Math 8

Unit

Congruency, Similarity,

3 Circle, Solid and their Nets

3.1 Congruent Figures

Similar in shape, different in size Similar in shape, equal in size

∴ First pair are not congruent triangles and the second pair are congruent triangles.

Two geometric figures having the same shape and size are called congruent figures.

When two triangles have the same shape and size then they are called congruent
triangles. Symbol for congruency is ≅.

If two triangles are congruent it means that the sides and angles of the one are equal
to the corresponding sides and angles of the other and their areas are equal.

Conditions for Congruency of Triangles

I. Two sides and the included angle (S.A.S. axiom) A P
In ∆ABC and ∆PQR,

AB = PQ = 5cm (S) 5cm
5cm

∠B = ∠Q = 600 (A) B 600 CQ 600 R
BC = QR = 4 cm (S) 4cm
4cm

∴∆ABC ≅ ∆PQR (By S.A.S axiom)

Corresponding sides (Opposite to equal angles)
AC = PR (Opposite to ∠B and ∠Q)

Corresponding angles (Opposite to equal sides)
∠A = ∠ P (Opposite to BC and QR)
∠C = ∠R (Opposite to AB and PQ)

If any two sides of a triangle and angle between them are equal to any two sides and
angle between them of another triangle, then the two triangles are congruent.

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Note:
• Sides opposite to equal angles of congruent triangles are corresponding sides.
• Angles opposite to equal sides of congruent triangles are corresponding

angles.
• Corresponding sides and corresponding angles of congruent triangles are

equal.

II. Two angles and included side (A.S.A. axiom)

In ∆ABC and ∆PQR AP

∠B = ∠Q = 600 (A)

BC = QR = 5cm (S) B 600 500 C Q 600 500 R
∠C = ∠R = 500 (A) 5cm 5cm

∴∆ABC ≅ ∆PQR (By A.S.A. axiom)

Corresponding sides (Opposite to equal angles)

AB = P Q ( Opposite to ∠C and ∠R)

AC = PR (Opposite to ∠B and ∠Q)

Corresponding angles (Opposite to equal sides)

∠A = ∠P (opposite to BC and QR)

If two angles and their adjacent side of one triangle are respectively equal to
two angles and their adjacent side of another triangle, then the two triangles are
congruent.

III. Three sides (S.S.S. axiom) AP
In ∆ABC and ∆PQR,
AB = PQ = 5cm 5cm 6cm 5cm 6cm
BC = QR = 4cm
AC = PR = 6 cm B 4cm C Q 4cm R
∴∆ABC ≅ ∆ PQR (By S.S.S. axiom)

Corresponding angles (opposite to equal sides)
∠A = ∠P (Opposite to BC and QR)
∠B =∠Q (Opposite to AC and PR)
∠C = ∠R (Opposite to AB and PQ)
If three sides of a triangle are respectively equal to three sides of another triangle,

then the two triangles are congruent triangles.

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IV. Right angle, hypotenuse and any other side (R.H.S. axiom) A P
In ∆ABC and ∆PQR, BR Q

∠B = ∠Q = 90° (R)

AC = PR (H) C
AB = PQ (S)

∴∆ ABC ≅ ∆ PQR (By R.H.S. axiom)

Corresponding sides (Opposite to equal angles)
BC = Q R (Opposite to ∠A and ∠P)

Corresponding angles (Opposite to equal sides)
∠A = ∠ P (Opposite to BC and QR)
∠C = ∠ R (Opposite to AB and PQ)

If the hypotenuse and one of two remaining sides of a right angle triangle are
respectively equal to hypotenuse and one of remaining two sides, then two triangles
are congruent.

V. Two angles and a side (A.A.S. axiom) A P
In ∆ABC and ∆PQR,

∠C = ∠R (A)

∠B = ∠Q (A) B 4cm C Q R
AB = PQ (S)

∴∆ABC ≅ ∆ PQR (By A.A.S. axiom)

Corresponding sides (Opposite to equal angles)
AC = P R (Opposite to ∠B and ∠Q)
BC = Q R (Opposite to ∠A and ∠P)

Corresponding angles (Opposite to equal sides)

∠A = ∠ P (Opposite to BC and QR)

If two angles of a triangle and a side of a triangle are respectively equal to two an-
gles and a side of another triangle, then the two triangles are congruent.

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Worked Out Examples



Example: 1

From the given condition, prove that ∆ABC ≅ ∆PQR. A P
Solution:

In ∆ABC and ∆PQR

∠BAC = ∠QPR (A) CQ R

B

AB = PQ (S)

∠ABC = ∠PQR (A)

∴ ∆ABC ≅ ∆PQR, by A.S.A. axiom.

Example: 2

If the following pair of triangles are congruent, find the value of x.

Solution: A P

In ∆ABC and ∆PQR (x+3)cm
(2x-1)cm
AB = PQ(S)
CQ R
B

∠ABC = ∠PQR (A)

BC = QR (S)

∴ ∆ABC ≅ ∆PQR, [by S.A.S. axiom.]

AC = PR [Corresponding sides of congruent triangles]
or, 2x – 1 = x + 3

or, 2x – x = 3 + 1

∴ x = 4

Example: 3

In the given figure, AB //CD. Make ∆ABO ≅ ∆CDO and find the value of x and y.

Solution: A C

In ∆ABC and ∆PQR 3x-2 2y+1

∠BAO = ∠ODC [Alternate angles] O

AB = CD [Given] B y+3 x+2

∠ABO = ∠OCD [Alternate angels] D

∆ABO ≅ ∆CDO [By A.S.A. axiom]

AO = OD [Corresponding sides of congruent triangles]

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or, 3x – 2 = x + 2
or, 3x – x
or, = 2 + 2
or, 2x
∴ x = 4
Again, x
BO = 4
2

= 2

= OC [Corresponding sides of congruent triangles]

or, y + 3 = 2y + 1

or, y – 2y = 1 – 3

or, –y = –2

or, y = 2

∴ x = 2, y = 2

Example 4: A

In the given triangle, AB = AC, AD⊥BC, prove that-

(i) ∆ADB ≅ ∆ADC, (ii) ∠ABD = ∠ACD.

Given: In the given figure, AB = AC and AD⊥BC

To prove : ∠ABD = ∠ACD

Proof: B DC
Statements
Reasons

1. In ∆ADB and ∆ADC 1.
(i) ∠ADB = ∠ADC (R) (i) Being both right angles
(ii) AB = AC (H) (ii) Given
(iii) AD = AD (S) (iii) Common sides

2. ∆ADB ≅ ∆ADC 2. By R.H.S. axiom
3. ∠ABD = ∠ACD 3. Corresponding angles of congruent triangles

Proved.

Exercise 3.1

1. From the given conditions, use necessary axiom and make ∆ABC ≅ ∆PQR.

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

B CQ RB CQ RB CQ R
P
(d) A P (e) A

B CQ RB CQ R

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2. In each of the following, make ∆ABC ≅ ∆PQR and also write corresponding sides
and angles.

(a) A P (b) B 6cm C P
5cm 5cm 5cm 5cm
3cm 4cm
3cm 3cm 4cm
80º
B CQ R A Q 6cm R

(c) A P (d) P C
4cm 4cm
B 5cm 60º B
Q P 60º
C A
5cm 500 RQ 80º R (2x-1)cm

(e) A 3cm (3y+1)cm
5cm
500 5cm

B 600 C Q 600 R

3. (a) In the given figure, prove that A D 5cm E

600 700

(i) ∆ABC ≅ ∆DEF (x+3)cm (2y+3)cm

(ii) Find the value of 'x' and 'y'. B 600 700 C F
5cm

(b) In the given figure, XP (x+10)0 Q
Prove that:
(i) ∆XYZ ≅ ∆PQR. (3a+2)cm (2x-10)0 (4a–2)cm
(ii) find the value of 'a' and 'x'.
Y Z R

(c) In the given figure, A (2y+5)cm D
Prove that:
(i) ∆ABC ≅ ∆ADC (3x+2)cm (2x+7)cm

(ii) Find the value of 'x' and 'y'. B (4y–1)cm C

B (3y-x)cm D
(d) In the given figure, AB = CD, 2 cm
C
AB//CD, OA = (2x–1)cm, OB = (3y–x)cm, A (2x-1)cm O x cm A
OC = xcm and OD = 2cm.
Find the value of x and y. D

4. (a) In the given figure, AB = AC and AD⊥BC
prove that ∆ABD ≅ ∆ACD.

P SB C
QR
(b) In the given figure, PS = QR and PQ = SR,
prove that ∆PQR ≅ ∆PRS.

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(c) In the given figure, AC//DB and AO = BO. AC
Prove that : (i) ∆ACO ≅ ∆BDO, (ii) CO = OD O

DB

(d) In the given figure, AB//CD and AB = CD. A B

Prove that : (i) ∆AOB ≅ ∆COD, (ii) AO = OD. O

CD A

(e) In the given figure, AB = AC and AD⊥BC.

Prove that : BD = DC. B DC
(f) In the given figure, YW = WZ and XW⊥YZ. X

Prove : (i) ∆XYW ≅ ∆XWZ, (ii) XY = XZ

Y WZ

(g) In the given figure, AB = DC and AD
∠ABC = ∠DCB. Prove that:

(i) ∆ABC ≅ ∆DBC, (ii) AC = BD. BC
Answer

1. Consult your teacher 2. Consult your teacher

3. (a) x = 4cm, y = 2cm, (b) x = 200, a = 4cm, (c) x = 5cm, y = 3cm (d) x = 3/2cm, y = 1 cm

3.2 Similar Figures P Q
B
Let's compare the following pair of figures.

D
A

A

B CE FD CS R


The above pair of figures are similar in shape but different in size. Hence, above

pairs of figures are similar figures.

Two geometrical figures having same shape are called similar figures. It is denoted
by ~ sign.

Note: Similar figures may not be of the same size.

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Condition of similarity of triangles A P

In the adjoining figures; 600

∠A = ∠P = 600 600
∠B = ∠Q = 700 and
Q 700 500 R

∠C = ∠R = 500 B 700 500 C

Hence, ABC and PQR are equiangular triangles. Such equiangular triangles are sim-
ilar triangles.

It is written as ∆ABC ~ ∆PQR.

Two triangles are similar if three angles of the first triangle are respectively equal to
the three of the other.

What happens when two triangles are similar?

If two triangles are similar, then their corresponding sides are proportional.

How to show corresponding sides proportional?

Here, ∆ABC ~ ∆PQR.

Take the ratio of opposite sides of equal angles P

Opposite side of ∠A = BC A

Opposite side of ∠P QR

Opposite side of ∠B = AC Q R
Opposite side of ∠Q PR C
Opposite side of ∠C = AB
B

Opposite side of ∠R PQ

∴ QBCR = APRC = PAQB


Note: If two angels of the two triangles are equal, the third angle is automatically
equal. Therefore, two triangles are similar by AA Test.

Worked Out Examples



Example: 1 AD

In the given figure, ∆ABC ~ ∆DEF. Make their
corresponding sides proportional.

Solution: E F
Here, ∆ABC ~ ∆DEF [Given] BC

∴ AC = BC = AB [Corresponding sides of similar triangles are proportional)
DF EF DE

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Example: 2

In the given figure, ∆ABC ~ ∆DEF, find the value of x.
Solution:

Here, ∆ABC ~ ∆DEF [Corresponding sides of similar triangles are proportional]

∴ AC = BC = AB AD
DF EF DE

or, 142 = 2 = AB 12 cm
x DE 4 cm

Taking first and second ratios, BC

4 2 2 cm x F
12 x
E

or, =

or, 1 = 2
3 x

or, x = 6

∴ x = 6 cm

Example: 3 A

In the given figure, PQ//BC, prove that ∆APQ ~ ∆ABC. P Q
Solution: B C
Given : In ∆ABC, PQ//BC
To prove : ∆APQ ~ ∆ABC
Proof :

Statements Reasons
1. In ∆APQ and ∆ABC, 1.

(i) ∠APQ = ∠ABC (A) (i) Corresponding angles
(ii) ∠AQP = ∠ACB (A) (ii) Corresponding angles
(iii) ∠PAQ = ∠BAC (A) (iii) Common angles
2. ∆APQ ~ ∆ABC 2. By A.A.A.

Proved.

Example: 4

In the adjoining figure DE //BC, find the value of x and y.

Solution: B

Given : In ∆ABC, DE//BC 20cm

To find: Value of x and y. D y
Procedure:
10cm
6cm

In ∆ABC and ∆ABC, A x E 18cm C

(i) ∠DAE = ∠BAC (A) [Common angle]

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(ii) ∠DEA = ∠ACB (A) [Corresponding angles, DE//BC]

(iii) ∠ADE = ∠ABC (A) [Corresponding angles, DE//BC]

∴ ∆ADE ~ ∆ABC [By A.A.A. test]

Then, AACE = DE = AADB [Corresponding sides of similar triangles]
BC

or, x +x 18 = 6 10
y = 10+20

or, x = 6 10
x+18 y = 30

or, x = 6 = 1
x+18 y 3

Taking the first and the third ratios,

x = 1
x+18 3

or, 3x = x + 18

or, 3x–x = 18

or, 2x = 18

x = 9 cm

Again, taking the second and the third ratios.

6 = 1
y 3

or, y = 18 cm

∴ x = 9 cm, y = 18 cm

Exercise 3.2

1. In the given figure, ∆ABC~∆PQR. Make their corresponding sides proportional.
(a) (b)

(c) A AP (d) A QP
R
R B
AQ C

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2. (a) In the given figure ∆ABC~∆DEF, D
find the value of x. EF


(b) In the given figure ∆PQR~∆XYZ,
find the value of x and y.

AP

(c) In the given figure ∆ABC~∆PQR, 5cm
find the value of x and y. 10cm
8cm
y
Q 3cm R
B xC

3. Prove that following pair of triangles are similar and make their

corresponding sides proportional.

(a) A D (b) A P 650 Q (c) A X 500 Y

250 250 350 B 500 600 C 700

B 700 700 F Z
B
E R
C C

A

4. (a) In the given figure XY//BC. Prove that ∆ABC~∆AXY. xy

A C BC
D A
(b) In the given figure AB//CD.

Prove that ∆AOB~∆COD. B O

5. (a) In the given figure, prove that 2cm x
(i) ∆APQ~∆ABC. (ii) Find the value of 'x' and 'y'. y 5cm
P 4cm Q
B 8cm C

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(b) In the given figure, prove that ∆ABC~∆DCE
and find the values of x and y.

(c) In the adjoining figure, prove that
∆PQR ~ ∆PST. If PR = 30 cm, QS = 8 cm,
PQ = 24 cm and ST = 24 cm, find RT and QR.



(d) In the given figure, AB and CD are perpendiculars

drawn on BE. AC=2cm, AE = 8 cm, CD = 3 cm,
DE = 4.5 cm.

(i) Prove that ∆ABE~∆CDE.

(ii) Find the length of AB and BD.

Answer

1. Consult your teacher.
2. (a) 14 cm (b) x = 12 cm, y = 7.5 cm (c) x = 6 cm, y = 4 cm
3. Consult your teacher. 4. Consult your teacher.
5. (a) x = 5 cm, y = 2cm (b) x = 5 cm, y = 3.3 cm
(c) RT = 10 cm, QR = 18 cm (d) AB = 4 cm, BD = 1.5 cm

3.3 Circle TS

Circle is a locus of a point which moves such that V R

it remains equidistant from the fixed point. O

The given figure is a circle. P Q Radius

Fixed point is O and OP = OQ = OR = OT = ...... . Centre ODiameter
Fixed point is the centre of the circle and the fixed distance is its radius.

Remember !

Diameter of a circle = 2×radius
i.e. d = 2r

Note: half part of the circle is called semi circle.

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Circumference of the Circle Circumference
Diameter
The length of outer boundary which rotates around the fixed point and
which is equidistant from that point is the circumference of the circle.

In every circle, the ratio of circumference to its diameter always remains

same. Such constant value is called pi (π).
Circumference
∴ Diameter = constant (π)

Circumference = π or Circumference = π
Diameter 2r

∴ Circumference = 2πr.
The value of π is approximately 272.

Remember !

π= Circumference ≈ 272 .
Diameter



Worked Out Examples



Example 1:

Find the circumference of a circle whose radius is 7cm.

Solution:

Given, Radius of a circle (r) = 7cm

Circumference (C) = ?

We have, Circumference (C) = 2πr

= 2 × 22 × 7 cm
7

= 2 × 22 cm
= 44 cm


Example 2:

Find the radius and diameter of a circle whose circumference is 176 cm.

Solution:

Given, Circumference (C) = 176 cm

Radius (r) = ?

Diameter (d) = ?

We have, Circumference (C) = 2πr

176 = 2 × 22 × r
7

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or, 176 × 7 = 2 × 22 cm × r

or, r = 28 cm

Again, diameter (d) = 2r = 2 × 28 cm = 56 cm.

Example 3:

Diameter of a circular track is 70m. Find the distance covered by a man who ran around
the track 5 times.

Solution:

Given, diameter of the circular track (d) = 70 m

Radius (r) = 35 m.

We have, Circumference (c) = 2πr

= 2 × 22 × 35 m = 220 m
7

Circumference of the track means, distance covered by a man in 1 round.

Then, distance covered by him in 5 rounds = 5 × 220 m = 1100 m.

Example 4:

Find the perimeter of the given figure.

Solution:

Given, radius of the given semi circle (r) = 7 cm

Its curved boundary = πr 7cm

= 22 × 7
7

= 22cm

Its diameter = 2r = 2 × 7 cm = 14 cm

Perimeter of the given figure = 22 cm + 14 cm = 36 cm.

Exercise 3.3

1. Find the circumference of given circles.

(a) 14cm (b) (c) 3.5cm
(c) r = 7 cm
14cm

2. Find the circumference of a circle whose.

(a) r = 35 cm (b) d = 28 cm

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3. Find the radius of a circle whose circumference is

(a) 44 cm (b) 220 cm (c) 176 cm

4. Find the diameter of the circle whose circumference is

(a) 660 cm (b) 440 cm (c) 352 cm

5. Find the perimeter of the given figures.

(a) (b) (c) 4.9cm

14cm 3.5cm



6. (a) Radius of a circular ground is 21m. Find.

(i) its circumference.

(ii) length of the wire required to fence the ground.

(b) Diameter of a circular field is 28m. Find the distance covered by a man of
walking around the field 3 times.

(c) A man runs around a circular field of diameter 112 m. Find the distance covered
by him running around 5 times.



Answer

1. (a) 88 cm (b) 44 cm (c) 22 cm 2. (a) 220 cm (b) 88 cm (c) 44 cm
3. (a) 7 cm (b) 35 cm (c) 28 cm 4. (a) 210 cm (b) 140 cm (c) 112 cm
5. (a) 36 cm (b) 18 cm (c) 12.6 cm
6. (a) 132 m, 132 m (b) 264 m (c) 1760 m

Activity

Area of circle: Look and learn the following:

Take a circular piece of paper. If we bisect each of these sectors, we get 16 equal
sectors. These sectors can also be arranged as shown in the figure.

πr

Remember!

r Area of a circle = πr2
Area o=f a21πsre2mi
circle

Area of circle = Area of rectangle ABCD = length × breadth = πr × r = πr2 sq. unit


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Worked Out Examples



Example: 1

Find the area of the circle having radius 42 cm.

Solution: Here, Radius (r) = 42 cm

We have, Area of a circle (A) = πr2

= 22 × (42)2
7

= 22 × 42 × 42cm2 = 5544cm2.
7

Example: 2

If the area of a circle is 154 cm2, find its radius and circumference.

Solution:

Here, Area of a circle (A) = 154 cm²

Radius (r) = ?

We have, Area of circle = πr²

or, 154 = πr2
or,
∴ r2 = 154 = 154 × 7 = 49 cm²
π 22

r = 49 = 7 cm

Hence, Radius (r) = 7 cm

Again, circumference of a circle (C) = 2πr

= 2 × 22 × 7cm = 44cm.
7
Example: 3

If the circumference of a circle is 88cm, find its radius and area.

Solution: Circumference (C) = 88 cm
Here,

We have, 2πr = 88cm
or,
or, 2 × 22 × r = 88cm
or,
7

44r = 88 × 7

r = 88 × 7
44

∴ r = 14cm

Again, Area of a circle (A) = πr2

= 22 × (14)2cm2
7

= 22 × 14 × 14cm2 = 616 cm2.
7

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Example: 4

Find the area of the unshaded part.

Solution:

Radius of the circle inside square (r) = 14 cm.
2

We have, area of a circle (A) = πr²

= 22 × (7)² cm²
7

= 154 cm²

Area of square = l ²

= (14)²

= 196 cm²

∴ Area of the unshaded part = 196cm² – 154 cm²

= 42 cm²

Exercise 3.4

1. Find the area of the following circles.

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

14cm d = 42 cm

r = 7 cm d = 70 cm

2. Find the area of the following circles.
(a) Radius (r) = 14 cm (b) Radius (r) = 35 cm
(c) Diameter (d) = 5.6 cm (d) Diameter (d) = 4.2 cm
3. Find the radius and area of each of the following circles.
(a) Circumference (C) = 88 cm
(b) Circumference (C) = 440 cm
(c) Circumference (C) = 308 cm

4. Find the radius and circumference of circles having
(a) Area (A) = 154 cm2 (b) Area (A) = 616 cm2

5. (a) Area of a wheel is 308cm2. Find.
(i) its radius (ii) its circumference (iii) distance covered by it in 10 revolutions.
(b) Area of circular wheel is 616cm2. Find
(i) its radius (ii) its circumference (iii) distance covered by it in 100 revolutions.

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6. Find the area of the shaded region of the given figures.

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

Answer
1. (a) 154 cm2 (b) 616 cm2 (c) 3850cm2 (d) 1386 cm2 2. (a) 616 cm2 (b) 3850 cm2
(c) 24.64 cm2, 17.6 cm (d) 13.86cm2 3. (a) 14 cm, 616 cm2 (b) 70 cm, 15400 cm2
(c) 49 cm, 7546 cm2 4. (a) 7 cm, 44 cm (b) 14 cm, 88cm
5. (a) (i) 9.9 cm ii) 62.2 cm (iii) 622 cm (b) (i) 14 cm (ii) 88 cm (iii) 8800 cm

6. (a) 42 cm2 (b) 707.14 cm2 (c) 23.625 cm2 (d) 168 cm2

Activity

Relation between diameter and circumference of a circle

It can be verified by experiments that, for every circle,

circumference = π (constant) where π = 22 or 3.14 or 3.1416
diameter 7

Now, draw three circles with radii 2 cm, 3 cm and 4 cm respectively. Measure the

circumference (C) of each circle by pieces of string. Also measure its diameter (d) and

tabulate them as follows.

Figures Circumference (c) Diameter (d) Circumference (C)Diameter (d)
i. 12.56 cm 4 cm
ii. 18.84 cm 6 cm 12.56 = 3.14
iii. 25.12 cm 8 cm 4

18.84 = 3.14
6

25.12
8 = 3.14

2 cm 3 cm 4 cm

Fig. (i) Fig. (ii) Fig. (iii)


From above experiment, we state below the circumference diameter formula.

C = π, C = πd
d

Also, C = π(2r) = 2πr. [∵ d = 2r]

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3.4 Solids and Their Nets

We have already studied about solid figures and their nets in previous classes. These
are made up with the combination of different planes. Let us review some solid figures
such as cube, cuboid, cylinder, prism, pyramid, triangular prism, etc. and their nets.

Solid objects Net Solid Objects Net

Cube Prism

Cuboid Pyramid

Cylinder Cone

Exercise 3.5

1. Draw the net of following solid figures:

(a) Cube (b) Cuboid (c) Cylinder

2. Draw the net of : (a) Prism (b) Pyramid (c) Cone

3. Name the solid objects having the following nets:

(a) (b) (c)

Answer
Consult your teacher.

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Assessment Test Paper

Time : 35 minutes Full marks : 22
Pass Marks : 9

Attempt all the questions E B
Group A [2×1=2] G D

1. (a) In the given figure, write the corresponding angle of ∠EGB. A H
F
C

(b) Find the circumference of a circle whose radius is 7cm.

Group B [6×2=12] 1060
2. (a) Find the value of x and y in the given figure.
x
y

(b) In the given figure, ABCD is a parallelogram. A B
If ∠ADC = 650, find C
the value of ∠ABC and ∠BCD. 650

D

(c) In the given figure, AB//CD, BO = 8cm, OC = 6cm Ax 6cm C
and OD = 3cm. Find the value of x. D
B 8cm O 3cm

3. (a) In the given figure, AB//DC and AD//BC. A B
Use necessary axiom to prove ∆ADC ≅ ∆ABC. D
C

(b) Find the circumference of a circle whose area is 308cm2.
(c) Draw the net a cube.

Group C [2×4=8]
4. Verify experimentally that opposite angles of a parallelogram are equal.
5. Construct a regular hexagon having each side of 6cm.

64 Oasis School Mathematics-8

Coordinates

6Estimated Teaching Hours

Contents
• Pythagoras Theorem
• Distance between Two Points
Expected Learning Outcomes
At the end of this unit, students will be able to develop the following
competencies:

• Identify whether the given triangle is right angled triangle or
not.

• Verify Pythagoras theorem experimentally.
• Find the missing side of right angle using Pythagoras theorem.
• Identify the Pythagorean triplets.

Teaching Materials

• Models of right angled triangle, Graph sheet, etc.

Oasis School Mathematics-8 6655

Unit

4 Co-ordinates

4.1 Pythagoras Theorem perpendicular C

The Right Angled Triangle A base B
In the given figure, ABC is right-angled triangle in
hypotenuse
which ∠C = 90° and ∠B be the angle of reference.
The side opposite to 90° is the hypotenuse (h).

The side opposite to the angle of reference is the perpendicular (p).

The side adjacent to the angle of reference is the base (b).

Remember !

• Side opposite to right angle is 'h'.
• Side opposite to angle of reference is 'p'.
• Side adjacent to the angle of reference is 'b'.

The Pythagoras Theorem

ABC is a right angled triangle, right-angled at ∠C. Q

Here, AC = 4 units and BC = 3 units and AB = 5 units. Y

Squares on side AB, BC and AC are ARSB, BXYC and P C

APQC of ∆ABC respectively. X

45 3
We know that, AB

Area of a square = (side)2

∴ Area of square on side AB = AB² = 52 = 25
Area of square on side AC = AC2 = 42 = 16 R S

Area of square on side BC = BC2 = 32 = 9

Now, 25 = 16 + 9

i.e. Area of square ARSB = Area of square APQC + Area of square BXYC

or, AB² = AC2 + BC2

i.e. square of its hypotenuse = sum of the squares on two sides of a right angled triangle.

Hence, Pythagoras theorem states that in a right-angled triangle, the area of the square
described on the hypotenuse is equal to the sum of the areas of the squares described
on the other two sides.

66 Oasis School Mathematics-8

The adjoining figure shows a right–angled triangle ABC in which A
∠C = 90° and AB is the hypotenuse.

By Pythagoras' theorem:

AB2 = AC² + BC2 C B

Note: The longest side of the right angled triangle is hypotenuse.

The angle opposite to the longest side of a right- angle triangle is a right
angle.


Experimental Verification

The square of the hypotenuse of a right-angled triangle is equal to the sum of the
squares of perpendicular and base.

Draw three right angled triangles ABC with different shapes and sizes and right–
angled at B as shown in figures.

A BA
B

BC C A
C
Fig. (i) Fig.(ii)
Fig. (iii)

To Verify : AC2 = AB² + BC2

Verification: Measure the lengths AB, AC and BC and tabulate them as follows:

Figure AB BC AC AB2 BC2 AC2 AB2+BC2 Results

i. AC2 = AB2 + BC2

ii. AC2 = AB2 + BC2

iii. AC2 = AB2 + BC2

Conclusion: The square of the hypotenuse of a right-angled triangle is equal to the
sum of the squares of perpendicular and base.

Pythagorean Triplets
If three numbers a, b and c are such that a² + b² = c2, then the numbers a, b and c are

called Pythagorean triplets.
e.g. 3, 4 and 5 are Pythagorean triplets as 32 + 42 = 9 + 16 = 25 = 52
Some Pythagorean Triplets

5, 12, 13 7, 24, 25 14, 48, 50 9, 12, 15 9, 40, 41
8, 15, 17 20, 21, 29 12, 35, 37 3, 4, 5

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Worked Out Examples



Example: 1 A

Find the value of 'x' in the given figure.

Solution:

In the given right angled triangle ABC, 13 cm
x
h = AC = 13cm
B 12 cm
p = AB = x C

b = BC = 12cm

Using Pythagoras theorem, h2 = p2 + b2

AC2 = AB2 + BC2

or, (13)2 = x2 + (12)2

or, 169 = x2 + 144

or, x2 = 169 – 144

or, x2 = 25

or, x = 5cm

Example: 2

Identify whether 20cm, 16cm and 10 cm form the sides of a right angled triangle.

Solution: Now, h2 = p2 + b2 Longest side is
(20)2 = (16)2 + (10)2 the hypotenuse
400 = 256 + 100

or,

or, 400 = 356 which is not true.

∴ 20cm, 16cm and 10 cm do not form a right-angled triangle.

Example: 3

Find the value of x in the given figure. A 3cm B

Solution: 12 cm 4cm

ABC is a right angled triangle where, AB = 3cm, BC = 4 cm. xC

Using Pythagoras theorem, AC2 = AB2 + BC2 D

or, AC2 = (3)2 + (4)2

or, AC2 = 9 + 16

or, AC = 25
∴ AC = 5cm.

Again, ADC is a right angled triangle where AD = 12cm, AC = 5 cm, CD = x cm

68 Oasis School Mathematics-8

Using Pythagoras theorem, CD2 = AD2 + AC2


or, x2 = (12)2 + (5)2
or, x2 = 144 + 25
or, x2 = 169
or, x = 169

∴ x = 13 cm.

Example: 4

A ladder 25 m long rests against a vertical wall. If the foot of the ladder is 24m far from

the foot of the wall, find the height of the wall.

Solution: A
In the figure, AB is a vertical wall i.e. ∠ABC = 900
AC = length of ladder = 25 m Ladder=25m

BC = the distance from the foot of the wall = 24 m C 24m B

AB = height of the wall

In the right angled ∆ABC,

AC2 = AB2 + BC2 [using Pythagoras theorem]

or, (25)2 = AB2 + (24)2
or, 625 = AB2 + 576
or, AB2 = 625 – 576

or, AB2 = 49
or AB = 49 = 7m.

∴ Height of the wall = 7m

Exercise 4.1

1. Identify perpendicular (p), base (b) and hypotenuse (h) in each of the following
right angled triangles with reference to the blank marked angle.

(a) (b) P (c) X (d)

A D E

Z

B CQ R F
Y

Oasis School Mathematics-8 6699

2. Find the value of 'x' in each of the following figures.

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

3cm x x 13cm 24cm 12cm

15cm C
25cm
x
B 4cm CB C Bx C
.12cm

B

3. (a) Identify whether the following triangles are right angled or not.

(i) (ii) 3 c m (iii)

4cm 5cm 12cm 6cm 9cm
13cm

3cm 2cm

Using Pythagoras theorem determine whether each of the following triangles
(b) with given sides is right angled triangle or not .

(i) 12 cm, 5 cm, 10 cm (ii) 6 cm, 8 cm, 10cm

4. Find the length of the diagonal of the following squares.

(a) A B (b) P Q

8 cm

D 4 cm C SR



5. Find the length of the diagonal of the following rectangles.

(a) E F (b) P 9 cm Q

6 cm
12 cm

H 8 cm G SR

6. Find the value of 'x' in each of the following figures.

(a) A x D (b) P x S

9 cm 8 cm 4 2cm 6 cm

B 12 cm C Q 4 2cm R

(c) A (d) A 9cm C x D

15cmx 37cm
10 cm
B 6cm C 9cm D
B

70 Oasis School Mathematics-8

7. (a) In the adjoining figure, find the length of the ladder=?
ladder that goes 8 ft up from the foot of the wall.

8 ft.

6 ft.

(b) A ladder 5m long rests against a vertical wall. If the foot of the ladder is 3m far
from the foot of the wall, find the height of the vertical wall.



Answer

1. Consult your teacher. 2. (a) 5 cm (b) 5 cm (d) 7 cm (d) 9cm

3. (a) (i) Yes (ii) No (iii) No (b) (i) No (ii) Yes 4. (a) 4 2cm (b) 8 2cm

5. (a) 10cm (b) 15cm 6. (a) 17cm (b) 10cm (c) 17 cm (d) 35cm 7. (a) 10 ft. (b) 4m

4.2 Rectangular coordinate axes

We are familiar with number line which neutral number 'zero', positive numbers to its
right, 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
is divided into four parts which are called
2
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 the first -5
quadrant. Any point in this quadrant -6
contains both positive coordinates.
Y'

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

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(d) Fourth quadrant: The XOY' plane which is determined by positive part of X–axis
and negative part of Y –axis is called the 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
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. The sign of each coordinate
in different quadrants are summarized in following table:

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

Third – ve

Fourth + ve

Distance between two points

Let A (x1, y1) and B (x2, y2) be any two points, and d be the distance between them.

Draw AM and BN perpendicular to x-axis. Y B (x2, y2)

Here, OM = x1, AM = y1 (x , y )
ON = x2, BN = y2
1 1

A P

Draw AP⊥BN.

Now, AP = MN = ON – OM OM NX

= x2 – x1
Again, BP = BN – PN

= y2 – y1
In right angled ∆ABP, Using Pythagoras theorem,

AB2 = BP2 + AP2

d2 = (y2 – y1)2 + (x2 – x1)2

or, d = (x2 – x1)2 + (y2 – y1)2
∴ Distance between two points.

d = (x2 – x1)2 + (y2 – y1)2


Note:

X - coordinate on Y-axis = 0

Y- coordinate on X-axis = 0

Any point on X-axis = (x, 0)

Any point on Y-axis = (0, y)

72 Oasis School Mathematics-8

Equilateral triangle : All sides are equal.

Isosceles triangle : Any two sides are equal.

Scalene triangle : No sides of triangle are equal.

Right angled triangle : Pythagoras theorem is satisfied.

Parallelogram : Both pairs of opposite sides are equal or parallel.

Rhombus : All sides are equal.

Rectangle : Opposite sides and diagonals are equal.

Square : All sides and diagonals are also equal.

Important formula:
Distance between two points (d) = (x2 – x2)2 + (y2 – y1)2

Worked Out Examples



Example: 1

Find the distance between the points A (1, 2) and B (–2, 3).
Solution: Here,
A (1, 2) and B (–2, 3) are given two points.
Let A (1, 2) be (x1, y1) and B (–2, 3) be (x2, y2).
By distance formula,
d = (x2 − x1 )2 + (y2 − y1 )2

= (−2 − 1)2 + (3 − 2)2
= (−3)2 + (1)2
= 9+1
∴ = 10 units
Hence, distance between the points A(1, 2) and B(–2, 3) is 10 units.

Example: 2

Prove that the points P(3, – 7), Q( – 3, 3) and R(7, 9) are the vertices of an isosceles
triangle.
Solution: Here,
P(3, – 7), Q( – 3, 3) and R(7, 9) are the three vertices of ∆PQR.
For PQ,
Let P(3, – 7) be (x1, y1) and Q( – 3, 3) be (x2, y2)

Oasis School Mathematics-8 7733

So, by distance formula.

PQ = (x2 − x1 )2 + (y2 − y1 )2
= (−3 − 3)2 + (3 + 7)2

= (−6)2 + (10)2

= 36 + 100

∴ PQ = 136 units

For QR,

Let Q( – 3, 3) be (x1, y1) and R(7, 9) be (x2, y2).
So, by distance formula,

QR = (x2 − x1 )2 + (y2 − y1 )2
= (7 + 3)2 + (9 − 3)2

= 102 + 62

∴ QR = 100 + 36
=
For PR, 136 units

Let P(3, – 7) be (x1, y1) and R(7, 9) be (x2, y2)
So, by distance formula,

PR = (x2 − x1 )2 + (y2 − y1 )2 A(2, -1) D(-3,-2)
= (7 − 3)2 + (9 + 7)2 B(3, 4) C(-2,3)
= 42 + 162
= 16 + 256

∴PR = 272 units
We have, PQ = QR = 136 units
∴ ∆ PQR is an isosceles triangle.

Example: 3

Prove that: A(2, – 1), B(3, 4), C(–2, 3) and D( – 3, –2) are vertices
of a rhombus.
Solution: Here,
A(2, – 1), B(3, 4), C(–2, 3) and D( – 3, –2)
For AB,
A(2, –1) = (x1, y1), B (3, 4) = (x2, y2)

74 Oasis School Mathematics-8

We have (x2 − x1 )2 + (y2 − y1 )2
AB = (3 − 2)2 + (4 + 1)2
12 + 52
= 1 + 25
26 units
=
=
∴AB =

Similarly,
Again, A(2, –1) = (x1, y1), D (–3, –2) = (x2, y2)
We have,

d = (x2 − x1 )2 + (y2 − y1 )2
= (−3 − 2)2 + (−2 + 1)2

= (−5)2 + (−1)2

= 25 + 1
= 26 units

Again, B(3, 4) = (x1, y1), C (–2, 3) = (x2, y2)
d = (x2 − x1 )2 + (y2 − y1 )2

= (−2 − 3)2 + (3 − 4)2

= (−5)2 + (−1)2

= 25 + 1
∴BC = 26 units

Again, C(–2, 3) = (x1, y1), D (–3, –2) = (x2, y2)

d = (x2 − x1 )2 + (y2 − y1 )2

= (−3 + 2)2 + (−2 − 3)2

= (−1)2 + (−5)2

= 25 + 1

∴CD = 26 units

Here, AB = BC = CD = AD = 26 units

∴ Given vertices are the vertices of a rhombus.

Oasis School Mathematics-8 7755

Example: 4

Find a point on X–axis whose distance from a point (5, 4) is 5 units.

Solution: Here,

Given point is (5, 4).

Let the point (5, 4) be (x1, y1) and the point (x, 0) be (x2, y2) [∵ point on the X-axis is (x, 0]

Then, by distance formula,


d = (x2 − x1 )2 + (y2 − y1 )2

or, 5 = (x− 5)2 + (0 − 4)2

Squaring on both sides, we get

5² = (x – 5)2 + (0 – 4)2

or, 25 = x2 – 2x . 5 + 52 + 16

or, 25 = x2 – 10x + 25 + 16

or, 0 = x2 – 10x + 25 + 16 – 25

or, x2 – 10x + 16 = 0

or, x2 – (8 + 2)x + 16 = 0

or, x2 – 8x – 2x + 16 = 0

or, x(x – 8) – 2(x – 8) = 0

or, (x – 8) (x – 2) = 0

Either, x – 8 = 0 or, x – 2 = 0

∴ x = 8 x = 2

Hence, the required points on X–axis are (8, 0) or (2, 0).

Exercise 3.4

1. Find the distance between the following points.
(a) (3, –2) and (–7, 5) (b) ( – 2, 3) and (6, 1)
(c) (0, – 3) and ( – 7, 0) (d) ( – 6, 7) and ( – 1, – 5)
(e) (a, 2a) and (4a, 4a) (f) (–8, 7) and (–3, 4)
2. (a) Prove that the following points are the vertices of an isosceles triangle.
(i) (4, –1), (2, –5), (8, 3) (ii) (1, 1), (–4, 4), (4, 6)
(b) Prove that the points (9, 0), (3, 3) and (12, 21) are the vertices of a right angled

triangle.

76 Oasis School Mathematics-8

(c) Prove that the triangle with vertices (2, 1), (3, 0) and (1, 0) is a right angled
isosceles triangle.

(d) Prove that the triangle with vertices (1, – 1), ( 3 , 3 ) and ( – 1, 1) is an equilateral
triangle.

3. (a) Prove that the following points are the vertices of a rhombus.
(i) (5, 0), (8, 5), (2, 3) and (0, – 3) (ii) (7, 5), (10, 11), (4, 8) and (1, 2)
(b) Prove that the following points are the vertices of a parallelogram.
(i) (0, 0), (3, 2), (7, 7) and (4, 5) (ii) (– 4, – 4), (2, –4), (4, 4) and ( – 2, 4)

(c) Prove that the points ( – 7, 2), ( – 3, 2), ( – 3, – 1) and ( – 7, – 1) are the vertices of
a rectangle.

(d) Prove that the points (1, 5), (3, 7), (5, 5) and (3, 3) are the vertices of a square.

4. (a) Find the value of 'a' if the point (a, 0) is at a distance of 5units from the point (2, – 3).

(b) Find a point on X–axis which is at a distance of 5 units from the point (5, 4).
(c) Find a point on Y –axis which is at a distance of 17 units from the point ( – 8, 3).
(d) Find the value of 'a' if the distance between (0, 6) and (a, 0) is 6 units.

Answer
1. (a) 149 units (b) 2 17 units (c) 58 units (d) 13 units (e) a 13 units
(f) 34 units 4. (a) 6 (b) (8, 0) (c) (0, 18) (d) 0

Time : 20 minutes Assessment Test Paper Full Marks : 12
Pass marks : 5
Attempt all the questions: Group A [4×1=4]
A
1. (a) In a right angled triangle, which side is the longest side?

(b) In the given right angled triangle, identify perpendicular (p),

base (b) and hypotenuse (h) with reference to ∠C.

2. (a) On the X-axis, which co-ordinate is 0? BC

(b) A point lies on Y-axis, which is 5 units above the origin. What are its co-ordinates?

Group B [4×2=8] A

3. (a) Prove that 9, 40 and 41 are Pythagorean triplets. x 13cm

(b) In the given figure, find the value of x. B 12cm C

4. (a) Find the distance between the points (2, –3) and (–3, 1). B (0,3)

(b) In the given figure, co-ordinates of A and B are (a, 0) 5uits x
and (0, 3) respectively; If AB = 5 units, find the A (a, 0)
value of 'a'.

Oasis School Mathematics-8 7777

Mensuration

8Estimated Teaching Hours

Contents

• Area of Plane Figures
• Volume of Cube and Cuboid
• Volume of Combined Solid of Cube and Cuboid

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• Find the area of plane figure like triangle, parallelogram, rectangle,

square, rhombus, trapezium
• Find the volume of cube and cuboid
• Find the volume of combined solid of cube and cuboid

Teaching Materials

• Graph sheet, models of cube, cuboid, etc.

78 Oasis School Mathematics-8

Unit Perimeter, Area
and Volume
5

5.1 Perimeter and Area of Plane Figures

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.

Area

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

It has an area of 1 square centimeter (cm²). A rectangle of sides 7cm by 4 cm
can be divided 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

Oasis School Mathematics-8 7799

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

SN Name Figure Area Perimeter

1. Triangle A b 1 × base (b)×height (h) = 1 AB+BC+AC =
ch 2 2 c+a+b and Semi-
2. Equilateral B base D a height perimeter (S)
Triangle × BC × AD
C = a+b+c
3. Scalene 2
Triangle
A 3 3a
BC 4
a2

A s(s-a) (s-b) (s-c) P = a +b + c
cb
s = a+b+c
2

BaC

4. Rectangle A D length(l) × breadth (b) = 2 (l + b)
b b BC × AB

B lC

5. Parallelogram Ab base(b) × height (h) = BC 2(a+b)
a D × AE
height
B a
E C

b(base)

6. Square Al (side)2 = (l)2 4l
D

ll

B lC

7. Rhombus A D 12× product of diagonals

d1

d2 = 1 dl×d2
2

BE C = 1 AC × BD Or,
2

base×height = BC × AE AB+BC+CD+AD

80 Oasis School Mathematics-8

8. Quadrilateral A D 1
2
P1 N × diagonal × (sum of
dM
perpendiculars)

B P2 = 1 × d × (P1 + P2) AB+BC+CD+AD
2
C

9. Trapezium A aD 1 × height×(sum of //sides)
2

h h = 1 × h(a + b)
B Eb 2
C AB+BC+CD+AD
1
= 2 ×AE (AD+BC)

10. Kite A 1 × product of diagonals
B d1 d2 2

C D 1 AB + BC + CD + AD
2
= × d1 × d2

= 12 × AC × BD

Worked Out Examples



Example: 1

Find the area of the following figure:

Solution:

(i) In the given figure, ABCD is a rhombus DC

where, d1 = AC = 12 cm

d2 = BD = 18 cm 12cm 18cm

∴ Area of rhombus = 1 × d1× d2 A B
2

= 21 ×12×18 cm²

= 108 cm²

(ii) In the given figure, ABCD is a parallelogram. AB

where, DC = 12 cm, AE = 8 cm 8cm
8cm
∴ Area of parallelogram = base × height D E 12cm C

= DC ×AE

= 12 × 8

= 96 cm².

Example: 2 AF D

Find the area of shaded region.

Solution:

(i) Here, B 14cm CE

Oasis School Mathematics-8 8811

The area of the parallelogram ABCD = base × height

= BC × DE = 14 × 8 = 112 cm2

And the area of the triangle BCF = 12 × base × height

= 21 × BC × DE

= 21 × 14 × 8 = 7 × 8=56 cm2

∴A rea of the shaded region = Area of parallelogram ABCD – Area of ∆BCF

= 112 – 56 = 56 cm2

Example: 3

A rectangular field is 30m long and 10m broad. Find (i) its area and (ii) its cost of

plastering the field at Rs. 20 per square metre.

Solution:

Here, length of rectangular field (l) = 30 m,

Breadth of rectangular field (b) = 10m

(i) ∴ Area of rectangular field = l × b = 30 × 10 = 300 m2 = 300 sq.m.

(ii) Here cost of plastering 1 sq.m area = Rs. 20.
∴ Cost of plastering 300 sq.m area = Rs. 20 ×300 = Rs. 6000.

Exercise 5.1

1. Find the area of the following triangles.
AQ
(a) (b) (c) Y 6 cm (d) E

8 cm
6 cm
6 cm
8 cm

B CP 5 cm R X Z F
D
2. (a) 6 cm 8 cm
(b)
Calculate the area of a triangle whose base is 8 cm and height is 9 m.

Calculate the perimeter and semi-perimeter of a triangle whose sides are
10 cm, 24 cm and 26 cm.

3. Find the area and perimeter of the following rectangles:

(a) (b) (c)

4 cm 3.5 cm
6 cm

6 cm 10.5 cm

4 cm

4. (a) Find the area and perimeter of a rectangle if its length and breadth are 25 cm
(b) and 12 cm respectively.

The area of a rectangle is 64 cm2 and its length is 16 cm. Find the breadth of the
rectangle.

82 Oasis School Mathematics-8

(c) The area of a rectangle is 450 cm2 and its breadth is 15 cm. Find its length.
(d) The area of a rectangle is 540 cm2 and its length is 27 cm. Find its breadth and

perimeter.
5. Find the area and perimeter of the following squares.
(a) (b) (c)

16 cm

4 cm 5 cm

6. (a) Find the area and perimeter of a square whose side is 6 cm.
(b) If the perimeter of a square field is 220m, find the side of this square field.
(c) The area of a square is 64 cm2. Find its perimeter.

7. Find the area and perimeter of the following parallelograms.

(a) A B (b) Q 3 c m R (c)P 6 cm Q

4 cm
5 cm
9 cm
4.2 cm
4.2 cm
T 2 cm
DE C PS S R
6 cm N

8. Find the area of the following rhombuses.

(a) A B (b) P Q (c) M

D CS RP O

BD = 6 cm, AC = 5 cm SQ = 4.6 cm, PR = 3.5 cm MO = 8 cm, PN = 10 cm

9. Find the area of the following quadrilaterals. D

(a) A (b) P (c) N
M
P B
Q
SY G E
X Q

DC F
R
GE = 44 cm, DM = 20 cm,
BD = 4 cm, AQ = 3 cm SQ = 33 cm, PX = 19 cm, and FN = 15 cm

and CP = 2 cm and RY = 11 cm

10. The sides of a quadrilateral field, taken in order are 26m, 27m, 7m and 24m
respectively. Find its perimeter.

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11. Find the area of the following trapeziums: Q (c) M Y 28 cm N
(a) A 6 cm B (b) P 30 c m

9 cm
13 cm
9 cm

DE 10 cm C S 40 cm R P X 40 cm O
W
12. Find the area of the following kites. (c)
X
(a) B (b) P QZ

A CS

DR

AC = 15 cm SQ = 10 cm Y

BD = 8 cm PR = 8 cm WY = 8.6 cm
XZ = 3.5 cm

A

BD
13. Find the perimeter of the kite ABCD in which 5cm 5cm
AB = AD = 5cm and BC = CD = 6 cm. 6cm6cm

14. Find the area of shaded regions in the following figures. C
D
(a) F (b) (c) A 10cm E
15cm C S
A DD P 6cm

6 cm 5 cm
4 cm 14 cm

B 20cm C E E A 8cm B Q R C
15cm
B

15. A rectangular field is 20 m long and 12 m broad.
(a) Find its area.
(b) Find the cost of plastering the field at Rs. 10 per square metre.

16. The perimeter of a square room is 60 m.
(a) Find its length.
(b) Find the area of its floor.
(c) Find the total cost for carpeting the floor at Rs. 20 per sq. m.

84 Oasis School Mathematics-8

Answer

1. (a) 18cm2 (b) 14.98cm2 (c) 24cm2 (d) 16 3 cm2
2. (a) 36cm2 (b) 60cm, 30cm 3. (a) 24cm2, 20cm (b) 24cm2, 20cm (c) 36.75cm2, 28cm
4. (a) 300cm2, 74cm (b) 4cm (c) 30cm (d) 20cm, 94cm
5. (a) 16cm2, 16cm (b) 25cm2, 20cm (c) 128cm2, 32 2 cm
6. (a) 36cm2, 24cm (b) 55m (c) 32cm
7. (a) 24cm2, 22cm (b) 18cm2, 24cm (c) 25.2cm2, 20.4cm
8. (a) 15cm2 (b) 8.05cm2 (c) 40cm2 9. (a) 10cm2 (b) 495cm2 (c) 770cm2
10. 84m 11. (a) 72cm2 (b) 455cm2 (c) 306cm2 12. (a) 60cm2 (b) 40cm2 (c) 15.05cm2
13. 22cm 14. (a) 60cm2 (b) 16cm2 (c) 145cm2 15. (a) 240cm2 (b) Rs. 2400
16. (a) 15m (b) 225m2 (c) Rs. 4500

5.2 Volume of Cube and Cuboid a cm
a cm
Given figure is a cube.

Its length, breadth and height are equal.

a cm

Given figure is a cuboid. Its length, breadth and height are not
equal.

Length, breadth and height of a cube are equal whereas
length, breadth and height of cuboid are not equal.

Volume of a cube 1 cm
Length, breadth and height of the given cube is 1 cm. What is

the volume of this cube?

Volume of this cube is 1 cm3.

∴ Volume of the cube = l 3 1 cm
Volume of the cuboid = l × b × h
1 cm

Worked Out Examples



Example: 1

Length, breadth and height of a cuboid are 6 cm, 5 cm and 4 cm respectively. Find its
volume.

Solution:

Here, Length of the cuboid (l) = 6 cm

Breadth (b) = 5 cm

Height (h) = 4 cm

Oasis School Mathematics-8 8855

We have,

Volume (V) = l × b × h

= 6 cm × 5 cm × 4 cm

= 120 cm3

Example: 2

Find the volume of a cube having a side 10 cm.

Solution:

Here,

Length of a side of the cube (l) = 10 cm

Volume (v) = ?

We have,

Volume (v) = l3

= (10)3

E xam ple: 3 = 1000 cm3

Find the volume of the given solid objects.

a) b)

8 cm

3 cm 12 cm 8 cm

5 cm 8 cm



a) Solution:

Given object is a cuboid

where, l = 12 cm b = 5 cm and h = 3 cm

We have, Volume of cuboid = l × b × h

= 12 cm × 5 cm × 3 cm = 180 cm3

b) Solution:

Given object is a cube

where, l = 8 cm

We have, Volume of the cube = l³

= (8 cm)3 = 216 cm3

Example: 4 16 cm 8 cm 8 cm
9 cm8 cm 8 cm
Find the volume of the given object.
Solution: 20 cm
Given figure is the combined solid of two cuboids.
For upper cuboid,

86 Oasis School Mathematics-8

length (l) = 8 cm

breadth (b) = 8 cm

height (h) = 16 cm – 9 cm = 7 cm

We have, Volume of cuboid = l × b × h

= 8 × 8 × 7 cm³

= 448 cm3

Again, for lower cuboid, l ength (l) = 20 cm

breadth (b) = 8 cm

height (h) = 9 cm

Again, Volume of cuboid = l × b × h

= 20 cm × 8 cm × 9 cm

= 1440 cm3

Volume of given combined solid = 448 cm3 + 1440 cm3 = 1888 cm3

Example: 5

Length, breadth and height of a cuboid tank are 20m, 15m and 10m respectively. Find the
volume of the tank in litre. [1m3 = 1000l]

Solution:

Here, length (l) = 20 m, b readth (b) = 15 m, height (h) = 10 m

We have,

Volume of the tank = l × b × h

= 20 m × 15 m × 10 m

= 3000 m3

= 3000 × 1000 litres [ 1m3 = 1000 litres]

E xam ple: 6 = 30,00,000 litres

Find the number of pieces stones of size 30 cm × 20 cm × 10 cm required to construct a
wall of 15m × 50 cm × 2 m.

Solution:

For stone,

l = 30 cm = 30 m = 0.3 m
100

b = 20 cm = 30 m = 0.2 m
100

h = 10 cm = 13000 m = 0.1 m
We have,

Volume of a stone (v) = l × b × h

= 0.3 × 0.2 × 0.1 m3

= 0.006 m3

Oasis School Mathematics-8 8877

Again,

For wall length (l) = 15 m

breadth (b) = 50 cm = 50 m = 0.5 m
100

height (h) = 2 m

Volume of the wall (V) = l × b × h

= 15 × 0.5 × 2

= 15 m3

We have, Number of stones pieces (N) = V
v
15
= 0.006

= 15000
6

= 2500

∴ 2500 pieces of stones are required to build the wall.

Exercise 5.2

1. Find the volume of cuboids having following dimensions.
a. length (l) = 15 cm, breadth (b) = 8 cm, height (h) = 5 cm
b. length (l) = 20 cm, breadth (b) = 12 cm, height (h) = 10 cm
c. length (l) = 2 m, breadth (b) = 50 cm, height (h) = 80 cm
d. length (l) = 1.2 m, breadth (b) = 0.8 m, height (h) = 20 cm

2. Find the volume of cubes having

a. a side 6 cm b. a side 4.5 cm c. a side 9 cm d. a side 6.5 m

3. Find the volume of the given solid objects.

a. 5 c m b.


15 cm 10 cm

10 cm d. 8 cm

c. 12 cm

17 cm12 cm
12 cm
24 cm

10 cm 8 cm

88 Oasis School Mathematics-8

4. a. Volume of a cuboid is 360 cm3. If its length and breadth are 18 cm and 10 cm
respectively, find its height.

b. Volume of a cuboid is 420 cm3. If its breadth and height be 10 cm and 14 cm
respectively, find its length.

c. Capacity of a tank is 180 litre. If its length and height be 2 m and 50 cm
respectively, find its breadth.

d. Volume of a cubical tank be 216m3. Find the length of its one side.
e. Volume of cubical tank be 125000 litre. Find its length.

[1 litre = 1000 cm3]

5. Find the volume of the given combined objects.

(a) 10 cm (b) (c) 10 cm

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

16 cm 8 cm 6 cm 3 cm 8 cm8 cm
18 cm 4 cm 3 cm 3 cm

4 cm 10 cm

(d) (e) (f)

50 cm

5 cm 4 cm 20cm
8cm
6 cm 6 cm 8cm
18 cm 6 cm
8cm
6 cm 5 cm 6 cm5 cm 20cm
5 cm 10 cm 5 cm 8 cm5 cm 6cm
4cm 10cm 4cm
8 cm

4 cm

6. a. How many pieces of stone each of size 20 cm × 30 cm × 10 cm are required to
b. construct a wall of 20 m × 60 cm × 2m?

Find the number of stone pieces required to construct a wall of 15 m × 40 cm
× 1 m with the brick of size 12 cm × 5 cm × 8 cm.

c. How many boxes each having 10 cm × 5 cm × 2 cm can be put into the box
20 cm × 10 cm × 4 cm?

d. How many cubical boxes having each side 6 cm can be put into the bigger
cubical box having each side 12 cm?



Oasis School Mathematics-8 8899

Answer
1. (a) 600cm3 (b) 2400 cm3 (c) 800000 cm3 (d) 192000cm3
2. (a) 216 cm3 (b) 91.125cm3 (c) 729cm3 (d) 274.625 cm3
3. (a) 750cm3 (b) 960 cm3 (c) 2040cm3 (d) 2304cm3
4. (a) 2cm (b) 3cm (c) 18cm (d) 6cm (e) 5m or 500 cm
5. (a) 232cm3 (b) 832cm3 (c) 752cm3 (d) 1200cm3 (e) 7000cm3
(f) 1840cm3 6. (a) 4000 (b) 12500 (c) 8 (d) 8

Assessment Test Paper

Time: 20 minutes Full marks : 12
Pass marks : 5

Attempt all the questions:
Group A [2 × 2 = 4]

1. (a) Find the area of a triangle whose three sides are 5cm, 6cm and 7cm.
(b) Capacity of a tank is 1000 litre. If its length and breadth are 1m and 60cm, find

its height.

Group B [4 × 2 = 8] 10cm

2. Find the volume of the given solid. 10cm 10cm
10cm

20cm 15cm
10cm 15cm
20cm

3. How many cubical boxes having each side 5cm can be put into the bigger cubical
box having each side 10cm?

90 Oasis School Mathematics-8

Transformation

10Estimated Teaching Hours

Contents
• Transformation
- Reflection
- Rotation
- Translation
• Bearing
• Scale Drawing

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• Reflect the point and geometrical shape using co-ordinates.
• Rotate the point and the geometrical shape through ± 900 and ± 1800

taking origin as the centre using co-ordinates.
• Translate the point and the given figure using co-ordinates.
• Find the compass bearing and three digit bearing of two places.
• Find the distance between two places from the given map using scale

factor.

Teaching Materials

• Maps, graph paper, etc.

Oasis School Mathematics-8 9911

Unit

6 Transformation

Introduction

An object (or a geometrical figure) changes its position or size (or both) by definite
rule which is known as the 'transformation'. There are many kinds of geometrical
transformations. They are (i) Reflection, (ii) Rotation (iii) Translation (or displacement)
(iv) Enlargement. etc.

6.1 Reflection – Review

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

(i) Distance of the image.

(ii) Nature of the image.

Distance of the image from the mirror is equal to the O P I
Object Image
distance of the object from the mirror.

The image so formed is inverted.

This phenomenon is the reflection. Mirror

I. Properties of reflection P A P'
Q Q'
The following are the properties of reflection. B
B M B'
(i) The object and its image are equidistant from A A'
the axis of reflection.
A M A'
Here, in the figure, PA = P'A and QB = Q'B.

(ii) The areas of object and its image are equal.
Here in the figure, area of ∆ABC = area of ∆ A'B'C'.

(iii) Object and image are opposite to each other. M D
C M
(iv) The points on the axis of reflection are invariant points
i.e., their images are the same points.

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

92 Oasis School Mathematics-8

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

A

II. The Method of Finding the Image Under Reflection

In the given figure, ABC is a triangle and M is the axis of B C
reflection. M

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

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

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

• Draw CZ perpendicular to the axis of reflection X' Y X ZX
C'
'M' and produce it to C' such that CZ = ZC'.
B'
• Join A'B', A'C' and B'C' respectively. Y' A'

Hence, ∆A'B'C' is the image of the ∆ABC after Y
reflection in the line 'M'.

III. Reflection of Geometrical Shapes Using
Co-ordinates

a. Reflection on x–axis (y = 0):

Take a point A(3, 4) and reflect it on x–axis. A(3, 4)

From A, draw AM perpendicular to x–axis and X' MX
produce it to A' such that AM = MA'. Then the
co-ordinates of A' will be A' (3, – 4). A'(3, –4)
Y'
∴ A(3, 4) Reflection A'(3, –4)
x-axis Y

∴ Reflection of P(x, y) on x–axis is

P(x, y) Reflection P'(x, –y)
x-axis



b. Reflection on y–axis (i.e. x = 0):

Take a point P(3, 1)

Plot the point P(3, 1) on graph paper. From P, draw P'(-3,1) M P(3,1)
PM perpendicular to y–axis and produce it to P' such

that PM = MP'. X' X

Then, the co-ordinates of P' be (–3, 1)

i.e. P(3, 1) Reflection P'(– 3, 1)
y-axis

i.e. P(x, y) Reflection P'(– x, y) Y'
y-axis

Oasis School Mathematics-8 9933

In symbol, the reflection on y–axis is denoted by

P(x, y) Reflection P'(–x, y).
y-axis

For example,

(i) P(1, 3) Reflection P'(–1, 3) (ii) Q(–3, –1) Reflection Q'(3, –1)
y-axis y-axis

Worked Out Examples R

P
Q
Example: 1
M
Find the image of the ∆PQR
under the reflection on the line M.
Solution:
Here,


Example: 2

Find the images of the following points under the reflection on x–axis (i.e. y = 0).

(i) A(5, 2)) (ii) B(4, –3) (iii) C(–7, 8) (iv) D(–4, –5)

Solution:

We have, P(x, y) Reflection P'(x, –y)
x-axis

(i) A(5, 2) Reflection A'(5, –2) (ii) B(4, –3) Reflection B'(4, 3)
x-axis (iv) x-axis

(iii) C(–7, 8) Reflection C'(–7, –8) D(–4, –5) Reflection D'(–4, 5)
x-axis x-axis

Example: 3

Find the images of the following points under the reflection on y–axis (i.e. x = 0).

(i) N(–3, 6) (ii) Q(4, –5) (iii) R(–9, –3) (iv) M(4, 7)

Solution:

We have, P(x, y) Reflection P'(–x, y).
y-axis

(i) N(–3, 6) Reflection N'(3, 6) (ii) Q(4, –5) Reflection Q'(–4, –5)
y-axis y-axis

94 Oasis School Mathematics-8