

Y
Y

RN P P

X’ Q X X’ S
OM
X

AQ

Y’ Y’

Teaching Materials:
Graph chart, geoboard, scale, pencil and different colour sign pens

Activities:
It is better to
● Demonstrate the graph with axes and origin in the graph.
● Display the graph chart with axes and origin to give the concept of plot-

ting the points on a graph paper.
● Give idea to find the coordinates of a point on a graph paper.

2.1 Introduction of Quadrants

In the adjoining figure, two number lines XOX´ Y
and YOY´ intersect each other perpendicularly 2nd 1st
at a point O. These number lines are called the
co-ordinate axes.

The horizontal number line XOX´ is called the X´ o X
x-axis and the vertical number line YOY´ is called
the y-axis. The point ‘O’ where the two lines XOX´ 3rd 4th

and YOY´ intersect each other is called the origin.
The point ‘O’ is also called the point of Reference.
The plane of two axes is called the X - Y plane. The Y´

X - Y plane is divided into four parts by two axes and each part is in the form of open
rectangle which is called the quadrant. In the above figure, XOY, X´OY, X´OY´, XOY´
are four quadrants.

1st quadrant XOY OX-positive OY- positive (+,+)
2nd quadrant X´OY OX-negative OY- positive (-, +)
3rd quadrant X´OY´ OX-negative OY- negative (-, -)
4th quadrant XOY´ OX-positive OY- negative (+, -)

Co-ordinates of the points in the graph

The position of a point on the plane is completely fixed if we know its distances
from X-axis and Y-axis . So, to find the position of a point on the plane, we draw the
perpendiculars on the X-axis and Y-axis from the point.

In the adjoining graph, for the position of a point Y

P, we draw the perpendiculars PM on the x-axis

and PN on the y-axis from the point P. Then the

distance PN=OM along the x-axis is called the RN P
X-cordinate of the point P. It is also called the Q

abscissa of the point P. The distance PM=ON X´ O M X
along the y-axis is called the y-co-ordinate A

of the point P. It is also called the ordinate of

the point P. The abscissa and ordinate together

are called co-ordinates. In the above graph,

OM = 5units and on ON = 4units. Thus the Y´
X-coordinate and Y-cordinate of the point P are 5

and 4 respectively. Therefore, the coordinates of the point P is (5,4).

Find the co-ordinates of the point Q by the above activities.
Are the co-ordinates of the point R as (-2,4) ?
What will be the co-ordinates of the point A?
From the above graph, the co-ordinates of the points are
P(x,y) = p(5,4), Q(x,y) = Q (3,2), R(x,y) =R(-2,4) and A(x,y) = (4,-3).

56 Coordinate

Thus , we find the co-ordinates of a point on the basis of the value of the number line
along the x-axis and y-axis .

The co-ordinates system gives the accurate position of the point in the plane.

Note: The co-ordinates of the origin are(0,0).

The y-co-ordinate of any point on the x-axis is o.
The x-co-ordinate of any point on the y-axis is o.

Example 1: Find the coordinates of the points A, B and C Y
from the given adjoining figure. Also write the name of the
figure which is formed by joining the points A, B and C.

Solution: A
The co-ordinates of the points A, B and C from the given BC
figure are as follows. X´ X

A(x,y)= A(3,4), B(x,y)=(-1,-2) and C(x,y)= C(4,-2). Y´

After joining the points A, B and C one after another
respectively, a closed figure bounded by three lines is
formed . So the name of the figure is a triangle ABC.

Exercise 2.1
1. Fill in the blanks:

a) If the co-ordinates of a point is (-2,5), the point lies on ……….quadrant.
b) If the x-co-ordinate of a point is 3 and y-co-ordinate is -2, the position of the

point is………………….
c) If the position of a point is (6,-2), the abscissa of the point is……….
d) If the co-ordinates of the point (0,4) ,the the point lies on…………..

2. Find the co-ordinates of the points A, D, M, P, S, T and Z Y
from the given graph.
T A X
X´ Z
O D
3. Find the co-ordinates of the points P, Q, R and S from S M
the given graph. Also write the name of the figure which
is formed by joining the points P, Q, R and S respectively. P

X´ Y´
Y

SP X

O
RQ

Y´ 57

Prime Mathematics Book - 7

4. Write down the coordinates of the vertices of the given different geometrical
shapes in the following graphs.

YY

SP X X´ P X
X´ O
S
RQ
O

Q

Y´ Y´

2.2 Plotting the given points in the graph.

To plot a point in the graph with the help of its coordinates we draw the coordinate
axes XOX´ and YOY´ perpendicularly at point O. Then we take a suitable scale to mark
on the axes, Positive values are marked along OX and OY, similarly negative values are
marked along OX´ and OY´.

We will plot a few points to illustrate the process . Y
Plot the points P(3,2) ,Q (-4,2), R(-4,-5)
and S(3,-5).

The scale used in the graph is 1 small X´ Q P X
division equals 1 unit. for plotting a S
point P whose position is (3,2). First we O
take the number 3 of the coordinates
(3,2) and move 3 units along ox to the right R
of the origin ‘O’. Then we take the number
2 of (3,2) and move 2 units upwards parallel Y´
to YOY´ from the x-axis which is shown in
the adjoining above graph.

By the same process as plotting the point
P, the points Q(-4,2) ,R(-4,-5) and S(3,-5) are also plotted in the same graph. In the
graph, the points P, Q, R and S are joined respectively. A closed figure is formed . The
name of the closed figure is a square PQRS. The numbers of boxes occupied by the
Square PQRS are 49. So the area of the square PQRS is 49 sq. units.

58 Coordinate

Example 1: Plot the points A(3, 5), B(-4, 2), C(-3, 5), D(3, -4) and E(6, 0) in the same
graph paper. Also write the points which are equidistant from the x-axis
and y-axis respectively. Which of these points lie on the same quadrant?

Solution: The Given Points A(3, 5) B(-4, 2) C(-3, 5)
D(3, -4) and E (6, 0) are plotted in the
adjoining graph. Y

From the graph, the points A(3, 5) and CA

C(-3, 5) are equidistant from the x-axis. X´ B E X
Similarly, the points A(3, 5), C(-3, 5) and
O

D(3, -4) are equidistant from the y-axis. D

The points B(-4, 2) and c(-3, 5) lie on the Y´
same quadrants which is 2nd quadrant.

Exercise 2.2

1. Plot the following points in the same graph. D(4, 7)
A(-1, 2) B(4, 0) C(-2, -4) S(-5, 0) E(6, -2)
P(0, 6) Q(-3, -2) R(9, 7) S(2, 2) T(-2, 10)
A(-3, -3) E(5, 5)
2. Plot the following points in the same graph. B(-5, -4)
P(3, -2) Q(6, 0) R(-1, 2)
U(-5, 0) V(7, -2) W(0, 4)

a) Which of these points lie on the same quadrant?
b) Which of these points are equidistant from the x-axis?
c) Which of these points are equidistant from the y-axis?
d) Which of these points lie on the x-axis ?
e) Which of these points lie on the third quadrant ?
3. The points A(-5, 3),B(-5, -2),C(0, -2) and D are the vertices of a square ABCD. Plot
these points on the graph. Also find the co-ordinates of the point D and area of the
square ABCD.
4. Plot the points A(4, 6) and B(8, 10). Join the points A and B and find the co-
ordinates of the mid-point of AB.
5. Plot the following points and join them in order by using ruler. Also write the name
of the figures so formed.

a) A(4, 5), B(6, 1) and C(2, 1)
b) P(-6, 4), Q(0, 4), R(0, 0) and S(-6, 0)

c) A(3,5), B(2,5), C(-2,5), D(0,5), E(-6,5) and F(6,5)
6. Points P(5,3) Q(-2,3) R(-2,-4) and S are the four vertices of a square.

a) Plot these points on a graph and then find the co-ordinates of the point S.

b) Find the area of the rectangle PQRS.

Prime Mathematics Book - 7 59

Unit Revision

1. Fill in the blanks:
a) If the coordinates of a point is (-3,2), the point lies on ............ quadrant.
b) If the y-coordinate of a point is 4 and x coordinate is -12, the position of the
point is ............ .
c) If the position of a point is (-3,-5), the ordinate of the point is ....... .

2. Find the coordinates of the points P, Q, R and S from the given graph. Also
write the name of the figure which is formed by joining the points P, Q, R and S
respectively.

Y
Q

P

R
X´ X

S

Y´

3. The points M(-2,3), N(-2,-3),O(4,3) and P are the vertices of a square MNOP. Plot
these points on the graph . Also find the co-ordinates of the point P and area of
the square MNOP.

4. Plot the points P(-4,6) and Q(4,-2) . Join the points P and Q and find the co-
ordinates of the mid-point of PQ.

5. Write the coordinates of the corner of the given flag in the adjoining graph.

Y
A

X´ CB X

ED

Y´

Answers

Show to your teacher. Unit:2 Co-ordinate Geometry
Show to your teacher. Exercise: 2.1

60 Coordinate Exercise: 2.2

UNIT Objectives:

3 MENSURATION
Estimated periods 10

At the end of this unit student will be able to

● find the perimeter of plane figures.
● find the perimeter of squares and rectangles using formula.
● find the area of squares and rectangles using formula.
● gain idea of finding area of some more special figures.
● gain the idea of volume and find the volume of simple solids.

5 cm

Teaching Materials:

Square papers, stat paper, attractive diagrams (simple bars, multiple bars,

subdivided bars).

Activities:

It is better to

● involve the students to measure the perimeter of objects within class room
and find their perimeter.

● discuss about formula to find perimeter of squares and rectangles.
● discuss about the formula to find area of rectangles and squares.
● discuss about the formula to find the area of triangles and the special plane

figures.

Perimeter, Area and Volume

Introduction:

The branch of mathematics that deals with measurement of lengths, areas and volume
of plane and solid figures is called mensuration.

The figures which can be drawn on a plane such as on the paper of your notebook,
on the blackboard etc. are called plane figures whose surface is a plane.
The measure of the length of the boundary of a plane figure is called perimeter of
the figure.
Measure of the space bounded by the figure is called the area.

3.1 Perimeter of a plane figures

Perimeter of triangle : ABC be a triangle then the side BC=a,
AC=b and AB=c

\ Perimeter = sum of all side = a+b+c

Note: - Perimeter is also length.

- Units of all the sides must be same.

Perimeter of irregular plane figure with straight sides
= sum of the lengths of sides

\Perimeter of the given figure (p)

= AB + BC + CD + DE + EF + FA

Perimeter of equilateral triangle
ABC be an equilateral triangle with AB = BC = AC = a(say).
then
Perimeter (p) =AB+BC+AC

= a + a + a = 3a

Perimeter of a rectangle :
We know opposite sides of a rectangle are equal.
Length = AB = DC = l
Breadth = BC = AD = b

\Perimeter (p) = AB + BC + CD + DA

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

\Perimeter of rectangle =2(l+b)

62 Mensuration

Note: Since a parallelogram has also opposite
sides equal.
Let AB ==noACtDDp==earap21 endicular,
But sides BC we do not
being
take as length and breadth.
Perimeter (p) = AB+BC+CD+DA
= 22a(a1a1+1+a+22aa+22)a1 + a2
=
=

Perimeter of a square:
We know that all the sides of a square are equal.
ABCD be a square where AB = BC = CD = DA = l(say).
then,
Perimeter (p) = AB+BC+CD+DA

=l+l+l+l
= 4l = 4(length of a side)

Note: A rhombus has also all the sides equal. Then perimeter of rhombus = 4(length of a sides)

Circumference of a circle:

A circle is a closed figure bounded by a regular curved line diameter
D
where each point of whose boundary is at a constant distance

from a fixed point. The fixed point is called centre and the O
r
fixed distance is called radius. A
radius
In figure, O is the centre, OA is the radius and the boundary

of perimeter of the circle. And CD is the diameter. Diameter C

is 2 times of radius,

Circumference of circle = 2pr or pd.

[For practical purpose p = 22 or 3.14]
7

Example 1: Calculate the perimeter of the plane figure given alongside.
Solution:
Here in the figure, AB = BC = 3cm ; CD = 4cm, DE = 2cm

\EF = AB + CD = 3cm + 4cm = 7cm

AF = BC + DE = 3cm + 2cm = 5cm

\Perimeter of the figure (p) = AB + BC + CD + DE + EF + FA

= 3cm + 3cm + 4cm + 2cm +
7cm + 5cm

= 24cm

Prime Mathematics Book - 7 63

Example 2: Find the perimeter of the DABC given in the figure.
Solution:
In the given triangle ABC,
AB = 5cm, BC = 6cm, CA = 3cm

\Perimeter (p) = AB + BC + CA

= 5cm+6cm+3cm = 14cm

\ The perimeter of the triangle ABC is 14cm.

Example 3: Find the length of side of an equilateral triangle having perimeter 18cm.
Solution:
Perimeter of the equilateral triangle (p)=18cm
If side is a, then

3a = 18cm

a = 18 cm = 6cm \The side of the triangle is 6cm.
3

Example 4: Calculate the perimeter of the rectangle of length 5cm and breadth 4cm.
Solution:
Here,
Length of rectangle (l)=5cm
Breadth of the rectangle (b)=4cm
Now, we have
Perimeter of rectangle (p) = 2(l + b)

= 2(5cm + 4cm)
= 2 × 9cm = 18cm

\ The perimeter of the rectangle is 18cm.

Example 5: Perimeter of a rectangle is 36cm. If its breadth is 8cm, find its length.
Solution.
For a rectangle, perimeter (p)= 36cm

breadth (b) = 8cm
length (l) = ?

We have, for a rectangle
p = 2(l+b)

or, 36cm = 2(l+8cm)

or, 36 cm = l+8cm
2

or, 18cm = l+8cm
or, l = 18cm-8cm

\ l = 10cm

64 Mensuration

Example 6: Calculate the perimeter of the square of side 7 cm.
Solution:
length of side (l)=7cm
we have,
Perimeter of square = 4l = 4 × 7cm = 28cm

\ The Perimeter of the square is 28cm.

Example 7: Find the length of wire needed to fence a rectangular garden of length 32m
and breadth 24m, five fold.

Solution:
Perimeter of the garden (p) = 2(l+b)

= 2(32m+24m)
= 2 × 56m = 112m

Now length of wire for fencing the garden five fold

= 5p
= 5 × 112m = 560m

\ The required length of the wire is 560m.

Example 8: Find the circumference (Perimeter) of the circle given below.
Solution:
Here,
Radius of the given circle (r) = 28cm
Now, we have, O 28cm A
circumference of circle = 2pr

= 2. 22 .28 4
7

= 44 × 4
= 176 cm.
\ The required circumference of circle is 176 cm.

Example 9: Find the cost of fencing 4 times around a rectangular garden of length 20m
and breadth 12m at the rate of Rs 95 per metre.

Solution: Here,
length of rectangular garden (l) = 20m
breath of rectangular garden (b) = 12m
rate of fencing = Rs 65 per metre
Now,
perimeter of rectangular garden (P) = 2(l + b)

= 2(20 + 12)
= 2× 32
= 64m

Prime Mathematics Book - 7 65

Total length of fencing when the garden is fencing 4 times around it = 4p
= 4 × 64m
= 256 m

And total cost of fencing 4 times around the rectangular garden = Rs. 256 × 95
= Rs 24,320

Exercise 3.1 c)
1. Find the perimeter of the following figures.
a) b)

d) e) f)

3 cm
g) h)

2. Find the perimeter of the triangles with given sides.
a) 3.5cm, 4.5cm and 5.5cm b) 2.7cm, 4.3cm and 5cm

3. Find the perimeter of the square of side

a) 4.2cm b)6.5cm c)3.2cm d)8cm

4. Find the perimeter of the rectangle, the length and breadth of which are.

a) 12cm, 8cm b) 5cm, 4.8cm c) 6.2cm, 3.5cm d) 7cm,6cm

66 Mensuration

5. Find the circumference (Perimeter) of the circle following radius and diameters.
a) radius = 3.5 b) radius = 7cm
c) diameter = 14cm d) diameter = 21cm

6. a) Perimeter of a rectangle is 20cm. If its length is 6cm, find its breadth.
b) If the perimeter of a square is 26cm, find the length of its side.
c) Length of a rectangle is twice its breadth and its perimeter is 54cm. Find its length.
d) Find the length of the wire required to fence a rectangular garden 24m long
and 18m broad, making five complete rounds.
e) The length and breadth of rectangular ground 20m and 18m respectively. A
man moves around the ground 3 times early in the morning. Find the distance
covered by him.

7. a. The circumference (perimeter) of circular pond is 176cm. Find its radius.
b. The circumference (Perimeter) of a circular pond is 132cm. Find its diameter.

c. Find the circumference of circle of radius 7cm. Also find the length of wire
required to fence aound it two times.

8. a. The length of rectangle exceeds its breadth by 3cm and it’s perimeter is 30cm.
Find the length and breadth of rectangle.

b. The breadth of rectangle is 2cm. less then its length and it’ perimeter is 24cm.
Find the length and breadth of rectangle.

9. a. If breadth of rectangle is one-third of its length and perimeter is 32cm. Find
length and breadth of rectangle.

b. If breadth of rectangle is two-third of its length and perimeter is 40cm, find
length and breadth of rectangle.

c. The perimeter of square is equal to the perimeter of rectangle of length 16cm
and 4cm. Find the length of the side of square.

d. Find the cost of fencing around a square of length 30m at the rate of Rs. 15 per m.
e. Find the cost of fencing two times around a rectangular garden of length 25m.

and breadth 15m. at the rate of Rs. 75 per m.

3.2 Area of plane figures:

Area of Rectangle.

Consider a rectangle APQR and divide into unit squares of side 1 cm each, the area
which is supposed to be 1 cm2 or 1 square cm.
Now count the unit square along length(l) and breadth (b)

\ No of unit square = 8 × 4 = 32
\ Area of rectangle APQR = 32 cm2

Area of rectangle. = l × b
In case of square, length (l) = breadth (b)

\ Area of square = l × b

=l×l
= l2

Prime Mathematics Book - 7 67

Area of Triangle.

Perform the following activity

Take a paper in triangle shape

Let PQR is the triangle where QR = base (b)
Fold along QS perpendicular to QR.
Let PS = height (h)
Fold along XY bringing P to S.

Thus PO = OS = h
2

Cut and adjust DPOX to DQMX and DPOY to DRNY,
(DS being congruent).
Area of DPQR = Area of rectangle MQRN.
= (QR) × (OS)

=b× h
2

\ Area of D = 1 × b × h
2

Also, observe this. ABC be a triangle with base,

BC = b and altitude AD = h.

From A draw XY//BC and complete a rectangle XYCB.

We get.

Area of DABD = 1 area of rectangle XBDA. [\ diagonal of rectangle bisects it]
2
1
Area of DADC = 2 area of rectangle ADCY.

Area of DABD + Area of DADC = 1 (Area of rectangle XBDA + Area of rectangle ADCY).
2

\ Area of DABC = 1 Area of rectangle XBCY.
2

\ Area of DABC = 1 (BC × AD) = 1 ×b×h B
\ Area of DABC 2 2

= 1 bh
2

Remember:

- AD is altitude over base BC
BE is altitude over base AC
CF is altitude over base AB.

- Altitudes of a triangle are
concurrent.

68 Mensuration

Area of Right Angled Triangle.

In a right angled triangle, the perpendicular side can be

considered as base and altitude. 1
2
Thus, area of right angled triangle = × BC × AB

= 1 ×b×p
2

Area of parallelogram = 1 pb.
2

ABCD be a parallelogram.
Join A and C and draw AE perpendicular to BC
We get

Area of ABCD = 2 Area of DABC.

=2× 1 × BC × AE.
2

\ Area of ABCD = BC × AE.
\ Area of parallelogram = base × height

Remember:

DX is height of parallelogram over base AB.
DY is the height of parallelogram over base BC.

Area of Quadrilateral A

ABCD is a quadrilateral. BD is its diagonal. P1

AF and CF are drwn-perperidiculars on BD. F D
1 E
Now, Area of quadrilateral ABCD = 2 BD (AE + CF) Q
or P2 P2 R
1
2 d (P1 + P2) C

Area of Trapezium P P1
P2
PQRS is a trapezium, where PQ || SR, and PT ^ SR then
1 h
Area of trapezium PQRS = 2 PT(PQ + SR) T
or
1
2 h(P1 + P2) S

Prime Mathematics Book - 7 69

Area of Kite A
d1
ABCD is a Kite digonal AC and BD are intersect at right O d2
angle at 0.
1 B C D
Area of Kite ABCD = 2
or AC x BD

1 (d1 x d2)
2

Area of circle

In the given figure, 0 is the centre of circle. Divide the given
circle into small pieces of triangles as shown in the given
figure. Cut each part and re-arrange them rectangle shape.
Where length of this rectangle is equal to half of circumfer-
ence and breadth is same as radius of circle.

Area of circle = Area of rectangle
=LxB

= pr x r r
= pr2

\ Area of circlee (A) = pr2

pr

Example 1. Calculate the area of a rectangle the length and breadth of which are 9cm

Solution: and 7cm respectively.
Here,

Length of rectangle (l) = 9cm.
Breadth of the rectangle (b) = 7cm.
We know, area of rectangle (A) =l×b
= 9cm × 7cm = 63cm2
\ The area of rectangle (A) is 63cm2

Example 2. Calculate the area of square one of whole side is 12 cm.
Solution:
Here,

length of side of square (l) = 12cm
we have, Area of square (A) = l2

= (12 cm)2 = 144 cm 2

\ The area of the square is 144cm2

70 Mensuration

Example 3. Find the area of the triangle given alongside.

Solution
Here, in the figure
Length of base of the triangle (b) = 15cm.
Altitude (height) of the triangle (h) = 8cm.
We have,
Area of triangle = 1 ×b ×h
2

= 1 × 15cm × 8cm
2
= 15cm × 4 cm = 60 cm2
\ The area of the triangle ABC is 60 cm2

Example 4. Calculate the area of DABC in the figure given along side.
Solution:
Here, in the figure, Length of base of triangle
(b) = BC = 10cm.
Altitude of the triangle (h) = AD = 6cm.
We have,
Area of a triangle = 1 ×b×h
2

= 1 × 10cm × 6 cm
2
= 5cm × 6cm = 30 cm2

\ The area of the given triangle is 30cm2.

Note: Altitude of the triangle should not necessarily fall on the base.

Example 5. Calculate the area of the triangle given in the figure.
Solution:

Here, ABC is a right angled triangle, where
perpendicular sides are
AB = p = 6cm
AC = p = 8cm
We know,
Area of right angled triangle = 1 pb
2
1
= 2 × 6cm × 8cm

= 3cm × 8cm
= 24cm2

\ The area of the given right angled triangle is 24cm2

Prime Mathematics Book - 7 71

Example 6. Calculate the area of the parallelogram ABCD in the figure given along side.
Solution:
Here, ABCD is a parallelogram where length of
base (BC) = b = 10cm
Altitude of the parallelogram over base BC is
h = 8cm. 8 cm

Now, we have
Area of parallelogram = base × height
= 10cm × 8cm
= 80 cm2
\ The area of the parallelogram is 80 sq.cm.

Example 7. In the given figure, ABCD is a parallelogram in which BC = 8cm
and CD = 12cm. If AM^BC, AN^CD and AM=6cm, find the length of AN.

Solution:
Here, in the figure,
BC = 8cm and CD = 12cm.
length of altitude over base BC i.e. AM = 6cm.
length of altitude over base CD i.e. AN = ?

Since, area of parallelogram = b × h.

\ Area of parallelogram ABCD = BC × AM

Also area of parallelogram ABCD = CD × AN.

\ CD × AN = BC × AM

or, CD × 12cm = 8cm × 6cm

or, AN = 8cm × 6cm
12cm

or, AN = 48cm2
12cm

\ AN = 4cm

Example 8. From the given figures. Find the area of shaded region.

A D 10cm D
P S AE C

Q 10cm 8cm 8cm
B B
R
C

(i) Solution :
Now, Area of rectangle
ABCD (A1) = Length x breadth
= 10cm x 8cm = 80cm2

72 Mensuration

Again, Area of rectangle PQRS (A2) = Length x breadth
= 5cm x 4cm
= 20cm2
at last
Area of shaded region (A) = A801 c-mA22 - 20cm2
=
= 60cm2
\ Area of shaded region is 60cm2
(ii) Solution :
Now, Area of rectangle ABCD = length x breadth
(A1) = 10cm x 8cm
= 80cm2
again,
Area of triangle BCE (A2) = 1 base x height
= 12 x 10cm x 8cm
2

at last = 40cm2

Area of shaded region (A) = 8A01 c-mA22 - 40cm2
=
= 40cm2
\ Area of shaded regioin is 40cm2

Example - 9 Find the area of circle with radius 7cm (p = 272)

Solution :
Radius of circle (r) = 7cm.
Area of circle (A) = ?

We know, Area of circle (A) = pr2
22
= 7 x 7 x 7 = 154cm2

\ Area of circle is 154 cm2

Example 10. Find the cost of carpeting a rectangular room 8cm by 6m at the rate of

Rs 480 per square meter.
Solution:
Here, length of room (l) = 8m First find area of room,
Breadth of room (b) = 6m then
Cost of carpet per m2 i.e. rate = Rs 480/m2 cost = area × rate.

\ Area of room = l × b

= 8m × 6m = 48 m2
As area of carpet = area of room.

\ Area of carpet = 48m2

Prime Mathematics Book - 7 73

Now, = Area × rate = 48m2 × Rs 480 = Rs 23040
Cost of carpet m2

\ The cost of carpeting the room is Rs 23040

Exercise 3.2
1. Calculate the area of given figure under the following condition.
a) b)

c) P d) A B

4cm.

Q R D 8cm. C
e) D C
f) P 5cm. Q

4cm. R
S T 8cm.
A B
g) P 8cm. h) A B
10cm. Q 6cm.
6cm.
T 10cm.4cm.

S 4cm. R D C
P
P

i) j) R 16cm. Q
S
3.2cm. 8cm. 4.8cm. Q 20cm.

S

R
74 Mensuration

k) l)
7cm. 28cm.

2. Find the area of the triangle in which
(a) base = 8cm and height = 6.4cm
(b) length of base = 16cm and altitude over the base = 8cm.

3. Find the area of the following.
(a) A rectangle with length = 3cm and breadth = 6.2 cm
(b) A rectangle in which length = 18.5cm and breadth 7.2 cm
(c) Square with length of side 4.25 cm.
(d) A square in which length of its side is 12.5cm.

4. Find the area of parallelogram in which
(a) base = 14cm and height over the base = 7.2 cm
(b) base = 40 cm and height over the base = 15.6cm.

5. Find the missing part of the following triangle.
(a) base = 8cm, area = 48cm2, height = ?
(b) height = 10cm, base = 18cm, area = ?
(c) area = 9cm2, height = 4.5cm, base = ?
(d) area = 7cm2, base = 3.5cm, height = ?

6. Find the missing part of the following parallelograms.
(a) base = 14cm, area = 56cm, height = ?
(b) height = 5cm, area = 12.5cm2, base = ?
(c) height = 7.5cm, base = 4.5 cm, area = ?

7. Find the area of the shaded region in the following figures.

a) b) c)

d) e) 18cm.

f)

Prime Mathematics Book - 7 75

g) h)

14cm

O 7cm O 7cm

22cm

8. Find the value of x in each of the following figures.

a) b)

x

6 cm

9. (a) A rectangular garden is twice as long as its breadth. If its perimeter
is 90m, find the area of the garden.

(b) The perimeter of a square ground is 124m, find its area.

(c) A rectangular room is 14m long and 12m broad. Find the area of carpet required
to cover its floor at Rs 120 per square meter.

(d) A rectangular park is 26m long and 25m broad. Find its area and cost of paving
it with blocks costing Rs 38 per m2.

10. a) The area of square is 64cm2. Find its perimeter.

b) Find the perimeter of a rectangle, if its area and breadth are 120cm2 and 8cm.
respectively.

11. a) Area of rectangle is 12cm2 and breadth is one third of the length find length
b) and breadth of rectangle.

Area of rectangle is 48cm2 and breadth is three fourth of the length. Find the
perimeter of rectangle.

76 Mensuration

3.3 Surface area of cuboids and cubes

Observe the following objects.

A brick A match box A dice An ice cube

You have learnt that the solids shown above have six faces (rectangular or square), 12
edges and 4 vertices. The first two solids have different length, breadth and height.
So, they are cuboids. The next two have equal length, breadth and height and each
face is a square, So these are cubes.

Surface area of a cuboid.

Consider a cuboid with length (l), breadth (b) and height (h)
Total surface area of the solid = Area of top with
buttom faces + Area of front and back faces + area of
side faces.

= (Area of rectangle ABCD + Area of rectangle
A´B´C´D´)+(Area of rectangle CDD´C´
+ Area of rectangle ABB´A´)+(Area of
rectangle ADD´A´+Area of rectangle BCC’B’)

= (l × b + l × b) + (l × h + l × h)+(b × h + b × h)
= 2lb + 2lh + 2bh = 2 (lb+lh+bh)

\ Total surface area of a Cuboid = 2(lb + bh + hl)
Note: Surface is the part of solid which is in contact with air.

Surface area of a cube.

You know that a cube consists six square faces. If the length of
side the of cube is l, then area of a face = l × l = l2

\ Total surface area = 6l2

Example 1. Find the total surface area of a cuboid 8cm long, 6cm
broad and 4cm high.

Solution:
In the cuboid, length (l) = 8cm, breadth (b) = 6cm and height (h) = 4cm
We have,
Total surface area (T.S.A) = 2 (lb + bh + hl)

Prime Mathematics Book - 7 77

\ T.S.A of the cuboid = 2(8cm × 6cm + 6cm × 4cm + 4cm × 8cm)
= 2 (48cm2 + 24cm2 + 32cm2)
= 2 × 104cm2 = 208 cm2
= 208 cm2.

Example 2. Total surface area of a cuboid is 688 cm2. If its breadth and height are 9cm
and 8cm respectively, find its length.

Solution:
Here, in the cuboid, Total Surface Area = 688 cm2, breadth (b) = 9cm, height (h) = 8cm,
length (l) = ?

We have,
T.S.A. of cuboid = 2(lb + bh + hl)

or 688 cm2 = 2(l × 9cm + 9cm × 8cm + 8cm × l)

or 688 cm2 = 9lcm + 72cm + 8lcm
2

or 344 cm2 - 72cm2 = 17l cm

or 272 cm2 =l
17 cm2

\ l = 16 cm

\ The length of the cuboid is 16 cm.

Example 3. Find the total surface area of the cube of side 11cm.

Solution:

Here, length of side of cube (l) = 11 cm.

We have, = 6l2

T.S.A. of cube = 6 × (11cm)2

= 6 × 121 cm2
= 726 cm2
\ The total surface area of the cube is 726 cm2

Example 4. The total surface area of a cube is 864cm2. Find the length of its side.

Solution: 2 144 \ 144 = 22 × 22 × 32
Here, T.S.A of a cube = 864 cm2. 2 72 =2×2×3
length of side of cube(l) = ? 2 36
We have, 2 18
T.S.A of cube = 6l2 39

or, 864 cm2 = 6l2 3

78 Mensuration

or, l2 = 864 cm2
6

or, l2 = 144 cm2

\ l = 144 cm2

or l = 12 cm

\ The length of side of the cube is 12 cm.

Exercise 3.3
1. Find the total surface area of the following cuboids or cubes
a. b.

c. d.

2. Calculate the total surface area of the cuboid having.
(a) l = 9.6 cm, b= 6.4cm and h=4.8cm. (b) l = 8cm, b= 7cm and h=6cm.
(c) l = 20cm, b= 15cm and h = 8cm.

3. Find the total surface area of the cube of side (d) 8m
(a) 32cm (b) 1m 40cm (c) 19.2cm

4 (a) Length and breadth of a cuboid are 18cm and 12cm respectively. If the total
surface area of the cuboid is 1272 cm2, find its height.

(b)The total surface area of a cuboid is 1710m2. If it is 15m broad and 12m high,
what is its length ?

(c) The total surface area of a cube is 1176cm2. Find its length of a side.
(d) Find the length of a cuboid room if its total surface area is 384m2.

5. a) The total surface area of rectangular is 320cm2. It has square base of area 64cm2,
find its height.

b) The length and perimeter of a room are 10m and 36m respectively. If its height is
4m. Find the total surface area of the room.

Prime Mathematics Book - 7 79

6. a) The length, breadth and height of rectangular box are 10cm and 8cm and 4cm
respeceively. Find the total cost of painting, its total outer surface at the rate of Rs.
7 per cm2.

b) The length, breadth and height of rectangular room are 15m, 13m and 10m
respectively, find the total cost of painting its total faces escpt its floor at
rate 12per m2.

3.4 Volume of Cuboids and Cubes.

The total space occupied by an object is called volume. Volume is measured in cubic
units or (unit)3 e.g. cm3, m3 e.t.c. A cube of length 1 cm is supposed to have volume
1cm3 i.e. 1cm × 1cm × 1 cm = 1cm3.
You have already learnt that
Volume of a cuboid = length × breadth × height
or volume of a cuboid (v) = l × b × h.
Area of base (A) = l × b so, volume of cuboid = A× h.
Since a cube has all the sides equal say ‘l’.

\volume of cube (v) = l × l × l = l3.

Note: Space occupied by the material of a solid is called its volume.

Volume of liquid that a vessel can hold is called the capacity of the vessel. The
capacity of vessels are measured in cm3, m3, litre, millilitre (ml) e.t.c.

1 litre = 1000 cm3 = 1 m3. i.e. 1000 litre = 1m3.
1000

Example 1. Find the volume of cuboid when the length, breadth and height of which
are 6cm, 4.5cm and 4cm respectively.

Solution:
Here, length of cuboid (l) = 6cm, Breadth of cuboid (b) = 4.5cm and
Height of cuboid (h) = 4cm
We have,

Volume of cuboid (v) = l × b × h = 6cm × 4.5cm × 4cm = 108 cm3.

\ The volume of the cuboid is 108cm3.

Example 2. The area of the base of a rectangular tank is 120m2 and its capacity is
840m3. Find its height.
Solution:
Here, base area (A) = 120m2 , Volume of the tank (V) = 840m3
Height of the tank (h) = ?
We have,
Volume of a rectangular tank (v) = A × h
or 840m3 = 120m2 × h

or h = 840m3 \ h = 7m
120m2
\ The height of the tank is 7m.

80 Mensuration

Example 3. Find the volume of a cube where the length of a side is 18cm.
Solution:
Here, length of side of cube (l) = 18cm.
Volume of the cube (v) = ?
We have,

Volume of cube (v) = l3 = (18cm)3 = 5832 cm3

\ Volume of the cube is 5832 cu.cm.

Example 4. If the volume of a cube is 1728cm3, calculate the length of its side.
Solution,
Here, Volume of the cube (V) = 1728 cm3 2 1728
length of the side of cube (l) = ? 2 864 \ 1728 = 23 × 23 × 33
We have, 2 432
2 216 = 123
Volume of cube (v) = l3 2 108
1728cm3 = l3 2 54

or, (12cm)3 = l3 \ l = 12 cm 3 27

\ The length of a side of the cube is 12 cm. 39
3

Example 5. Find the capacity of a cuboidal tank of length 8m, breadth 7m and
height 6m in litres.

Solution:

Here, length of the cuboid tank (l) = 8m, Breadth (b) = 7m and Height (h) = 6m .

We have,

Volume of cuboid (V) = l × b × h = 8m × 7m × 6m = 336m3
= 336 × 1000 l [1m3 = 1000 litres] = 336000 l

\ The capacity of the tank is 336000 litres.

Example 6. The surface area of a cubical block is 96cm2.Find its volume.
Solution.
Here, T.S.A of cube = 96cm2
Volume of cube = ? - Using T.S.A at first to
We have find side (l)

T.S.A of cube = 6l2 - Then find its volume
96cm2 = 6l2

or 96cm2 = l2 or 16cm2 = l2 or l = 4cm.
6

Now,
Volume of the cube (V) = l3 = (4cm)3 = 64 cm3

\ The volume of the cubical block is 64cm3

Prime Mathematics Book - 7 81

Example 7. A cuboid is thrice as long as its breadth. If its height and volume are 8cm
and 864cm3 respectively, find its total surface area.
Solution.
Here, volume of cuboid = 864cm3, l = 3b
Length = thrice the breadth i.e. l = 3b - Using l × b × h, find b
Height (h) = 8cm.
We have. and l
Volume of cuboid = l × b × h - Then find T.S.A
or, 864 cm3 = 3b × b × 8 cm
or 864 = 24 b2

or b2 = 864 or b2 = 36 \ b = 6cm
24

\ l = 3b = 3 × 6 cm = 18cm

Now, Total surface area of cuboid
= 2(lb+bh+hl)
= 2(18cm × 6cm + 6cm × 8cm + 8cm × 18cm)
= 2 (108cm2 + 48cm2 + 144cm2)
= 2 × 300cm2
= 600 cm2

\ The total surface area of the cuboid is 600 cm2.

Exercise 3.4

1. Find the volume of the given solids. b)
a)

c) d)

82 Mensuration

2. Find the volume of the following cuboids, whose length, breadth and height are

(a) 10cm, 9cm, 8cm (b) 15cm, 12cm, 9.5cm.

(c) 8m20cm, 6m30cm, 4m50cm (d) 1m20cm, 80cm, 65cm.

3. Find the volume of the cubes having length of a side

(a) 10.5cm (b)4.5cm (c) 9.8m (d) 3m 60cm,

4. (a) The area of the base of a cuboid is 43.2cm2. If its height is 11cm, find its
volume.

(b) Volume of a cuboid is 4032cm3. If its height is 14cm, find the area of its base.
(c) Volume of a cube is 551.368 cm3. Find the length of its side.
(d) Volume of a cubical box is 512cm3. Find its total surface area.
(e) The volume of a cubical tank is 4913m3. Calculate its surface area.

5. (a) Find the capacity of a cuboidal box of measures 51m × 48m × 23m. Find its
capacity in litres.

(b) Length of a cuboidal box is twice as long as its broad. Its height is 20cm and
volume 1440 cm3. Find its length and breadth.

(c) Area of base of a rectangular tank is 900 cm2. Find the height of the water
level when it contains 36 litres water.

6. a) The volume of a cuboid is 512cm3. If the length and breath of cuboid are 16cm

and 8cm respectively. Find its height.
b) The volume of a cuboid is 240cm3. If the breath and height of cuoid are 6cm

and 5cm. Find its length.
c) If total surace of cube is 864cm2, find its volume.
d) Area of the base of a rectangular box and its volume are 24cm2 and 60cm3, find

its height.

7. a) The volume of a cube is equal to the volume of a cuboid of length 16cm,
breadth 12cm and height 9cm. Find the total surface area of the cube.

b) The rectangular tank of length, breadth and height are 12cm, 10cm and 6cm
respectively is full of water. If this water is poured into a smaller tank. How
many tanks of length 5cm, breadth 4cm and height 3cm are required ?

Prime Mathematics Book - 7 83

Unit Revision Test

1. Find the perimeter of the following figure

2. Find the perimeter of the rectangle, the length and breadth of which are 14cm
and 9.5cm respectively.

3. Perimeter of a rectangle is 28cm. If its length is 8cm, find the breadth of the
rectangle.

4. Calculate the area of the triangle given in the figure.

5. Find the area of a parallelogram in which base = 14cm and altitude over the
base = 8.4 cm.

6. In the given figure, ABCD is a parallelogram. Find the area of the shaded part.

7. A rectangular garden is thrice as long as its breath. If its perimeter is 120m, find
the area of the garden .
84 Mensuration

8. Find the total surface area of the cuboid having l = 9.6cm, b= 6.4 and
h = 4.8 cm

9. Length and breadth of a cuboid are 18cm and 12 cm respectively. If the total
surface area of the cuboid is 1272 cm2, find its height.

10. The volume of a cubical box is 512 cm3. Find its total surface area.

Answers

Unit:3 Mensuration

Exercise: 3.1

1. a) 19 cm b) 19 cm c) 28 cm d) 27 cm
e) 40 cm h) 36 cm
f) 38 cm g) 42 cm
2. a) 13.5 cm d) 32 cm
3. a) 16.8 cm b) 12 cm d) 26 cm
4. a) 40 cm d) 66 cmmm
5. a) 22 cm b) 26 cm c) 12.8 cm d) 420 cm
6. a) 4 cm
b) 19.6 cm c) 19.4 cm
e) 288 m
7) a) 28 cm b) 44 cm c) 44 cm
8) a) 9 cm, 6 cm
9. a) 12 cm, 4 cm b) 6.5 cm c) 18 cm

b) 42 cm c) 88 cm
b) 7 cm, 5 cm c) 10 cm d) Rs. 1800 e) Rs. 12000
b) 24 cm, 12 cm

Exercise: 3.2

1. a) 42 cm2 b) 10 cm2 c) 150 cm2 d) 32 cm2
e) 64cm2 h) 50cm2
i) 32cm2 f) 26cm2 g) 42cm2 l) 616cm2

2. a) 25.6 cm2 j) 160cm2 j) 154cm2 d) 156.25 cm2
3. a) 18.6 cm2
4. a) 100.8 cm2 b) 64 cm2 d) 4 cm
5. a) 12 cm
6. a) 4 cm b) 133.2 cm2 c) 18.06 cm2 d) 160 cm2
7. a) 135 cm2 h) 330cm2
b) 624 cm2
e) 423 cm2 d) 650 m2, Rs. 24700
8. a) 8 cm b) 90 cm2 c) 4 cm
9. a) 450 cm2
10. a) 32cm b) 2.5 cm c) 33.75 cm2
11. a) 6cm, 2cm
b) 100 cm2 c) 240 cm2

f) 165.6 cm2 g) 462cm2

b) 6.25 cm

b) 961 cm c) Rs. 20160

b) 46cm

b) 28cm

Prime Mathematics Book - 7 85

1. a) 124.8 cm2 b) 360 cm2 Exercise: 3.3 d) 34.56 cm2
2. a) 276.48 cm2 b) 292 cm2 c) 365.04 cm2 d) 384 m2
3. a) 6144 cm2 b) 11.76 m2 c) 1160 cm2 d) 8 m
4. a) 14 cm b) 25 cm c) 2211.84 cm2
5. a) 6cm b) 304 m2 c) 14 cm d) 166.375 cm3
6. a) Rs. 2128 b) Rs. 9060 d) 0.624 m3
d) 46.656 m3
Exercise: 3.4 d) 384 cm2

1. a) 40 cm3 b) 432 cm3 c) 343 cm3 d) 2.5 cm
2. a) 720 cm3
3. a) 1157.625 cm2 b) 1710 cm3 c) 232.47 m3
4. a) 475.2 cm3
b) 91.125 cm2 c) 941.192 m3
e) 1734 cm2
5. a) 56304000 l b) 288 cm2 c) 8.2 cm

6) a) 4 cm b) 12 cm, 6cm c) 40 cm
b) 8cm c) 1728 cm3
7) a) 864 cm2 b) 12

86 Mensuration

4UNIT TRANSFORMATION
Estimated periods 16 Objectives:

At the end of this unit, students will be able to
● give introduction of reflection, rotation and translation.
● draw the image of given figure after reflection on the given

line in simple plane.

● draw the image of given figure after rotation about given
point through given angle in simple plane.

● draw the image of given figure after translation by given
vector in simple plane.

● get the idea of use of formula to find the coordinates of images of points
under reflection on (x-axis, y-axis) and rotation of given point about origin

through ±90oand ±180o.

● get the introduction of line symmetry and point symmetry.
● calculate the order of rotational symmetry.
● design by tessellating polygons.
● use the scale in drawing and find the bearing in different positions.

Teaching Materials:
Models of different polygons, instrument box, graphs,
different symmetrical figures, maps.

Activities:

It is better to
● demonstrate the reflection, rotation and translation in simple plane.
● explain the formula to find the coordinates of the image of a point in

the coordinate plane (reflection on x-axis and y-axis; rotation about ‘O’ through on

(x-axis, y-axis) and rotation of given point about origin through ±90o, ±180o.

● involve the student to draw the axes of symmetry and to find the order of
rotational symmetry

● involve the students to make tessellated designs.
● involve the students to measure and find distances and draw bearing in maps.

4.1 Transformation

Transformation simply means change in geometrical figures. Change may be in shape,
size and position. You are familiar with the following activities.
(i) (ii)

(iii) (iv)

These are some examples of transformation. In transformation, the initial point or
figure is called object and the corresponding point or the figure formed latter is called
image.

If A´ is the image of A, we write A→A´ (read

as A maps to A´)
Similarly, A´B´ is the image of AB and we write

AB→A´B´
DA´B´C is the image of DABC.

i.e. DABC →DA´B´C´

If the figure and their images are congruent, the transformation is called
isometric transformation. If the size of the object and image are not same (reduced or
enlarged) the transformation is called non-isometric.
During the transformation some points remain unchanged (object and image
points coincide) Such points are called invariant points. If all the points remain
unchanged (Figure and its image coincide), the transformation is an identify
transformation.

88 Transformation

There are different types of transformation. But there are four fundamental
transformations.
(i) Translation (ii) Reflection (iii) Rotation (iv) Dilation

Translation:

Translation is a transformation in which each point of a figure displaced by certain
distance and direction is called translation vector.

Characters of translation.
1. Translation is an isometric transformation.

2. Each and every point get displaced, so there exists no invariant point.

Translation in simple plane: P

Given a translation vector PQ and DABC Q

Draw lines parallel to vector PQ from A, B and C
using set squares.

Take AA´ = BB´ = CC´ = PQ
Join A´and B´; B´and C´; A´and C´.

Thus DABC →DA´B´C´ under translation by PQ.

Reflection: Reflection of a point P about a line MN is the 89
transformation which maps P to P´ such that MN is perpendicular
bisector of PP´.

P´ is the image of P. MN is called line of reflection, or axis of
reflection or mirror.

Character of Reflection:

1. It is an isometric transformation.
2. The point which lies on the mirror

remain invariant
3. Figure and its image form a symmetri-

cal pattern about the axis.
4. The position of image is reversed (op-

posite i.e. right to left and left to
right)

Reflection on Simple Plane:
Given figure is DABC and axis of reflection is MN.

Draw AP, BQ, CR Perpendicular to MN.

Prime Mathematics Book - 7

Produce AP, BQ, CR to A´, B´ and C´
respectively such that AP = PA´,
BQ = QB´ and CR = RC´

Join A´and B´; B´ and C´’; A´ and C´

Thus D ABC→DA´B´C´ under reflection on MN.

Rotation

The pole star lies just above north pole of the
earth. So it seems fixed on the northern sky.
The pole star can be located by the help of the
constellation called Ursa major
(Saptarishi) or plough which is a
group of 7 stars forming nearly like a
plough as shown. In a straight line to
the stars devoted by 2 and 1,
we get a star which is the pole star.
Movement of stars about the pole
star is an example of rotation.

Rotation of a point P about a fixed

point O through an angle q is the

transformation which map P to P´

such that OP = OP´ and POP´ = q.

Under rotation about O through an

angle q is P→P´.

The fixed point O is called centre

of rotation and POP´ = q is called

angle of rotation. POP´ measured
in anti-clockwise direction is taken as positive and clockwise direction is taken as
negative.

Character of Rotation. q

1. Rotation is an isometric transformation.
2. If the point or figure lies at the centre it is invariant.
3. The figure and its image form a radially symmetrical pattern.

90 Transformation

Rotation in Simple Plane
(i) To rotate DABC about O through + 90o

Join A,O ; B,O; C ,O
Draw AOA´ = 90o in +ve direction and
take OA = OA´
Similarly BOB´ = COC” = 90o and
OB = OB´ and OC = OC´

Join A´and B´; B´and C´; C´ and A´

Then DABC →DA´B´C´ under rotation about O thorough +90o.

(ii) To rotate figure ABCD about O thorough -180o A’ D
C’ CB
Join A and O; B and O; C and O; D and O.
D’ OA
Draw AOA´ = -180o and take AO = OA´ B’

similarly BOB´ = COC´ = DOB´ = -180o and take
BO = OB´, CO = OC´ and DO = OD´

Join A´and B´; B´and C´ C´ and D´; A´ and D´.

Thus Figure ABCD→Figure A´B´C´D´ under rotation through -180o about O.

Note: Because of its complexity, we
will learn about dilation in latter
classes.

Transformation in Coordinate Plane

Transformation in co-ordinate plane (graph) is easier than transformation in
simple plane with devised construction. In this case image of the points are directly
found with simple formulae and plotted in the graph.

Reflection on co-ordinate Plane

Reflection on X-axis:

Let P(x, y) be the point. The line of reflection is
OX. So, draw PM^OX and produce to p´ such that
PM = MP´. Here MP´ = -PM = -y and OM = x

\Co-ordinates of p´ is (x, -y)
\Under reflection on x-axis p(x, y)→p´(x, -y)

Prime Mathematics Book - 7 91

Example 1. Find the image of the point A(4,-3) under reflection on x-axis.
Solution :
Given point is A(4,-3)
We have,

under reflection on x-axis p(x,y)→p´(x,-y)

\A(4,-3)→ A´(4,-(-3)) A B
C’ B’
= A´(4,3)
C
Example 2. Find the co-ordinates of the images A’

of the vertices of DABC with vertices

A(-3,3), B(3,4) and C(-2,-2) under

reflection on x-axis and plot the DABC
and image DA´B´C´ on graph.

Solution:
Given vertices A(-3,3), B(3,4) and C(-2,-2)
we have,

under reflection on x-axis, p(x,y)→p´(x,-y)
A(-3,3)→A´(-3,-3)
B(3,4)→B´(3,-4)
C(-2,-2)→C´(-2,2)

Reflection on Y-axis: M

Consider a point P(x,y), preferably on 1st
quadrant OY being line of reflection, draw

PM ^OY and produce to p´ such that PM = MP´.

Here OM = y
and MP´ = -MP = -x

\Coordinate of p´ is (-x,y)
\ Under reflection on Y-axis p(x,y)→p´(-x,y)

Example 3. Find the co-ordinates of the point A(-3,4)
Under Reflection on Y-axis.
Solution.
Given point A(-3,4)
Line of reflection is y-axis.
We have,

under reflction on y-axis P(x,y)→P´(-x,y)
\ A(-3.4)→A´(-(-3),4) = A´(3,4)

92 Transformation

Example 4. Vertices of DABC are A(2,4), B(4,2) and C(3,-2). Find the coordinates of
the image DA´B´C´ Under Reflection on y-axis.

Solution.
Given vertices of DABC are A(2,4), B(4,2) and C(3,-2).
Line of reflection is Y-axis (YY´)
We have,
Under reflection on y-axis.

P(x,y)→P´(-x,y)
\ A(2,4)→A´(-2,4),

B(4,2)→B´(-4,2).
C(3,-2)→C´ (-3,-2)

Note:
Points or figures can be reflected on any general line. As an introduction here we
show the reflection on X-axis and y-axis only.
Geometrical proof of the formulae are not important for the students. Teacher are
requested to demonstrate only.

Rotation in co-ordinate plane

Rotation about O through +90o (quarter turn about anti clockwise)
Let P(x,y) be a point on coordinate plane. Joining O and P. Let POX = q, rotate OP

about O through +90 o to OP´.
Here, POP´ = 90o
We get YOP´ = q and thus OP´M´ = q
and DOP´M´@DOPM.
qq
Where OM´ = -PM = -y
P´M´ = OM = x
\ Co-ordinates of P´ is (-y, x) q

\ Under rotation about origin through +90o (x,y)
P(x,y)→P´(-y,x) q

Rotation about origin through -90o quarter turn aout clockwise. qq
Consider a point P(x,y).
Joining O and P, rotate OP in clockwise direction
to OP´
such that

POP´ = -90o

Draw PM and P´M´^ OX. We get

POM = YOP´ = OP´M´.

and DOP´M´@DOPM.

Where OM´ = PM = y
P´M´ = -OM = -x.

\Co-ordinate of P´is (OM´, P´M´) = (y,-x)
\ Under rotation about origin through -90o
P(x,y)→P´(y,-x)

Prime Mathematics Book - 7 93