Prime Mathematics 8

Prime Mathematics Book - 8

(f) The given triangles ∆ABC & ∆PQR are Congruent triangles. Find the value of x and y.

A(3x - 0.6)cm. P
56° 3.1cm. 60°
C 64° (2y-1.7)cm.

Q (x 64°
3)cm
+ R

B

AD

4. (a) In the given figure, line segments AB and CD bisects O
B
each other at the point O. Prove that AD = CB. C
A

(b) In the given figure, ABC is an isosceles triangle with D C
AB = AC. If AD ⊥ BC, Prove that ∠B = ∠C and BD = DC.

B
(c) Prove that diagonals of a parallelogram bisect each other.

46 1 Geometry

Prime Mathematics Book - 8

Unit - 4 Similar Triangles Estimated period : 4

Introduction

Similar figures:
Two figures having same shape and same or different sizes are called similar figures.

Similar triangles:

Two triangles are said to be similar if three angles of one triangle are respectively

equal to three angles of other triangle and corresponding sides of the triangles are

proportional. D

A

B CE F

∆ ABC ∆ DEF such that AB BC AC
DE EF DF
∠A = ∠D, ∠B = ∠E and ∠C = ∠F and = = then ∆ ABC and ∆ DEF are similar.

We write ∆ ABC ~ ∆ DEF

Conditions for similarity of triangle.

A. Angle, angle similarity theorem. (Angle, angle similarity theorem)
If three angles of a triangle are respectively equal to the three angles of other, then
the two triangles are similar.
Also if two angles of a triangle are respectively equal to two angles of other triangle,
then the third pair of angles are also equal and hence, the two triangles are similar.

D

A

B CE F

If ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
or ∠A = ∠D, ∠B = ∠E then ∆ ABC ∼ ∆ DEF.

Similar Triangles 47

Prime Mathematics Book - 8

B. Side, side, side similarity theorem: D A
If the three sides of one triangle are
B
proportional to the corresponding F

sides in the other triangle then the C

two triangles are similar. E

If AB = BC = AC then ∆ ABC and ∆ DEF are similar.
DE EF DF

∆ ABC ~ ∆ DEF

C. Side, angle, side similarity theorem: A

If two sides of one triangle are proportional to B C
the two corresponding sides of other triangle D F
and the corresponding angle between these side
are equal, then the triangles are similar. E

If AB = BC and ∠ABC and ∠DEF, then
DE EF

∆ ABC ~ ∆ DEF

N- ote: Proofs of these theorems are beyond the course of class VIII. We use any of the
above conditions (theorems) to show the triangles similar.

- We also use an important theorem.
"The similarity between the triangles is a transitive relation."
i.e. if ∆ ABC ~ ∆ DEF and ∆ DEF ~ ∆ GHI, then ∆ ABC ~ ∆ GHI.

- In similar triangles, Angles opposite to the proportional sides are corresponding
angles and the sides opposite to the equal angles are corresponding sides.

D
A

B C AB BC AC E F
DE EF DF
If = =

then, ∠C = ∠F, ∠A = ∠D and ∠B = ∠E and if ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

then, BC = AC = AB
EF DF DE

48 1 Geometry

Prime Mathematics Book - 8

Example 1: In the given figure, ∆ABC ∼ ∆DEF. A D 2 cm
Find the length of EF. E
6 cm

Solution : B 9 cm C F
R
Here, ∆ABC ~ ∆DEF where ∠B = ∠E and ∠C = ∠F.
∴ ∠A = ∠D (remaining angles). AC = 6cm, BC = 9cm, DF = 2cm and EF = ?
Since corresponding sides of similar triangles are proportional,

DF = EF
AC BC

or 2 cm = EF
6 cm 9 cm

or EF = 1 × 9 cm
3
A P
∴ EF = 3 cm
7 cm 700 3 cm
Example 2: In the given figure, find the length of (x+3)cm

3.5cm
the sides AB and QR. Q 500
B 600 500 y cm
Solution: Here, C
∠A = 180o - (60o + 50o) = 70o 8 cm
∠R = 180o - (50o + 70o) = 60o
In ∆ABC and ∆PQR

∠A = ∠P = 70o
∠B = ∠R = 60o
∠C = ∠Q = 50o

∴ ∆ABC ~ ∆PQR (by AAA similarity theorem)

∴ AB = AC = BC (Corresponding sides of similar
PR PQ QR triangle are proportional)

or (x + 3) cm = 7 cm = 8 cm
3 cm 3.5 cm y cm

Taking first two ratios,

or (x + 3) = 7 Now,
3 3.5

or x = 6 - 3 8 = 7
or x = 3 y 3.5
∴ AB = x + 3 cm
or, 7y = 8×3.5
= 3 + 3
= 6 cm or, y = 4

∴ QR = 4 cm

Similar Triangles 49

Prime Mathematics Book - 8

Example 3: In the adjoining figure, XY//BC. If AX = 6cm, AY = 8cm and C

XY = 2 , find AB and AC. Y
BC 3 8 cm

Solution :

Here, in ∆ AXY and ∆ ABC, A 6 cm X B
∠XAY = ∠BAC (common angle)

∠AXY = ∠ABC (corresponding angles as XY//BC)

∴ ∆ AXY ~ ∆ABC (by A.A. similarity theorem)

XY = AX = AY (corresponding sides of similar triangles are proportional)
BC AB AC

or 2 = 6 cm = 8 cm
3 AB AC

Now, 2 6 cm
3 AB
=

∴ AB = 9 cm

And 2 = 8 cm
3 AC

or 2 × AC = 24 cm

or AC = 24 cm
2

∴ AC = 12 cm

Example 4: From the adjoining figure, find the length of CD. C
?
Solution : A 8 cm D

Here, in ∆ AOB and ∆DOC, 3.5 cm

AO = 3.5 cm = 1 2.7 cm 4 cm o 7 cm
DO 7 cm 2 B

BO = 4 cm = 1
CO 8 cm 2

and ∠AOB = ∠COD (Vertically opposite angles)

∴ ∆ AOB ~ ∆ DOC (By S.A.S. similarity theorem)

∴ AB = BO (Corresponding sides of similar triangles are proportional.)
CD CO

or 2.7 cm = 4 cm 1
CD 8 cm 2

∴ CD = 5.4 cm

50 1 Geometry

Prime Mathematics Book - 8

Exercise 4

1. State the conditions under which the following pairs of triangles are similar.

(a) A P

30o C Q 30o R

B

(b) A C (c) A
E 3 cm
6 cm E
B 2 cm C 9 cm
55o
F 35o D
B

D

(d) P X

1 cm
4 cm
8 cm 2 cm

R Z
Q 7.2 cm Y 1.8 cm

(e) M (f)
N
40o

Q 110o 30o
O
110o
P

Similar Triangles 51

Prime Mathematics Book - 8

2. Show that the following triangles are similar and write the corresponding sides in

each of the following pairs of triangles.

A

(a) A (b)

DE C
BD

B CF
(c) P E

Q
(d) A

R

T SB M C

3. Find the unknown sides of the following similar triangles: Y
C
(a) 6 cm

15 cm
3xcm
y
3 cm
12 cm 3 cm

(b) 18 cm 6 cm y
2 cm
x
4 cm
(c) A 15 cm (d)

N y
M 2.5 cm
12 cm x
x 12 cm
y 4 cm

B 8 cm C A

4. (a) In the figure given alongside, prove that: X 45o
∆ ABC ~ ∆ AXY. B 45o

52 1 Geometry

Prime Mathematics Book - 8

D

(b) In the adjoining figure, find the length of the 4 cm A 12 cm
side CD.
2 cm 8 cm
B
O C
(c) In the adjoining figure, ∠BAC = ∠BED. If A
AD = 9 cm, BD = 6 cm, EB = 7.5 cm, 3 cm D
BC = 12 cm and DE = 4.8 cm, find AC. x
9 cm 6 cm

4.8 cm B

CE 7.5 cm
12 cm

A

(d) In the adjoining figure, AB ⊥ BE and CD ⊥ BD, C ?
find the height of the tree (AB).

E 8m 6m

D 12 m B

(e) In the adjoining figure, ∠A = 90o A C
and AD ⊥ BC. Prove that BD

(i) AB2 = BC.BD
(ii) AC2 = BC.CD
(iii) AB2 + AC2 = BC2.

(f) In the adjoining figure, AB ⊥ AC and AD ⊥ BC. A

Prove that ∆ ADB ~ ∆ ADC and hence show D
Q
that AC . BD = AB . AD. B C
P
(g) In the adjoining figure, ML ⊥ NP, L
∠LMW = ∠NPQ = - 90, Prove that M N
∆ LMN ~ ∆ NPQ. Similar Triangles 53

Prime Mathematics Book - 8 Circle Estimated period : 5

Unit - 5

Introduction centreO radius A

A circle is a closed figure bounded by a regularly circumference
curved line where each point of whose boundary is at
a constant distance from a fixed point. The fixed point diaOmeter
is called centre and the fixed distance is called radius.

The boundary is called circumference. The straight line
segment drawn through the centre and ending at the
circumference is called diameter.

Relation between the circumference and diameter of a circle:

Draw circles with radius r1 = 5 cm
3 cm, r­2­­ = 4 cm and r3 = 5
cm, then the lengths of the 3 cm 4 cm

diameters of the circles are (i) (ii) (iii)

d1 = 6 cm, d2 = 8 cm and d3 Fig Diameters Circumference C
= 10 cm respectively. By the D
(i) d1 = 6 cm C1 = 18.86 cm 3.14
help of thread, Measure the (ii) d2 = 8 cm C2 = 25.14 cm
(iii) d3 = 10 cm C3 = 31.42 cm 3.14
length of the circumference
3.14
of each circle and compare

the ratios of the lengths of

circumference and diameter.

In each figure (circle), we get C = 3.14
D
C
∴ In any circle, D = 3.14

The constant number 3.14 is denoted by π (pi).

∴ C =π
D

or, C = π D ............. (i)

Since D = 2 r (Diameter is double the radius.)

So, C = π 2r

∴ C = 2πr
Thus, circumference (C) = 2πr ............... (ii)

54 1 Geometry

Prime Mathematics Book - 8

Note: (i) For the exact measures of C and D, the value of nDCumisbneorn. terminating
and non recurring decimal. So, π is an irrational

(ii) For practical purpose π is also taken as 22
7

Area of circle :

Draw a circle in chart paper. Divide it into a number of equal sectors, cut the sectors
and arrange them as in the figure.

AB

D C
A B

DC

If we divide the circle into considerably small segments and arrange in the same way,

the shape ABCD so formed becomes a rectangle with

Length (l) = 1 of Circumference = 1 × 2πr = πr
2 2

Breadth (b) = radius = r

Now area of the circle = area of the rectangle = l × b = πr × r = πr2

Therefore, area of circle = πr2

Example 1: Find the length of the circumference of the circle of radius 3.5 cm.

Solution: Here, radius (r) = 3.5cm

We have,

Circumference (c) = 2πr 0.5
× 3.5 cm
= 2× 22 = 22 cm 3.5 cm
7

∴ Circumference of the circle is 22 cm.

Circle 55

Prime Mathematics Book - 8

Example 2: If the length of the circumference of the circle is 44 cm, find its diameter.

Solution : Here,
Circumference (C) = 44 cm [ ∴ c = πd or 2 πr]

or πd = 44 cm

or 22 ×d = 44 cm
7
44 × 7
or d= 22 = 14 cm

∴ Diameter of the circle is 14 cm.

Example 3: Find the area of the circle with radius 14 cm.

Solution : Here,

Radius of the circle (r) = 7 cm

We have, = 22 × (7cm)2 = 22 × 7 154 cm2
Area of circle = πr2 7 7 49 cm2 =

∴ Area of the circle is 154 cm2.

Example 4: If the circumference of a circle is 88 cm, find the area of the circle.

Solution : Here, Circumference (C) = 88 cm [ ∴ C = 2πr]
or 2πr = 88 cm

or 2 × 22 × r = 88 cm
7 2

∴ r 4= 88 × 7 = 14 cm
2 × 22
Now,
Area of the circle = πr2 = 22 × (14 cm)2 = 22 2 = 616 cm2
7 7 × 14 cm × 14 cm

Therefore, the area of the circle is 616 cm2.

Example 5: Find the area of the shaded part of the figure given below. A B
C
Solution :
In the given figure, ABCD is a rectangle and unshaped

part is a half circle (semi -circle). D
Here, radius of the semi- circle (r) = 3.5 cm

length of the rectangle (l) = 2r = 2 × 3.5 cm = 7 cm 3.5 cm
breadth of the rectangle (b) = r = 3.5 cm

∴ Area of rectangle (A) = l × b = 7 cm × 3.5 cm D

24.5cm2
22
Area of semi circle (a) = 1 πr2 = 2 x7 (3.5cm)2 = 19.25cm2
2

Area of the shaded part = 24.5 cm2 - 19.25 cm2 = 5.25 cm2

∴ Area of the shaded part = 5.25 cm2

56 1 Geometry

Prime Mathematics Book - 8

Exercise 5

1. Find the circumference and area of the circle having;

(a) radius = 35 cm (b) radius = 21 cm (c) radius = 4.2 cm
(d) diameter = 14 cm (e) diameter = 5.6 cm (f) diameter = 6.3 cm

2. Find the radius of the circle having; (b) Circumference = 220 cm
(d) Area = 38.5 cm2
(a) Circumference = 132 cm (f) Area = 154 cm2
(c) Circumference = 63.8 cm
(e) Area = 1386 cm2

3. Find the area of the shaded parts of the following figures.

(a) (b) (c)

14 cm 28 cm 7 cm

(d) (e) (f)

7 cm

14 cm

2.8 cm 5.6 cm

4. Solve the following problems:

(a) The circumference of a circle is 6 π cm. Find the area of the circle. (π = 3.14)
(b) In a complete rotation, a bicycle wheel moves a distance of 66 cm. Find the
diameter of the wheel.
(c) How far will a wheel with radius 1.4 cm move in 20 revolution?
(d) A goat is tethered to a stake by a rope 8.4 m long. If the goat makes a complete
revolution keeping the rope straight, how far does it move?
(e) Area of a circular garden is 3850 m2. Find the length of the wire required to fence
the garden with 5 rounds.
(f) A wire in the form of circle of radius 42 cm is bent into the form of a square. Find
the area of the square.

Circle 57

Prime Mathematics Book - 8

Unit - 6 Solid Shapes Estimated period : 6

Introduction

We know that solids are the matters having fixed shape and volume. Solids have
different shapes, irregular and regular. Solid shapes of geometric importance are
spherical, cylindrical, conical, polyhedral etc.

Polyhedra : Solids having plane polygonal surfaces (faces) are called polyhedra. Prisms,
pyramids, cubes are some examples of polyhedra.

base

Prisms:

A prism is a solid (Polyhedron) containing any polygonal Lateral part
bases which are congruent and rectangular lateral faces. (rectangular)
The prism are named according to the shapes of the base
base (cross-section). (any polygon)



Triangular Square/rectangular Hexagonal
based prism based prism based prism

Note:
• In a prism two opposite base are congruent.
• If we make a cubic parallel to the bases the cross-section should be congruent to the bases.
• A cube and a cuboid are all side prism. i.e. any opposite faces can be considered or
bases and the faces between them as lateral faces.

Other prisms:

Lateral parts Lateral parts
L Shaped base U Shaped base

Lateral parts

H Shaped base Lateral parts base Lateral parts T Shaped base

58 1 Geometry

Prime Mathematics Book - 8

Nets of a prism:

i) Triangular prism

• Draw the net of equilateral

triangular based prism as shown 5cm 5cm 5cm

in a chart paper.

• Cut along the outlines (dotted 7cm

lines)

• Fold along the continues (bold)

lines, put que on shaded parts

and join respective parts.

ii) Square based prism


4cm

Note:
- Using the same idea, we can draw the nets of any other prism and make attractive models.
- Joining solid pieces (straw or wire) along the edges, we can get skeletal models.

Surface area of a triangular prism

Base of a triangular prism is a triangle. The formula for area of triangle depends upon
its shape.

If the base is a triangle with base (b) and height (h), area of base (A) = 1 × b × h
2

If the base is an equilateral triangle,

area of base (A) = 3 a2 where a is the length of side.
4

Solid Shapes 59

Prime Mathematics Book - 8

If the base is a right angles triangle,

area of base (A) = 1 p.b where p and b are lengths of perpendicular and base sides
2

If the base is of sides a, b, c,

area of base (A) = s(s−a)(s−b)(s−c) where s = a + b + c
2



Lateral surfaces area of prism.

As in the figure,

∴ Lateral surfaces area (L.S.A.) A

= Area of rectangles (ABB1A1 + BCC1B1 + ACC1A1) B C
= AB × h + BC × h + AC × h

= (AB + BC + AC) × h h

∴ L.S.A. = p × h (Where p is perimeter of base). A1

B1 C1

∴ Total surfaces area (T.S.A.)= 2 (base area) + L.S.A.

= 2 A + L.S.A.

C. Volume of a prism:
Consider a cuboid of sides l, b and h which is also a prism of base area l × b = A.

We know

volume (v) = l × b x h h

= (l × b) × h

=A×h b
l
∴ Volume of prism = base area × height

Cylinder

You are familiar with cylindrical shapes. A hollow right cylinder can be formed by join-
ing the opposite edges of a rectangular sheet of paper as shown in the figure.


60 1 Geometry

Prime Mathematics Book - 8




A. Net of Cylinder A 10cm B
Process

1. On a chart paper draw a rectangle

with length 22cm and desired width

(say 10cm). CD = AB = 10cm, BC = AD y 3.5cm 22cm
= 22cm

2. At any point on BC and AD draw

perpendicular XY and taking radius D C
XY = 3.5cm draw circles.

3. Cut along the outlines.

4. Bring the edges AB and CD together and join with cello tape.

5. Along the curved edge AD and BC gently join the circumference of the circles.

Then, a solid cylinder of base radius 3.5cm is formed.

B. Surface area of Cylinder Area= πr2
r

Curved Area = l×h
Surface h = 2πrh
Area= πr2
l = 2πr
r

Circular base

Two circular ends are called circular bases. Surface between Base area (A) = πr2
two circular bases is called Curved surfaces. If we defold the C.S.A = 2πrh
cylinder as shown above, the curved surface form a rectangle
of length (l) = 2πr and breadth(b) = height of cylinder (h) T.S.A = 2πr(h+r)

∴ Curved surface area = l × b = 2πrh.

Total surface area = Curved surface area + Area of two
circular bases.

or, T.S.A. = C.S.A + 2A = 2πrh + 2πr2 = 2πr(h + r)

Note: Surface is the part of solid which is in contact with the air.

Solid Shapes 61

Prime Mathematics Book - 8

Cone P Curved
o surface
A cone is a solid having regularly curved surface managed Vertical
from a point and a circular base. In the given cone, O is the height A
centre of the circular base, with OA its radius, OP is called Radius
vertical height and AP is called slant height.

A. Experiment: Circular

- Take a circular piece of paper or a filter paper from base
science lab.

- Make a fold and double fold.
- Separate a leaf and with your figure form

a cone.
The form so obtained is a cone.

B. Model of a cone: B
- On a chart paper, draw a circle of radius 7cm.

- Take half the circle (shaded part) and draw a circle O 7cm 3.5cm
P
of radius 3.5cm separately or little attached to the

semi-circle. A

- Cut along outlines. Join OA along OB with cello tape
and gently join the small circle along circumference of the base of the cone so
formed.

Thus, you get a model solid cone. 7cm P

C. Surface area and volume of a cone 3.5 hl
In the given cone, orA
Radius of base (OA) = r
Vertical height (OP) = h
Slant height (AP) = l
As, OP ⊥ OA, in rt ∆OAP
AP2 = OP2 + OA2
or, l2= h2 + r2

Base of the cone is a circle with radius r
∴ Base area (A)= πr2
Considering the above cone as a paper cone, defold the curved served surface to

form a sector of radius = l and length of arc 2πr.
62 1 Geometry

Prime Mathematics Book - 8
hl
l

or 2πr

Pyramids

Egyptians built Pyramids over tombs. The great Pyramid
of El Giza is one of the ancient wonders. Pyramidal solids
are the solids having polygonal base and triangular lateral
faces emanating from a common vertex. Pyramids are
named according to the shape of the base.





Triangular based Square based Hexagonal based
Pyramid Pyramid Pyramid

(Tetrahedron) P

A. Parts of a Pyramid

The given figure is a square based pyramid. ABCD is the D C
O Q
base which is square shaped. PAB, PBC, PCD, PDA are
B
the triangular faces. P is the common vertex, OP is the
vertical height of the pyramid and PQ is the height of A 6cm 8cm
8cm
the triangular face and is called slant height.

B. Net of a square based Pyramid:

- Draw the given net on sketch paper.
- Cut along the outlines (dotted lines)
- Fold along the continuous (bold) lines.

- Put glue on shaded parts and with glued parts inside
bring the vertices of the triangle together and join the
parts properly. You will get a nice paper model of a
Pyramid.

Solid Shapes 63

Prime Mathematics Book - 8 →

C. Net of a triangular based Pyramid
(Tetrahedron):

Similarly draw the net of regular tetrahedron as
shown and make a paper model.





Exercise 6

1. Write the name of the following solid objects.

a) b) c)



d) e) f)

2. Write the name of the base and surfaces of the following solid objects.

a) O b) O c) A1
AB
B B1 C1

A C

FC

CA

ED B

64 1 Geometry

Prime Mathematics Book - 8

3. Identify the objects whose net are given below.

b)
a)



e)
d)

Solid Shapes 65

Applied Mathematics Book - 8

Area Revision Test (Geometry)

1. (a) Find the angle which is two third of its complementary angle. zy
(b) If 5x and 4x are supplementary angles, find them. 4x

(c) From the given figure, find the values of x, y and z. 2x

2. (a) From the given figure, find the value of x. 1800 −x 400 115 0
x

(b) Find the unknown angles in the given figure. zy

3. Verify experimentally that base angles of an isosceles triangle are equal.

AQ B
D
4. In the given figure PQ and PR are bisectors of BQR and P

QRD respectively and AB//CD. Prove that QPR = 90°. C R

5. Prove that the diagonals of a rhombus bisect each other at right angles.

6. Construct a rectangle ABCD in which AB = 6cm and AD = 7cm. 1300 x
1500
7. (a) Find the size of the angle represented by x in the given figure.


A xcm BP (x−3)cm Q

(b) From the given congruent triangles,
(y+1)cm
find the values of x and y. 6cm

600 300

CR
8. Verify experimentally that the diagonals of a rhombus bisect each other at right angels.

A C

9. In the adjoining figure, AB ⊥ AC and AD ⊥ BC.
Prove that BC2 = AB2 + AC2
B D

10. Curved surface area of a cone having radius 4.2cm is 92.4cm2. Find its volume.

11. Construct a regular pentagon of side 4.5cm.

12. Find the total surface area of the given prism.

13. Find the area and the circumference of a circle having
diameter 21 cm.

14. Find the curved surface area, total surface area and volume of a cylinder having
the radius of the base 14cm and height 15cm.

66 1 Geometry

Area Answers

Exercise 1 (Lines angles angles)

1. (a) 58° (b) 22° (c) 45° (d) 15° (e) 90° (f) 0° (g) 40° (h) 80°

2. (a) 140° (b) 110° (c) 90° (d) 80° (e) 60° (f) 45° (g) 20° (h) 8°

3. (a) 30°,60° (b) 30°,150° (c) 20°,70° (d) 74°,106°

4. (a) 50° (b) 36° (c) 30° (d) 126°, 54°

5. (a) 2x = 20°, 3x = 30°, 4x = 40° (b) 108°, 72° (c) 60°, 120°,180°

(d) 2x = 72°, 3x = 108°, y = 108°, z = 72° (e) x = 70°, y = 110°, z = 70° (f) 60°, 30°

(g) 24°, 48°, 72°, 96°, 120° (h) 2x = 40° (i) a = 60°, b = 70°,c = 50° (j) x = 40°, y = 140°, z=40°

6. (a) c = e, d = f (b) a = e, b = f, d = h, c = g (c) c + f = 180°, d + e = 180°

7. (a) corresponding angles (b) co-interior angles (c) alternete angles

(d) corresponding angles (e) alternate angles (f) co-interior angles

8. (a) AB and CD are parallel (Exterior alternate angles are equal)

(b) AB and CD are not parallel (Sum of Co-interior angles is not 180°)

(c) MN and OP are parallel (∠MAB = ∠DBY = 52°, being corresponding angles are equal.

(d) AB and XY are parallel (being alternate angles are equal)

(e) CD and EF are not parallel (being ∠CIJ ≠ ∠EJH)
(f) IJ and KL are not parallel (being ∠MPI ≠ ∠PQK)

9. (a) 3x = 108°, 2x = 72°, y = 72°, z = 108° (b) x = 50°, y = 130° (c) x = 110°, y = 70°, z = 70°

(d) x = 45°, y = 45°, z = 135°, 3x = 135° (e) x = 85°, y = 95° (f) x = 65° (g) x = 65°, y = 115°

(h) x = 35°, y = 65°, z = 100° (i) x = 115° (j) x = 60°, y = 60°

(k) x = 27°, y = 35° (l) x = 70° (m) x = 70°

(n) x = 85°, y = 95°, z = 85° (o) x = 110°, y = 70°, z = 70° (p) x = 40°, y = 110°, z = 30°

Exercise 2.1 (Triangles)

1. (a) x = 60° (b) x = 50° (c) x = 40° (d) x = 20° (e) x = 85° (f) x = 30° (g) x = 65°

(h) x = 71° (i) x = 50°, y = 50° (j) x = 135° (k) x = 120°, y = 60° (l) x = 45°,y =135°,z = 45°

2. (a) x = 65°, y = 30° (b) a = 60°, b = 70°, c = 50°(c) a = 132°, b = 162° (d) a = 50°, b = 50°, c = 42°

(e) x = 108°, y = 42° (f) x = 54°, y = 54° (g) x = 65°, y = 60° (h) a = 60°, b = 60°, c = 60°

3. (a) 40°, 60°, 80° (b) 40°, 50° (c) 40°, 100° (d) 72°, 54°, 54°

Exercise 2.2 (Quadrilateral)

1. (a) a = 115°, b = 65°, c = 115° (b) a = 105°, b = 75°, c = 105°, d = 75°

(c) a = 60°, b = 60°, c = 120°, d = 60° (d) x = 70°, y = 70°, z = 110°

(e) x = 30°, y = 90°, z = 90°, w = 90° (f) x = 18°, y = 98°

(g) a = 5°, b = 60°, c = 70°, d = 60° (h) x = 45°, y = 75°

(i) a = 110°, b = 70°, c = 110°, d = 70°

2. (a) x = 5cm, y = 4cm (b) x = 4cm, y = 5cm (c) x = 5cm, y = 5cm (d) x = 6cm

Exercise 3 (Congruency of Triangles)

1. (a) RHS (b) SAS (c) AAS (d) SSS (e) ASA 2. (a) RHS (b) SAS (c) ASA (d) AAS

3. (a) x = 4cm, y = 2cm (b) x = 8cm, y = 5cm (c) x = 2.2cm (d) x = 4cm (e) x = 2 (f) x = 1.8, y = 2

Answers 67

Area Exercise 4.1 (Similar Triangles)

1. (a) AAA (b) AAA (c) SAS (d) SSS (e) AAA (f) AAA

2. (a) BC DE, AB AD, AC AE (b) BC EF, AB DE, AC DF
BM
(c) PQ TS, QR TR, PR RS (d) BC AB, AC AM, AB

3. (a) 7.5cm, 6cm (b) 5.6cm, 7.2cm (c) 5.6cm, 0.28cm
(d) x = 3cm, y = 20cm

4. (b) 16cm (c) 9.6cm (d) 15m

Exercise 5 (Circle)

1. (a) 220cm (b) 132cm (c) 26.4cm (d) 88cm, 616cm2 (e) 35.2cm, 98.56cm2 (f) 19.8cm, 31.185cm2

2. (a) 21cm (b) 35cm (c) 10.15 (d) 3.5cm (e) 21cm (f) 7cm

3. (a) 42cm2 (b) 168cm2 (c) 10.5cm2 (d) 1.68cm2 (e) 462cm2 (f) 18.48cm2

4. (a) 28.28cm2 (b) 21cm (c) 176cm (d) 52.8m (e) 1100m (f) 4356cm2

Exercise 6 (Solid Shape)

Show to your Teacher.

68 Answers

Area

Estimated periods - 8

At the end of this unit the student will be able to Objectives:

● know Pythagoras theorem.
● use Pythagoras theorem for solving right triangle and get idea of pythagorean triples.
● use Pythagoras theorem to find the distance between two points in co-ordinate plane.

Ladder Y
B

Base A

X' O X

Y'

Teaching Materials:

Graph, chart of pythagorean triples, geoboard, scale, pencil and different colour sign pens.

Activities:

It is better to
● discuss about Pythagoras theorem its historical significance.
● demonstrate the property in right angled triangle.
● discuss about pythagorean triples.
● derive geometrically the distance formula.

Prime Mathematics Book - 8 Coordinates Estimated period : 6

Unit - 1

1.1 Pythagoras Theorem

A. Historical Facts:

The first general proof of the theorem "the square on the hypotenuse of a right triangle

is equal to the sum of the square on the two legs" was given by Pythagoras (580 - 500

B.C., Greek). There has been much conjectures as to the proof Pythagoras might have

offered but it is generally felt that it was a direction type of proof,which follows as:

B

Let a, b denote the legs and c denotes hypotenuse of a c
right angled triangle ABC where c is right angle. a

Cb A

Two squares each with ab ab
sides a + b are taken

as illustrated in a1 3 b B C a
the adjoining figures. 4 c c

5 2 A c b
First square is dissected b a Ec D

into six places with two 6
squares on legs and four

right triangles congruent to ba

the given right triangle. The second square is dissected into five places with square on

hypotenuse and four right triangles congruent to the given right triangle. Subtracting

equals from equals we get square in hypotenuse = sum of square on legs. A

Areas 1+2+3+4+5+6 = Areas A+B+C+D+E c b
As areas denoted by 3, 4, 5, 6, B, C, D, E are equal C

∴ Areas 1+2 = Area A B a b2
a2 + b2 = c2

Introduction:

In a right angled triangle ACB with c = 900, c2 A
AB = c is called hypotenuse and BC = a, AC = b cb
are called legs.
B aC
Pythagoras theorem states that square of a2
hypotenuse of a right angled triangle is equal to
the sum of the squares of legs.
i.e. c2 = a2 + b2.

70 2 Coordinate Geometry

Prime Mathematics Book - 8

It can also be stated that area of square on the hyptenuse of a right angled triangle is
equal to the sum of the areas of squares on the legs.
i.e., c2 = a2 + b2.

Demonstration 1: A
BC
→ Construct a right angled triangle with legs 3 units and
4 units

→ Construct square on the legs and hypotenuse
→ Divide each squares into squares of unit length as in

the figure. We find, hypotenuse = 5 units and counting
the unit square on the legs and hypotenuse. Sum of
No. of units squares Legs = No. of units squares an on
hypotenuse,
9 + 16 = 25 ⇒ 32 + 42 =52 ⇒ a2 + b2 = c2

Experimental verification: c = 900 of

Experiment: Construct three right angled triangles namely ABC with
different measures.

(i) A (ii) B (iii) C A

CB AB
C

To verify : AB2 = BC2 + AC2
Verification : Measure the sides AB, BC and AC in each figure and complete the table.

Fig AB(c) BC(a) AC(b) AB2 (c2) BC2 (a2) AC2 (b2) BC2 + AC2 Result
= a2 + b2

(i) AB2 = BC2 + AC2
(ii)
(iii)

Conclusion : From this experiment we observed that in each right angled triangle
AB2 = BC2 + AC2. Hence, in a right angled triangle ABC with c = 900,
c2 = a2 + b2.

Coordinates 71

Prime Mathematics Book - 8 c a Q c
b
Demonstration 2: A
→ Construct a right angled triangle ABC Pb
with c = 900. b
→ Complete squares on each side of ∆ ABC. a C
→ On the square on c, construct two right
triangles as shown. c
→ Separate (cut) the squares. c

a
B

Q
P

P' Q' b2
a2

Also cut the triangles on c and adjust as shown and bring and fit the two squares
(a2 and b2) on it
∴ a2 + b2 = c2

Pythagorean triples
Observe the following triples of numbers

3, 4, 5 where 32 + 42 = 52 ⇒ 9 +16 = 25

5, 12, 13 where 52 + 122 = 132 ⇒ 25 +144 = 169
7, 24, 25 where 72 + 242 = 252 ⇒ 49 +576 = 625
6, 8, 10 where 62 + 82 = 102 ⇒ 36 +64 = 100 e.t.c.

Such triple numbers in which square of one is equal to the sum of squares of other
two are called Pythagorean triples.

Pythagoras formulated "Pythagorean triple generator" as 2n + 1, 2n(n + 1) and 2n(n + 1) + 1.
Where n is any positive number.

Value of n Pythagorean triples
1
2 2n + 1 2n(n + 1) 2n(n + 1) + 1
3
34 5
5 12 13
7 24 25

Find other sets of Pythagorean triples as many as you like.

72 2 Coordinate Geometry

Prime Mathematics Book - 8

Illustrative Examples

Example 1: Find the unknown sides of the following right angled triangles.

Solution: hypotenuse (c) = AB = 41cm
(i) In ∆ ACB, c = 900 other leg (a) = BC = ?

one of the leg (b) = AC = 9cm (i) A
we have, by pythagoras theorem
c2 = a2 + b2

or, (AB)2 = (BC)2 + (AC)2 41cm

or, (41cm)2 = a2 + (9cm)2 9cm

or, 1681cm2 = a2 + 81cm2

or, 1681cm2 − 81cm2 = a2 C B

or, 1600cm2 = a2

∴ a = 40cm
∴ BC = 40cm

(ii) In the given ∆ ABC, A = 900 length of hypotenuse (a) = BC = ?
length of other leg (b) = AC = 8cm
length of one leg (c) = AB = 6cm

we have, by pythagoras theorem (ii) A 6cm B

a2 = c2 + b2

or, (BC)2 = (AB)2 + (AC)2

or, a2 = (6cm)2 + (8cm)2 8cm

or, a2 = 36cm2 + 64cm2

or, a2 = 100cm2

∴ a = 10cm C

∴ BC = 10cm

Example 2: In the given figure, ABC = ACD = 900, AB = 9cm, BC = 12cm and AD= 25cm.

Find the length of CD. 25cm D
Solution:
In ∆ ABC, ABC = 900 A
9cm
So, AC is hypotenuse.

Now, (AC)2 = (AB)2 + (BC)2 = (9cm)2 + (12cm)2 B 12cm C

= 81cm2 + 144cm2 = 225cm2

∴ AC = 15cm

In ∆ ACD, ACD = 900

Coordinates 73

Prime Mathematics Book - 8

So, AD is hypotenuse

Now, (AD)2 = (AC)2 + (CD)2

or, (25cm)2 = (15cm)2 + (CD)2

or, 625cm2 = 225cm2 + (CD)2

or, 400cm2 = (CD)2

∴ CD = 20cm

Example 3: Check whether the triangle ABC, having lengths of sides AB = 2.1cm,
BC = 7.2cm and CA = 7.5cm is a right angled triangle or not.

Solution:
Here, in ∆ABC, AB = 2.1cm, BC = 7.2cm and CA = 7.5cm. If ∆ ABC is a right angled
triangle, then hypotenuse should be the longest side. CA = 7.5cm is longest side
in ∆ ABC.

So, (CA)2 = (AB)2 + (BC)2

or, (7.5cm)2 = (2.1cm)2 + (7.2cm)2

or, 56.25cm2 = 4.41cm2 + 51.84cm2

or, 56.25cm2 = 56.25cm2, which is true.

Pythagoras theorem h2= p2 + b2 being satisfied.

Hence, ∆ABC is a right angled triangle.

Example 4: Which of the following triplets of number is pythagorean triple?

(a) 7, 9, 10 (b) 8, 15, 17
Solution:
(a) Given triple 7, 9, 10

If the numbers are pythagorean triple, then

Square of the largest number = sum of square of other two numbers

Now, 102 = 72 + 92

or, 100 = 49 + 81

or, 100 = 130 which is false.

∴ 7, 9, 10 is not pythagorean triple.

(b) Given triple 8, 15, 17

If the numbers are pythagorean triple, then

Square of the largest number = sum of squares of other two numbers

So, 172 = 82 + 152

or, 289 = 64 + 225

or, 289 = 289 which is true

∴ 8, 15, 17 is pythagorean triple.

74 2 Coordinate Geometry

Prime Mathematics Book - 8

Exercise 1.1

1. Find the unknown sides of the given triangles.

(a) A (b) A 3cm B (c) (d) P
4cm x L 181cm 19cm

10cm x 5 2cm
x

B 6cm CC M NR xQ
5cm
A

2. (a) Find the height AD of the 5 25cm h 25cm
given isosceles triangle; 2cm
17cm
B D C A
14cm

(b) Find the length of base BC of the B 8cm
Q D
given isosceles triangle, ?
where AD is height of ∆ ABC.

C

P

(c) If the length of a diagonal of the S R 12cm D
given square is 5 2cm, A
find the length of its side.
6cm x

C
(d) Find the value of x in the
given adjoining figure.

B 8cm
B
(e) If the given figure, ABCD A
is a rectangle where 35cm 12cm
BC = 12cm, CD = 35cm. C
Find the length of the 2.9cm x
diagonal of the rectangle. D 2.1cm
9.5cm
3. (a) Given figure is a kite.
Find the values of x and y. y

Coordinates 75

Prime Mathematics Book - 8 ? 5.2cm
?
(b) A pole 5.2m tall is supported
with a wire from a point 3.9m ?
away from the foot of the pole
to its top. Find the length of the
supporting wire.

(c) From a point 3.6m away from the 3.9cm
foot of the wall, a man reaches
to the top of the wall 7.7m high 7.7m
by using a ladder which is shown
in the adjoining figure. Find the 3.6m
length of the ladder.

12m

(d) A 12m tall tree is broken by the wind but not completely 6m
detached and the upper part touches the ground 6m away
from the foot of the tree. Find the length of the broken
parts of the tree.

4. Check whether the following length are the sides of a right angled triangle or not.
(a) 8cm, 10cm, 12cm (b) 6cm, 8cm, 10cm
(c) 41cm, 40cm, 9cm (d) 13cm, 84cm, 85cm

5. Which of the following are the pythagorean triples? (c) 6, 8, 10
(a) 2, 3, 4 (b) 3, 4, 5 (f) 21, 220, 221
(d) 9, 12, 18 (e) 17, 15, 8

6. Generate any five pythagorean triples using pythagoras formula.

1.2 Distance between two points in co-ordinate plane:

Lets consider two points A (x1, y1) and B (x2, y2) in the co-ordinate plane. Draw AL and
BM perpendiculars to OX and AN ⊥ BM. We get
Y
OL = x1, AL = y1 , OM =x2 , BM = y2

∴ AN = LM = OM - OL = x2 - x1 B(x2, y2)

BN = BM - NM = BM - AL = y2 - y1 A(x1, y1)

Now, in right angled triangle ANB, by using N
pythagoras theorem

(AB)2 = (AN)2 + (BN)2 X’ O L MX
or, (AB)2 = (x2 - x1)2 + (y2 - y1)2

∴ AB = (x2 - x1)2 + (y2 - y1)2 Y’

∴ When two points (x1 , y1) and (x2 , y2) on co-ordinate plane are known then
distance between them (d) = (x2 - x1)2 + (y2 - y1)2

76 2 Coordinate Geometry

Prime Mathematics Book - 8

Illustrative Examples

Example 1: Find the distance between the points;

(a) A (3, 2) and B (6, -2) (b) P (-2, 5) and Q (3, 1)
Solution:
(a) Here, given points A (3, 2) = (x1, y1) and B (6, -2) = (x2, y2)
Now, we have

d = (x2 - x1)2 + (y2 - y1)2

AB = (6-3)2 + (-2-2)2 = (3)2 + (-4)2 = 9 + 16 = 25
∴ AB = 5 units.

b) Here, the given points P (-2, 5) = (x1, y1) and Q (3, 1) = (x2, y2)
Now, we have

d = (x2 - x1)2 + (y2 - y1)2
PQ = {3-(-2)}2 + (1-5)2 = (3+2)2 + (-4)2 = (5)2 + 16 = 25 + 16 = 41

∴ PQ = 41 units.

Example 2: If distance between the points (4, -3) and (-2, a) is 10 units, find the value of a.

Solution : Here, the given points (4,-3) = A (x1, y1) and (-2, a) = B (x2, y2)
And distance between them (AB) = 10 units

Now, we have by distance formula (d) = (x2 - x1)2 + (y2 - y2)2

AB = (-2 - 4)2 + {a - (-3)}2

or, 10 = (-6)2 + (a + 3)2

or, 10 = 36 + (a + 3)2

Squaring both sides, we have

or, (10)2 = ( 36 + (a + 3)2)2

or, 100 = 36 + (a + 3)2

or, 100-36 = (a + 3)2

or, 64 = (a + 3)2

or, (±8)2 = (a + 3)2

∴ a+3 = ±8

Taking positive sign Taking negative sign
or, a + 3 = -8
or, a + 3 = 8 or, a = -8 - 3
∴ a = -11
or, a = 8 - 3

∴a=5

∴ a = 5 or -11

Coordinates 77

Prime Mathematics Book - 8

Example 3: If a point A(9, 12) lies on the circumference of a circle with centre
O(1, 6), what will be the radius of the circle? Find it.

Solution: Here, O(1, 6) is the centre of the circle. Point r A (9,12)
A(9, 12) lies on the circumference of the circle. OA O (1, 6)
is the distance between two points O and A which
is the radius of the circle. Therefore, the radius of
the circle (r) = OA

Now, d = (x2 - x1)2 + (y2 - y1)2

= (9 - 1)2 + (12- 6)2 = (8)2 + (6)2 = 64 + 36 = 100 = 10 units

∴ The radius of the circle is 10 units.

Example 4: Show that the points (-4, 0), (-4, -4) and (2, -4) are the vertices of a
right angled triangle.

Solution: Let ABC be a triangle in which three vertices are A(-4, 0), B(-4,-4) and C(2,-4).

The length of AB from A(-4, 0) to B(-4, -4) is
AB = (-4 + 4)2 + (-4 - 0)2 = (0)2 + (-4)2 = 16 = 4 units.
The length of BC from B(-4, -4) to C(2, -4) is
BC = (2 + 4)2 + (-4 + 4)2 = (6)2 + (0)2 = 36 = 6 units.
The length of AC from A(-4, 0) to C(2, -4) is

AC = (2 + 4)2 + (-4 - 0)2 = (6)2 + (-4)2 = 36 + 16 = 52 units.

Here, 52 > 6 > 4 . So, AC > BC > AB

If the triangle is a right angled triangle, then square of the longest side = Sum of
squares of others two;

Now, AC2 = BC2 + AB2

or, ( 52 )2 = (6)2 + (4)2

or, 52 = 36 + 16

or, 52 = 52, which is true.

So, the triangle ABC satisfy pythagoras theorem.

Hence, the given three points are the verities of the right angled triangle.

78 2 Coordinate Geometry

Prime Mathematics Book - 8

Example 5: Show that points (5, 1), (3,2) and (1, 3) lie on the same line.

Solution: Here, the given three points are A(5, 1), B(3, 2) and C(1, 3).
The distance of AB from A(5, 1) to B(3, 2) is
AB = (3 - 5)2 + (2 - 1)2 = (-2)2 + (1)2 = 4 + 1 = 5 units
The distance of AC from A(5, 1) to C(1, 3) is
AC = (5 - 1)2 + (1 - 3)2 = (4)2 + (-2)2 = 16 + 4 = 20 = 2 5 units
The distance of BC from B(3, 2) to C(1, 3) is
BC = (1-3)2 + (3-1)2 = (-2)2 + (1)2 = 4 + 1 = 5 units
Now, 2 5 = 5 + 5
or, AC = AB + BC

So, the points A, B and C do not satisfy the property of triangle, AB + BC > AC.
Hence, the given three points lie on the same line.

Exercise 1.2

1. Find the distance between the following pair of points.

(a) (3,5) and (-5,-1) (b) (7,2) and (3,2) (c) (-4,2) and (4,-1)
(f) (-5,0) and (5,0)
(d) (-1,5) and (-1,-2) (e) (3,-2) and (0,-1)

2. Plot the points on graph and find the distance between them using graph.

(a) (2,7) and (-2,7) (b) (3,4) and (3,-4)

(c) (-1,3) and (-1,-4) (d) (0,-5) and (0,5)

3. If distance between the points (1, 4) and (-3, p) is 5 units,find the value of P.

4. If the distance between the points A(a, 2) and B(-5, 2) is 10 units, find the co-ordinates of A.

5. If a point B(10, 8) lies on the circumference of circle with centre O(4, 6), what will
be the radius of the circle ? Find it.

6. Show that the points A(7, 8), B(11, 4) and C(3, 4) are the vertices of an isosceles triangle.

7. Show that the points M(1, 6), N(4, 1) and P(-4, 3) are the vertices of an equilateral triangle.

8. Show that the points P(5, 3), Q(-2, 3), R(-2, -4) and S(5, -4) are the vertices of a rectangle.

9. If the points M(2, -1), N(3, 4), P(-2, 3) and Q(-3, -2) are the vertices of a rhombus
MNPQ, find the length of the diagonals of the rhombus MNPQ.

10. Show that the following points lie on the same line.

(a) (3,-1), (1,1) and (-2,4) (b) (1,7), (2,1) and (3,-5)

(c) (1,-2), (5,-6) and (-3, 2) (d) (2,1), (3, 2) and (4,3)

Coordinates 79

Applied Mathematics Book - 8

Area Revision Test (Co-ordinate Geometry)

1. (a) Find the unknown side of the right angled triangle ABC. 12cm A B
(b) Check whether 41, 40, 9 are a Pythagoras triple or not? 8cmC 17cm

(c) Find the distance between the points (3,4) and (3, -3).

(d) Write the formula to find the distance between the points A (x1, y1) and B(x2, y2)

2. (a) show that the points (3, -1), (1, 1) and (-2, 4) lie on a straight line.

(b) If the distance between the points P(a, 2) and Q(-5, 2) is 10 units, find the
value of a.
A

(c) Show that the points A(4, 1), B(-4, 3) and c(1,6)
are the vertices of an equilateral triangle.

(d) Find the height of the given isosceles
triangle.

B D C
30cm
3. If A(4, 6) is a centre of a circle and a point p(10, 8)
is a point on the circumference of the circle, find

the radius of the circle.

4. Show that the points (4, 3), (2, 1) and (3, 2) lie on a straight line by using

distance formula.

5. If A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus ABCD,

find the length of its two diagonals AC and BD.

Answers

Exercise 1.1 (Pythagoras theorem)

1. (a) x = 8cm (b) x = 5cm (c) x = 5cm (d) 180cm

2. (a) h = 24cm (b) BC = 30cm (c) 5cm (d) 6.63 cm
(e) 37 cm

3. (a) x = 2cm, y = 9.7cm (b) 6.5m (c) 8.5m (d) 7.5m, 4.5m

4. (a) No (b) Yes (c) Yes (d) Yes 5. (a) No (b) Yes (c) Yes (d) No (e) Yes (f) Yes 6. Show to your teacher.

Exercise 1.2 (Distance between two points)

1. (a) 10 units (b) 4 units (c) 73 units (d) 7 units (e) 10 units (f) 10 units

2. (a) 4 units (b) 8 units (c) 7 units (d) 10 units

3. P = 7 4. (5, 2) or (−15, 2) 5. 2 10 units 9. 4 2 units, 6 2 units

80 2 Coordinate Geometry

Area

Estimated periods - 8

At the end of this chapter the student will be able to Objectives:

● find the area of triangles.
● find the area of special quadrilaterals (square, rectangles, parallelogram, rhombus and

trapezium).
● find the area of quadrilaterals when a diagonal and perpendiculars drawn from opposite

vertices to it are given
● find the value of cubes and cuboids.

B C h
N D
b
M l
A

Teaching Materials:

Paper models of triangles, quadrilaterals and polygons, models of solids, geometry box and
geoboard.

Activities:

It is better to
● demonstrate paper models to find area of triangle.
● explain and make the students familiar with different triangles and quadrilaterals.
● show the paper model of cubes and cuboids.

Prime Mathematics Book - 8

Unit - 1 Perimeter, Area and Volume Estimated periods : 8

Introduction:

- Area of a plane figure is the surface enclosed by its boundary.

- Area is measured in square units like square centimetres(cm2), square metre(m2)
e.t.c.

- Mensuration deals with measurement of lengths, perimeter, area and
volume of plane and solid figures.

1.1 Area of rectangles, triangles and quadrilaterals :

A rectangle is a quadrilateral having opposite sides equal and all the angles 900.
Area of the rectangle = length × breadth = l × b

Area of a square:

A square is a rectangle having all sides equal.

i.e. if l = b = a then a
a
Area of the square = l × b = a × a = a2

If diagonal of a square be d, then

d2 = a2 + a2 = 2a2 ∴ a2 = d2
2
d2
∴ Area of the square = 2 [in terms of diagonal]

Area of triangle: A

Take a chart paper in triangular shape

ABC with base BC = b and XY

height (altitude) AP = h

Fold along XY, XM, YN bringing A, B and C on P. B MP N C
We get a rectangle XMNY where

MN = 1 BC = 1 b and XM = 1 AP = 1 h
2 2 2 2
Area of right angled triangle
Area of ∆ ABC = 2 area of rectangle MNYX. A
1
= 2.12 1 1 = 2 b.h
2 2
= 2 MN . XM b. h = .b.h 1
2
∴ Area of ∆ = 1 base × altitude = 1 BC . AB
2 = 2
BC

b.p [Where b = base, p = perpendicular]

82 3 Mensuration

Prime Mathematics Book - 8

Area of an equilateral triangle: A

Consider an equilateral triangle ABC with side AB = BC = AC = a.

Drawing AD ⊥ BC, we get

BD = a and AD = AB2 − BD2
2

( ( = a2 − a 2 = a2 − a2 BDC
2 4

= 4a2 − a2 = 3a2 = a3
4 4 2

Now,

Area of equilateral ∆ ABC = 1 BC. AD = 1 a. a3 = 3 a2
2 2 2 4

3
∴ Area of equilateral ∆ = 4 (side)2

Area of an isoscales triangle A

Consider an isosceles triangle ABC with side AB = AC = b and
BC = a. Draw AD perpendicular to BC, we get

BD = a and AD = AB2 − BD2
2

= ( (b2 −a2 BDC
2

= b2 − a2
4

= 4b2 − a2
4

= 4b2 − a2
2
1
Now, area of isosceles ∆ABC = 2 BC.AD

= 1 a. 4b2 − a2
2 2

= a 4b2 − a2
4

a
4
∴ Area of isosceles ∆ = 4b2 − a2

Perimetre, Area and Volume 83

Prime Mathematics Book - 8

Area of a parallelogram: A B
XC Y
Consider a parallelogram ABCD with base AB = CD = b
A B
and height AX = h.

Produce DC and draw BY ⊥ DC produced. D
We get ∆ AXD = ∆ BYC

∴ Area of ∆ AXD = Area of ∆ BYC

∴ Area of parallelogram ABCD = Area of rectangle ABYX

= XY × AX = AB . AX = b . h

∴ Area of parallelogram = base × height

Area of a rhombus:

(i) A rhombus being a parallelogram, C
if its base (side) is b and height is h, then D B
∴ area of rhombus = b × h
= base × height O
C
(ii) Diagonals of a rhombus bisect each other at right angles. A

If the length of two diagonals AC = d1 and BD = d2 then

∴ Area of rhombus ABCD = Area of ∆ ABD + Area of ∆ BCD

= 1 BD. AO + 1 BD. OC = A21CB] D=.(12AOd2+. OC)
= 12 BD. AC 2 OC = d1
2
[∴ AO + D

∴ Area of rhombus ABCD = 1 d1.d2
2

Area of a trapezium:

A trapezium is a quadrilateral having any two opposite sides parallel. Let ABCD be

a trapezium with AB // DC, AB = b1, DC = b2 and let distance between two parallel

sides i.e. height of trapezium BM = h then,

∴ Area of trapezium = Area of ∆ ABD + Area of ∆ BCD A b1 B

= 1 AB × BM + 1 DC × BM h
2 2 M
b2 C
= 1 b1 . h + 1 b2 . h D
2 2

= 1 h.(b1 + b2)
2

∴ Area of trapezium = 1 distance between parallel sides × sum of parallel sides
2

84 3 Mensuration

Prime Mathematics Book - 8

Area of a quadrilateral: B C
Here we will discuss about the area of quadrilateral when N D
one of its diagonal and lengths of perpendiculars drawn
from other vertices to the diagonal are given. M
A
ABCD is a quadrilateral where diagonal

AC = d, BM ⊥ AC and D N ⊥ AC. With BM = h1 and DN = h2

Then, Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ADC

= 1 AC × BM + 1 AC × DN
2 2

= 1 AC.(BM + DN) = 1 d. (h1 + h2)
2 2

1
∴ Area of quadrilateral = 2 d. (h1 + h2)

Illustrative Examples

Example 1: Find the area of the following figures.

(b)
(a)

5.2cm

9.4cm 4.8cm

(d)


(c)

4cm

Solution: 7.2cm 6cm

(a) In the given rectangle, length (l) = 9.4cm, breadth (b) = 5.2cm

∴ Area of rectangle (A) = l × b = 9.4cm × 5.2cm = 48.88cm2

∴ Area of rectangle is 48.88cm2

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Prime Mathematics Book - 8

(b) In the given square, side (a) = 4.8cm

∴ Area of the square (A) = a2 = (4.8cm)2 = 23.04cm2

∴ Area of square is 23.04cm2

(c) In the given triangle, length of base (b) = 7.2cm, altitude (h) = 4cm
1 1
∴ Area of ∆ = 2 .b.h = 2 × 7.2cm × 4 cm = 14.4cm2

∴ Area of the triangle is 14.4cm2

(d) In the given equilateral triangle, length of side (a) = 6cm

∴ Area of equilateral triangle = 3 a2 = 3 (6cm)2
4 4

= 3 × 36cm2 = 9 3cm2
4

∴ Area of the equilateral triangle is 9 3cm2

Example 2: Find the area of the given figures: (b)
(a)

17cm 12cm

16cm B 16.5cm
(d) B
(c) A

A N C
M
AC = 12cm
DC D BM = 4cm
AC = 7cm, BD = 10.4cm DN = 6cm

Solution:

(a) In the given right angled triangle,

base (b) = 16cm

perpendicular (p) = 17cm

∴ Area of right angled triangle = 1 .p.b = 1 x 17cm x 16cm = 136cm2
2 2

∴ Area of the given right angled triangle is 136cm2
(b) In the given parallelogram
length of base (b) = 16.5cm
height (h) = 12cm
we know
∴ Area of parallelogram = b × h = 16.5cm × 12cm = 198cm2
∴ The area of the given parallelogram is 198 cm2

86 3 Mensuration

Prime Mathematics Book - 8

(c) In the given rhombus

diagonal (AC) = d1 = 7cm
diagonal (BD) = d2 = 10.4cm

Now, we know

Area of rhombus = 1 .d1.d2 = 21= 3×67.4cmcm×2 5.2 cm
2
10.4

∴ Area of the given rhombus is 36.4cm2.

(d) In the given quadrilateral diagonal (AC) = 12cm, BM ⊥ AC and BM = 4cm and
DN ⊥ AC and DN = 6cm

Now, we know 1 1
2 2
∴ Area of quadrilateral = AC(BM + DN) = × 12cm × (4cm + 6cm)

= 6cm × 10cm = 60cm2

Area of the given quadrilateral is 60cm2.

Example 3: Find the area of the given figures:

(a) B (b) C 6cm D
3cm
A
CB

D 4cm

AC = 25cm, BD = 16cm

A

Solution:

(a) Given figure is a kite where length of diagonals, AC = 25cm and BD= 16cm

we know 1 1
2 2
Area of kite = × d2 × d1 = × AC × BD

8

= 1 × 25cm × 16cm
2

= 200cm2

∴ Area of the given kite is 200cm2

(b) In the given figure, the quadrilateral consists two right angled triangles.

In right angled ∆ ABC

AC = AB2 +BC2 [by using pythagoras theorem]

= (4cm)2 + (3cm)2 = 16cm2 + 9cm2 = 25cm2 = 5cm.

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Prime Mathematics Book - 8

Area of right angled ∆ ABC = 1 × AB × BC
2 × 2
1
= 2 × 4cm 3cm = 6cm2

And area of right angled ∆ ACD = 1 × AC × CD = 1 × 5cm × 3 = 15cm2
2 2 6cm

Now, area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD

= 6cm2 + 15cm2 = 21cm2

Example 4: Find the area of the shaded part shownA in the figure given below. B

Solution: E

In the given figure, 7.2cm
base of the parallelogram ABCD

and triangle CDE is CD = b = 14cm

and height AF = h = 7.2cm F D C

Area of Parallelogram ABCD = b × h 14cm

= 14cm × 7.2cm

= 100.8cm2 7
And area of ∆ CDE = 14cm
1 × b × h = 1 × × 7.2cm
2 2

= 50.4cm2

Now, area of shaded part = Area of Parallelogram ABCD - Area of ∆ CDE

= 100.8cm2 - 50.4cm2

= 50.4cm2

Example 5: A rectangular garden is twice as long as it is broad. If its perimeter is 120m,
find the area of the garden.

Solution:

Let the breadth of the garden is x then length of the garden is 2x

As, perimeter of the garden is 120m

or, 2(l + b) = 120m

or, 2(2x + x ) = 120m

or, 2 × 3x = 120m

or, 6x = 120m

∴ x = 20m

breadth (b) = x = 20m, length (l) = 2 x = 2 × 20m = 40m

Now, area of the rectangular garden = l × b = 40m × 20m = 800m2

∴ Area of the garden is 800m2.

88 3 Mensuration

Exercise 1.1 Prime Mathematics Book - 8

1. Find the area of the following figures. (c)
(a) (b)

20cm 24cm 16cm 8.4cm
18cm(d)

9.2cm
12cm
(e) C (f)

A B 32cm
12.4cm

16cm

2. Find the area of the following figures.
(a) (b)

12cm15.2cm 42cm
(c) 8cm
27cm(d)(e) A B
16.5cm A DC

BD AC=28cm, BD=16cm

C
AC = 40.5cm , BD = 60.2cm

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Prime Mathematics Book - 8 (g) 6cm (h) 20cm

(f) B C 17cm

A N 9cm
M

D 10cm 35cm

AC = 35cm, BM = 10.5cm,
DN = 8.2cm

3. Calculate the area of the following.

(a) A 20cm B (b) A

16cm 12cm B
D
41 cm 15cm
(d)
D 40 cm C C
8cm

22 cm
(c)

12 cm

8 cm 18 cm

4. Find the area of the shaded part of the following figures.

(a) A E B (b) 18cm

12cm 8cm 16cm

D C 9cm
90 3 Mensuration
15cm

Prime Mathematics Book - 8

(c) (d)

18.5 cm 12 cm

25.4 cm 9 cm
(e) 15 cm

(f)

4 cm 4 cm

4 cm 4 cm

4 cm 15 cm

4 cm 4 cm 4 cm

5. (a) Perimeter of a square is 40cm. Find its area.

(b) Length of a rectangular plot of land is thrice its breadth. If the perimeter of the plot
is 100m, find its area.

(c) What is the cost of plastering the floor of a square room of side 6.5m at the rate of
Rs. 200 per square metre ?

(d) Perimeter of a rectangular ground 32m long is 112m. Find the cost of turfing it at
Rs.25 per square metre.

6. From the adjoining figures find the value of x. c) P S

a) Area = 42cm2 b) AreAa = 31.36cm2 D
A

6cm. x

Q R
PR = 88cm
B Dx C BE C Area = 44cm2
11.2cm QS = x

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Prime Mathematics Book - 8

1.2 Area and Volume of Cube and Cuboid

For a cuboid:

Length = l, breadth = b, height = h

Diagonal (line joining opposite corners) h

d= l2 + b2 + h2

Total surface area (T.S.A)

= 2lb + 2bh + 2hl

= 2(lb + bh + hl) b

Volume(v) = l×b×h l

For a cube:

Length = breadth = height = l l
Diagonal(d) = 3 l
Total surface area (T.S.A)

= 6l2 l
Volume(v) = l3 l

Example 1: Find the length of the diagonal, total surface area and volume of a cuboid
12cm long, 8cm broad and 6cm height.

Solution:
Here, in the cuboid length(l)= 12cm, breadth(b) = 8cm and height(h) = 6cm

Now, we have = l2 + b2 + h2
Length of diagonal(d) = (12cm)2 + (8cm)2 + (6cm)2


= (144 + 64 + 36)cm2

= 244cm2

= 15.62cm

Total surface area (T.S.A) = 2(lb + bh + hl)

= 2 (12cm × 8cm + 8cm × 6cm + 6cm × 12cm)

= 2 (96cm2 + 48cm2 + 72cm2)

= 2 × 216cm2

= 432cm2

∴ Total surface area (T.S.A) of the cuboid is 432cm2

And, Volume of the cuboid(v) = l × b × h = 12cm × 8cm × 6cm = 576cm3

∴ Volume of the cuboid is 576cm3

92 3 Mensuration

Prime Mathematics Book - 8

Example 2: Total surface area of a cuboid is 1032cm2. If length and breadth of the
cuboid are 18cm and 12cm respectively, find the volume of the cuboid.

Solution: Here, length of cuboid(l) = 18cm, breadth(b) = 12cm, height(h) = ?
T.S.A. = 1032cm2, volume(v) = ?
we have
T.S.A. of cuboid = 2(lb + bh + hl)
or, 1032cm2 = 2 (18cm × 12cm + 12cm × h + h × 18cm)

or, 1032cm2 = 216cm2 + 30cm × h
2

or, 516cm2 − 216cm2 = 30cm × h

or, 300cm2 = 30cm × h

10
300cm2
or, 30cm =h

∴ h = 10cm
Now, Volume of the cuboid(v) = l × b × h = 18cm × 12cm × 10cm = 2160cm3

∴ The volume of the cuboid is 2160cm3

Example 3: Find the total surface area and volume of the cube of side 12cm.

Solution: Here, length of side of cube(l) = 12cm
we have,

Total surface area (T.S.A) = 6l2 = 6 × (12cm)2

= 6 × 144cm2 = 864cm2

∴ Total surface are of the cube is 864cm2

And volume of the cube (v) = l3 = (12cm)3 = 1728cm3

∴ Volume of the cube is 1728cm3. 12cm

Example 4: If the volume of a cube is 3375cm3, find its total surface area.

Solution: Here, volume of cube(v) = 3375cm3. If side of the cube is (l), then

or, l3 = 3375cm3

or, l3 =33 × 53 3 3375
3 1125
∴ l = 3 × 5 = 15cm 3 375
5 125
Now, 5 25

T.S.A of the cube = 6l2 5 \ 3375 = 33 × 53

= 6 × (15cm)2 = 6 × 225cm2 = 1350cm2 = 153

∴ Total surface are of the cube is 1350cm2

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Prime Mathematics Book - 8

Example 5: Find the capacity of a cistern 4m long 3m broad and 2m high in litres.

Solution:
Here, length of cistern (l) = 4m

breadth (b) = 3m

height (h) = 2m

Capacity of the cistern = l × b × h

= 4m × 3m × 2m

= 24m3

= 24 × 1000l

= 24000l

∴ Capacity of the cistern is 24000 litres.



Note:
( (→ Capacity of a vessel is the volume of liquid it can hold.
1 litre = 1000cm3 = 1000 × 1 m3 = 1000 m3 = 1 m3
→ 100 1000000 1000

→ 1m3 = 1000 litre



Example 6: Capacity of a tank is 750 litre. If its length is thrice its height and its

breadth is twice its height, find the total surface area of the tank.

Solution: Here,

Length (l) = thrice its height

= 3h

Breadth (b) = twice its height

= 2h

As capacity = 750litres

or, l × b × h = 750 × 1000cm3

or, 3h × 2h × h = 750 × 1000cm3

or, h3 = 125000cm3

or, h = 50cm = 0.5m

∴ l = 3h = 3 × 0.5m = 1.5m b = 2h = 2 × 0.5m = 1m

Now,

T.S.A. of the tank = 2(lb + bh + hl)

= 2 (1.5m × 1m + 1m × 0.5m + 0.5m × 1.5m)

= 2 (1.5m2 + 0.5m2 + 0.75m2)

= 2 × 2.75m2

= 5.50m2

94 3 Mensuration

Prime Mathematics Book - 8

Exercise 1.2

1. Find the volume of the following solids.

(a) 8cm 4cm (b)
6cm 12cm
8cm

2. Calculate the surface area of the given solids.

(a) (b)

12cm 10cm 8cm

3. Find the total surface area and volume of the cuboid having length, breadth and
height respectively.

(a) 15cm, 12cm, 8cm (b) 1m 20cm, 80cm, 60cm (c) 2.4m, 1.2m, 90cm

4. Calculate the surface area and volume of the cube of side.
(a) 10.4 cm (b) 6 in. (c) 6.8 cm (d) 3 m 40 cm (e) 8ft.

5. (a) Volume of a cubical box is 729cm3, find its total surface area.
(b) If the total surface area of a cube is 4056cm2, find its side.
(c) Volume of a cube is 74088cm3. Calculate the surface area of the cube.
(d) Surface area of a cuboid having length 22cm and breadth 21cm is 2644cm2,
find its volume.

6. (a) Height of a cuboidal box is one third of length and half of breadth. If the volume
of the box is 384cm3, find its surface area.

(b) Find the capacity of a cuboidal tank 2.4m × 1.8m × 1.5m in litres.
(c) A cuboid is thrice as long as it is high. If the volume and breadth of the cuboid are

1080cm3 and 10cm respectively, find the length and height of the cuboid.
(d) A drum has a square base. When it contains 72litres of liquid, height of the liquid

is 20cm. Find the length of base of the drum.

Perimetre, Area and Volume 95