6.
`
Mensuration
Introduction :
Mensuration is a special branch of mathematics that deals with the
measurement of geometric figures.
In previous classes we have studied certain concepts related to areas of
plane figures (shapes) such as triangles, quadrilaterals, polygons and circles.
Now we will study how to find some measurements related to circle and
the surface area and the volume of solid figures.
`

Circle : Arc, Sector, Segment :
Area of sector :
Sector of a circle is the part
Y
of the circle enclosed by two
radii of the circle and their
intercepted arc. (i.e. arc between
the two ends of radii)

× r 2
Area of the sector (A) =
360

O
Major arc
r
B

Central angle
Minor arc
A
X
Length of an arc :
Length of an arc of a circle (arc length) is the distance along the curved
line making up the arc.

× 2r
Length of the arc (l) =
360

Relation between the area of the sector and the length of an arc :
Area of the sector =
r
× length of arc
2
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 157)
1.
(i)
Sol.
The diameter of a circle is 10 cm. Find the length of the arc, when the
corresponding central angle is as given below : ( = 3.14)
(2 marks)
144º
Diameter of a circle = 10 cm
10
2
= 5 cm

Its radius (r) =

l =

× 2r
360

l =
144
× 2 × 3.14 × 5
360

l =

144
× 3.14 × 10
360
l = 12.56 cm
 The length of the arc is 12.56 cm.
286
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(ii)
Sol.
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(2 marks)
45º
Diameter of a circle = 10 cm
10

Its radius (r) =
2
= 5 cm
l =

× 2r
360

l =
45
× 2 × 3.14 × 5
360

l =
45
× 3.14 × 10
360

l =
5
314
×
4
100

l =
785
2  100

l =


392.5
100
l = 3.925
l = 3.93 cm
 The length of the arc is 3.93 cm.
(iii)
Sol.
270º
Diameter of a circle = 10 cm

10
2
= 5 cm
It radius (r) =
l =

× 2r
360

l =
270
× 2 × 3.14 × 5
360

l =

(2 marks)
3
× 3.14 × 10
4
l = 23.55 cm
 The length of the arc is 23.55 cm.
(iv)
Sol.
180º
Diameter of a circle = 10 cm

Its radius (r) =
=
l =
=
=
(2 marks)
10
2
5 cm

× 2r
360
180
× 2 × 3.14 × 5
360
15.70 cm
 The length of the arc is 15.70 cm.
S C H O O L S E C TI O N
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EXERCISE - 6.1 (TEXT BOOK PAGE NO. 157)
2.
(i)
Sol.
Find the angle subtended at the centre of a circle by an arc, given the
following :
22
)
(2 marks)
radius of circle = 5.5 m, length of arc = 6.05 m ( =
7

l =
× 2r
360
22


6.05 =
×2×
× 5.5
7
360
605
22


=
×
× 11
100
7
360
605 × 360 × 7
 100 × 22 × 11 = 

 = 63º
 Measure of the arc is 63º.
(ii)
Sol.
(2 marks)
radius of circle = 20 m length of arc = 78.50 m ( = 3.14)

l =
× 2r
360


78.50 =
× 2 × 3.14 × 20
360
785
314


=
×
× 40
10
360 100
785 × 9 × 100

= 
10 × 314

 = 225º
 Measure of an arc is 225º.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
3.
(i)
Sol.
(ii)
Sol.
The radius of the circle is 7 cm and m (arc RYS) = 60º, with the help of
the figure, answer the following questions :
(3 marks)
Name the shaded portion.
S
P-RYS
Find the area of the circle.
Area of circle = r 2
22
=
×7×7
7
= 154 cm2
X
P
60º
Y
R
 Area of a circle is 154 cm2.
(iii)
Sol.
288
Find A (P-RYS)

× r2
360
60
22
=
×
×7×7
360
7
77
=
3
= 25.67
Area of the sector P -RYS is 25.67 cm2
Area of the sector =
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(iv)
Sol.
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Find A (P-RXS)
Area of sector P-RXY = Area of circle – Area of sector P-RYS
= 154 – 25.67
= 128.33 cm2
 Area of sector P-RXY is 128.33 cm2.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
4.
(i)
Sol.
The radius of a circle is 7 cm. Find area of the sector of this circle if the
angle of the sector is :
30º
(2 marks)

Area of the sector =
× r2
360
30
22
=
×
×7×7
360
7
77
=
6
= 12.83
 Area of the sector is 12.83 cm2.
(ii)
Sol.
(2 marks)
210º
Area of the sector =
=
=
=

× r2
360
210
22
×
×7×7
360
7
539
6
89.83
 Area of the sector is 89.83 cm2.
(iii)
Sol.
(2 marks)
3 rt. angles
Area of the sector =
=
=
=

× r2
360
270
22
×
×7×7
360
7
231
2
115.50
 Area of the sector is 115.50 cm2.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
5.
Sol.
An arc of a circle having measure 36 has length 176 m. Find the
(2 marks)
circumference of the circle.
Length of arc (l) = 176 m

measure of arc () = 36º

l =
× 2r
360
36

176 =
× 2r
360
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
176


176 × 10
2r
But, circumference
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1
× 2r
10
= 2r
= 1760
= 2r
=
 Circumference of the circle is 1760 m.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
6.
Sol.
An arc of length 4 cm subtends an angle of measure 40º at the centre.
Find the radius and the area of the sector formed by this arc.(2 marks)
Length of arc (l) = 4 cm

measure of arc () = 40º

l =
× 2r
360
40

4 =
×2××r
360
4×9

= r
2

r = 18 cm.
l ×r
Area of the sector =
2
4 × 18
=
2
= 36cm 2
 Radius of the circle is 18 cm and Area of the sector is 36 cm2.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
7.
Sol.
If the area of the minor sector is 392.5 sq. cm and the corresponding
central angle is 72º, find the radius. ( = 3.14)
(2 marks)
Measure of arc () = 72º
Area of the sector = 392.5 cm2
Area of the sector
392.5




3925
10
3925 × 360 × 100
10 × 72 × 314
r2
r

× r2
360
72
=
× 3.14 × r2
360
72
314
=
×
× r2
360
100
=
= r2
= 625
= 25
[Taking square roots]
 Radius of the circle is 25 cm.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
8.
Sol.
290
Find the area of sector whose arc length and radius are 10 cm and 5 cm
respectively.
(2 marks)
Length of arc (l) = 10 cm
Radius of a circle (r) = 5 cm
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Area of the sector
=
r
×l
2
5
× 10
2
= 25 cm2
=
 Area of the sector is 25 cm2.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
9.
Sol.
If the area of minor sector of a circle with radius 11.2 cm is 49.28cm2,
find the measure of the arc.
(2 marks)
Radius of a circle = 11.2 cm
Area of the sector = 49.28 cm2
Area of the sector




49.28
4928
100
4928 × 360 × 7 × 10 × 10
100 × 22 × 112 × 112


× r2
360
22

=
×
× 11.2 × 11.2
7
360
22
112
112

=
×
×
×
7
10
10
360
=
= 
= 45º
 Measure of arc is 45º.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
10.
Sol.
Find the length of the arc of a circle with radius 0.7 m and area of the
sector is 0.49 m2.
(2 marks)
Radius of a circle = 0.7 cm
Area of the sector = 0.49 m2
r
Area of the sector =
×l
2
0.7

0.49 =
×l
2



49
100
49  20
100  7
l
=
7
×l
20
= l
= 1.4
 The length of the arc is 1.4 m.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
11.
Sol.
Two arcs of the same circle have their lengths in the ratio 4:5. Find the
ratio of the areas of the corresponding sectors.
(2 marks)
Ratio of lengths of two arcs is 4 : 5.
Let the common multiple be ‘x’
 Lengths of two arcs are (4x) units and (5x) units respectively
Let the lengths of two arcs be ‘l1’, and ‘l2’ and Areas of their corresponding
sectors be A1 and A2.
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

l1 = (4x) units
l2 = (5x) units
Both Arcs are of the same circle.
 Their radii are equal
Now,
r
× l1
.......(i)
A1 =
2
r
× l2
......(ii)
A2 =
2
Dividing (i) and (ii) we get,
A1
l1
=
A2
l2
A1
4x

A2 = 5x
 A1 : A 2 = 4 : 5
 Ratio of the areas of sectors is 4 : 5.
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
12.
Sol.
Adjoining figure depicts a racing track
7m
whose left and right ends are semicircular.
105 m
The distance between two inner parallel
70 m
line segments is 70 m and they are each
70 m
105 m long. If the track is 7 m wide, find
105 m
the difference in the lengths of the inner
7m
edge and outer edge of the track. (4 marks)
Diameter of inner circular edge (d 1) = 70 m
Width of the track = 7 m
Diameter of outer circular edge (d2) = 70 + 7 + 7
= 84 m
The inner and outer edges of the racing tracks comprises of two
semicircles and parallel segments of length 105 m each
1
1
d2 + 105 +
d2 + 105
2
2
= d2 + 210
= (84 + 210) m
 Length of outer edge =
1
1
d1 + 105 +
d1 + 105
2
2
= d1 + 210
= (70 + 210) m
Difference in the lengths of
= (84 + 210) – (70 + 210)
inner and outer edge
= 84 + 210 – 70 – 210
= 14
Length of inner edge =
= 14 ×
22
7
= 44 m
 The difference in the lengths of inner edge and outer edge of the
track is 44 m.
292
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EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
13.
(i)
Sol.
10 m
In the adjoining figure,
A horse is tied to a pole fixed
10 m
at one corner of a 30 m × 30 m square
field of grass by means of a 10 m long rope.
( = 3.14).
(3 marks)
Find the area of that part of the
field in which the horse can graze.
Side of a square = 30 m.
30 m
Length of the rope = radius of the sector

Radius of the sector (r) = 10 m
Measure of arc () = 90º
[Angle of a square]
Area of field that can be grazed = Area of sector
=

× r2
360
=
90
× 3.14 × 10 × 10
360
=
1
× 314
4
 Area of field that can be grazed = 78.5 m2
(ii)
Sol.
10 m 10 m
In the adjoining figure,
What will be the area of the part
of the field in which the horse can
graze, if the pole was fixed on a side
exactly at the middle of the side?
If the pole is fixed at the middle of
the middle of the side of a
square, then
30 m
Area of field that can be grazed = 2 × Area of sector
= 2 × 78.5
 Area of field that can be grazed = 157 m2
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
2.
Sol.
The area of a circle is 314 sq.cm and area of its sector is 31.4 sq.cm.
Find the area of its major sector.
(2 marks)
2
Area of a circle = 314 cm
Area of major sector = Area of a circle – Area of its minor sector
= 314 – 31.4
= 282.6
 Area of major sector is 282.6 cm2
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
3.
Sol.
1
C r, for a circle having radius, circumference and area r, C,
2
A respectively.
(2 marks)
L.H.S. = A
 L.H.S. = r 2
........(i)
Prove A =
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R.H.S. =
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1
Cr
2
1
× 2r × r
2
 R.H.S. = r 2
L.H.S. = R.H.S
1

A =
Cr
2
R.H.S =
........(ii)
[From (i) and (ii)]
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
4.
Sol.
The radius of a circle is 3.5 cm and area of the sector is 3.85 cm2. Find
the length of the corresponding arc and the measure of arc. (3 marks)
Radius of a circle (r) = 3.5 cm
Area of the sector = 3.85 cm2
Area of sector =





3.85 =
r
×l
2
3.5
×l
2
3.85 × 2
= l
3.5
3.85 × 2 × 10
= l
100 × 35
22
10
l = 2.2 cm
l =
Area of sector =

 r 2
360

3.85 =

22

 3.5  3.5
360
7

3.85 =

22 35 35



360
7
10 10



385

35
 11 
=
100
360
10
385 × 360 × 10
100 × 11 × 35 = 
 = 36º
 Length of arc is 2.2 cm and measure of an arc is 36º.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
6.
Sol.
294
Find the perimeter of each of
these sectors. (Give your answers
in terms of )
(3 marks)
Radius of the sector (r) = 8 cm
Measure of arc () = 40º
40º
8 cm
10
cm
120º
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
 2r
360
40
=
×2××8
360
16
=
 cm
9
Perimeter of the sector = r + r + l
Length of arc (l) =
= 8+8+
16

9
16
9


= 16 1 + 
9

= 16 
 Perimeter of the sector =
(b)
16 (9 +  )
cm
9
Radius of a sector (r)
Measure of arc ()
= 10 cm
= 126º

 2r
Length of arc (l) =
360
126
=
× 2 ×  × 10
360
= 7 cm
Perimeter of the sector = r + r + l
= 10 + 10 + 7
 Perimeter of the sector = (20 + 7 ) cm.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
7.
Find the area of the shaded part. (Give your answers in terms of )
8 cm
(b)
50º
(a)
m
3c
9c
m
70º
30º
(3 marks)
4 cm
Sol.
(a) Area of shaded part
=
Area of sector I + Area of sector II
=
 30
  90

 360    3  3    360    4  4  
 


 Area of shaded part
=
 3

 4  sq.units


4
(b) Area of shaded part
=
Area of sector I + Area of sector II
=
 70
  50

 360    9  9    360    8  8  
 


=
 63 80 


 sq. units
4
9 
 Area of shaded part
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203)
In the adjoining figure, seg QR is
a tangent to the circle with centre O.
Point Q is the point of contact.
O
Radius of the circle is 10 cm.
T
OR = 20 cm. Find the area of the
shaded region. ( = 3.14, 3 = 1.73 )
R (4 marks)
Q
In OQR,
[Radius is perpendicular to the tangent]
m OQR = 90º
2
2
OQ + QR = OR 2
[By Pythagoras theorem]
2
2
2

10 + QR = 20

QR 2 = 400 – 100

QR 2 = 300

QR =
300
10 cm
10.
Sol.

QR
=



QR
QR
QR
= 10 3
= 10 (1.73)
= 17.3 cm
Area of OQR
=
1
× Product of Perpendicular sides
2
=
1
× OQ × QR
2
=
In OQR,
m OQR
OQ
OR
100 × 3
=
=
=
=
1
 10  17.3
2
86.5 cm2
90º
10 cm
20 cm
1
OR
2
 By converse of 30º - 60º - 90º triangle theorem.

m ORQ = 30º

m QOR = 60º
[Remaining angle]
Now, For sector O-QXT
Measure of arc () = 60º
Radius (r) = 10 cm

× r2
Area of Sector O-QXT =
360
60
=
× 3.14 × 10 × 10
360
157
=
3
Area of sector O-QXT = 52.33 cm2
Area of shaded region = Area of OQR – Area of sector O-QXT
= 86.5 – 52.33
= 34.17 cm2

OQ
=
 Area of the shaded region is 34.17 cm2
296
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203)
11.
Sol.
In the adjoining figure,
PR = 6 units and PQ = 8 units.
P
Semicircles are draw taking
sides PR, RQ and PQ as diameters
as shown in the figure. Find out the area
of the shaded portion. ( = 3.14) (5 marks)
Diameter PR = 6 units
Q
R
 Its radius (r1) = 3 units
Diameter PQ = 8 units
 Its radius (r2) = 4 units
In PQR,
m RPQ = 90º ......(i)
[Angle subtended by a semicircle]
2
2
2
[By Pythagoras theorem]
QR = PR + PQ
2

QR = 62 + 82

QR 2 = 36 + 64

QR = 100

QR = 10 units
[Taking square roots]
Diameter QR = 10 units
 Its radius (r3) = 5 units
PQR is a right angled triangle [From (i)]
1
× product of perpendicular sides
2
1
=
× PR × PQ
2
1
×6×8
=
2
= 24 sq. units.
Area of shaded portion = Area of semicircle with diameter PR + Area
of semicircle with diameter PQ + Area of PQR
– Area of semicircle with diameter QR
A (PQR) =
=
1
1
1
r12 +
r22 + 24 –
r 2
2
2
2 3
1
1 2 1
2
2
=  r1  r2 – r3   24
2
2
2
=
1
 (r12 + r22 – r32) + 24
2
=
1
× 3.14 (32 + 42 – 52) + 24
2
=
1
× 3.14 × (9 + 16 – 25) + 24
2
1
× 3.14 (0) + 24
2
= 0 + 24
= 24 sq. units
=
 Area of shaded portion = 24 sq.units
S C H O O L S E C TI O N
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A
Area of segment of a circle :
A segment of a circle is the region
bounded by a chord and an arc.
O
sin  
2  
–
Area of segment = r 
2 
 360

P
Q
R
EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162)
1.
Sol.
In the adjoining figure,
A (O-AXB) = 75.36 cm2 and
radius = 12 cm, find the area
O
of the segment AXB. ( = 3.14). (2 marks)
Radius of a circle (r) = 12 cm
A (O-AXB) = 75.36 cm2
1 2
Area of OAB =
r sin 
2
1
=
× 12 × 12 × sin 60º
2
3
= 72 ×
2
= 36 3
= 36 × 1.73
= 62.28 cm2
Area of the segment AXB = A (O-AXB) – A (OAB)
= 75.36 – 62.28
= 13.08 cm2
A
60º
X
B
 Area of the segment AXB is 13.08 cm2.
EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162)
2.
Sol.
A
Calculate the area of the shaded
region in the adjoining figure where
X
ABCD is a square with side 8 cm each.
Mark point X as shown in the figure
ABCD is a square
[Given]
side = 8 cm
D
8 cm
Radius (r) = side of a square
 r = 8 cm
Measure of arc () = 90º
[Angle of a square]
B
(3 marks)
C
sin  
 

Area of the segment AXC = r2 
360
2 

 3.14  90 sin 90 

= 82 
2 
 360
1.57 1 
 
= 64 
2
 2
1.57  1
= 64 

2

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=
=
Area of shaded region =
=
=
64  0.57
2
36.48
cm2
2
2 × Area of segment AXC
36.48
2×
2
36.48 cm2
 Area of shaded region is 36.48 cm2.
EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162)
P
In the adjoining figure,
P is the centre of the circle with
radius 18 cm. If the area of the
PQR is 100 cm2 and area of the
segment QXR is 13.04 cm2.
Find the central angle . ( = 3.14) (3 marks) Q
R
Radius of a circle (r) = 18 cm
Area of PQR = 100 cm2
X
Area of the segment QXR = 13.04 cm2
Area of sector P-QXR = Area of PQR + Area of segment QXR
= 100 + 13.04
Area of sector P-QXR = 113.04 cm2

Area of sector =
× r2
360


113.04 =
× 3.14 × 18 × 18
360


11304 =
× 314 × 18 × 18
360
11304 × 360

314 × 18 × 18 = 

 = 40
18
cm
3.
Sol.
 Central angle is 40º.
EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162)
4.
Sol.
In the adjoining figure, the centre of the
circle is A and ABCDEF is a regular hexagon
of side 6 cm. Find the following :
( 3 = 1.73,  = 3.14)
B
(i) Area of segment BPF
(5 marks)
(ii) Area of the shaded portion
C
Side of hexagon = radius of a circle

Radius (r) = 6 cm
Measure of arc () = Angle of a regular hexagon

 = 120º

Area of sector A-BPF =
× r2
360
120
=
× 3.14 × 6 × 6
360
= 3.14 × 12
= 37.68 cm2
S C H O O L S E C TI O N
A
M
P
P
F
E
D
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In ABF,
seg AB  seg AF
 ABF  AFB
......(i)
mBAF + mABF + mAFB = 180º



120 + mABF + mABF =
2m ABF =
2m ABF =

m ABF =


m ABF =
m ABM =
[Radii of the same circle]
[Isosceles triangle theorem]
[Sum of measures of angles
of a triangle is 180º]
180
[Given, from (i)]
180 – 120
60
60
2
30º
30º
.......(ii)
[B - M - F]
In AMB,
m AMB = 90º
m ABM = 30º

m BAM = 60º
 AMB is 30º - 60º - 90º triangle,
 By 30º - 60º - 90º triangle theorem,
AM =
1
AB
2
[By construction]
[From (ii)]
[Remaining angle]
[Side opposite to 30º]
1
×6
2
 AM = 3 cm
 AM =
BM =
 BM =
3
× AB
2
[Side opposite to 60º]
3
×6
2
 BM = 3 3 cm
seg AM  chord BF
1
 BM =
BF
2
 3 3 =
[The perpendicular drawn
from the centre of circle to
a chord, bi sec ts the chord]
1
BF
2
 BF = 6 3
 BF = 6 (1.73)
 BF = 10.38 cm.
Area of ABF =
=
1
× base × height
2
1
× BF × AM
2
1
× 10.38 × 3
2
= 15.57 cm2
Area of segment BPF = Area of sector A-BPF – Area of ABF
= 37.68 – 15.57
=
 Area of segment BPF = 22.11 cm2
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(ii)
side
= 6 cm
Area of regular hexagon ABCDEF =
=
3 3
× (side)2
2
3 3
×6×6
2
= 54
3
= 54 × 1.73
= 93.42 cm2
Area of the shaded portion
= Area of regular hexagon ABCDEF
– Area of ABF
= 93.42 – 15.57
= 77.85 cm2
 Area of segment BPF is 22.11 cm2 and Area of shaded portion is
77.85 cm2.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
8.
Find the area of the shaded region.
( = 3.14,
3 = 1.73)
(3 marks)
12
cm
60º
Sol.
Radius of the sector (r)
Measure of arc ()
Area of segment
= 12 cm
= 60º
sin  
 
–
= r2 
2 
 360
sin 60 
2  3.14 × 60
–
= 12 
2 
 360
 3.14
3 1
–
 
= 144 
2
2
 6
 3.14
3
–

= 144 
4 
 6
 6.28 – 3(1.73) 
= 144 

12

 6.28 – 5.19 
= 144 

12

144 × 1.09
12
= 12 × 1.09
= 13.08 cm2
 Area of shaded region is 13.08 cm2.
=
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203)
9.
Sol.
In the adjoining figure,
m POQ = 30º and radius OP = 12 cm.
Find the following (Given  = 3.14)
O
(i)
Area of sector O-PRQ
(ii) Area of OPQ
M
(3 marks)
(iii) Area of segment PRQ
12 cm 30º
Radius of the circle (r) = 12 cm
Q
•
Measure of arc () = 30º
R
P

Area of sector O - PRQ =
× r2
360
30
=
× 3.14 × 12 × 12
360
= 37.68 cm2
In OMP,
m OMP = 90º
[Given]
m POM = 30º
[Given and O - M - Q]

m OPM = 60º
[Remaining angle]
 OMP is 30º - 60º - 90º triangle
 By 30º - 60º - 90º triangle theorem
1
PM =
OP
2
1
× 12
=
2

PM = 6 cm.
OP = OQ = 12 cm
[Radii of same circle]
1
× base × height
Area of OPQ =
2
1
× OQ × PM
=
2
1
× 12 × 6
=
2
= 36 cm2
Area of segment PRQ = Area of sector O-PRQ – Area of OPQ
= 37.68 – 36
= 1.68 cm2
(i) Area of sector O-PRQ is 37.68 cm2
(ii) Area of OPQ is 36 cm2
(iii)Area of segment PRQ is 1.68 cm2
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203)
12.
Sol.
302
In the adjoining figure,
P
PR and QS are two diameters of the circle.
If PR = 28 cm and PS = 14 3 cm, find
120º
(i) Area of triangle OPS
O
(ii) The total area of two shaded segments.
Q
( 3 = 1.73)
(4 marks)
Draw seg OM  side PS
1
 PR
OP =
[Radius is half of diameter]
2
1
 28
 OP =
2
 OP = 14 cm
S
R
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seg OM  chord PS
1
 PS

PM =
2
1
 14 3
2
= 7 3 cm

PM

PM
In OMP,
OMP = 90º
OM2 + PM2 = OP 2

=

 OM2 + 7 3

OM2

OM2

OM
[By construction]
[The perpendicular drawn from the
centre of a circle to a chord bi sec ts
the chord]
2
=
=
=
=
Area of OPS =
Area of OPS =
=
=
=
[By construction]
[By Pythagoras theorem]
142
196 – 147
49
7 cm
[Taking square roots]
1
× base × height
2
1
× PS × OM
2
1
× 14 3 × 7
2
49 3
49 (1.73)
 Area of OPS = 84.77 cm2
Area of sector OPS
=
=
=

× r2
360
120 22

 14  14
360
7
616
3
205.33 cm2
Area of sector OPS – Area of OPS
205.33 – 84.77
120.56 cm2
=
Area of segment PS =
=
=
Similarly we can prove,
Area of segment QR = 120.56 cm2
 Total area of two shaded segments = 120.56 + 120.56 = 241.12 cm2
Area of OPS is 84.77 cm2 and total area of two shaded segments is
241.12 cm2.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203)
S C H O O L S E C TI O N
cm
60º
cm
(Given  = 3.14, 3 = 1.73) (5 marks)
For a segment PMQ,
radius (r) = 10 cm
measure of arc () = 60º
O
10
Sol.
In the adjoining figure,
seg PQ is a diameter of semicircle PNQ.
The centre of arc PMQ is O.
OP = OQ = 10 cm and m POQ = 60º.
Find the area of the shaded portion
10
13.
P
Q
M
•
Q
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sin  
2  
–
Area of segment PMQ = r 
2 
 360
sin 60 
2  3.14 × 60
–
= 10 
2 
 360
 3.14
3 1
–
 
= 100 
2
2
 6
 3.14
3
= 100  6 – 4 


 6.28 – 3(1.73) 
= 100 

12







 6.28 – 5.19 
= 100 

12

100  1.09
=
12
109
=
12
Area of segment PMQ = 9.08 cm2
In OPQ,
seg OP  seg OQ
[Radii of same circle]
OPQ  OQP
[Isosceles triangle theorem]
Let, m OPQ = m OQP = x
m OPQ + m OQP + m POQ = 180º [Sum of the measures of
angles of a triangle is 180º]
x + x + 60 = 180
2x = 180 – 60
2x = 120
120
2

x = 60
 m POQ = m OPQ = m OQP = 60º
 OPQ is an equilateral triangle


x =
[An equiangular triangle is an
equilateral triangle]
OP = OQ = PQ = 10 cm [Sides of an equilateral triangle]
Diameter PQ = 10 cm
10
2
= 5 cm
Radius (r) =
Area of semicircle =
=
=
Area of the shaded portion =
=
=
1 2
r
2
1
 3.14  5  5
2
39.25 cm2
Area of semicircle – Area of segment PMQ
39.25 – 9.08
30.17 cm2
 The area of shaded portion is 30.17 cm2.
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CUBOID [RECTANGULAR PARALLELOPIPED]
A cuboid is a solid figure bounded by six
rectangular faces, where the opposite faces
are equal.
A cuboid has a length, breadth and height
denoted as ‘l’, ‘b’ and ‘h’ respectively as shown
in the figure,
b
l
h
In our day to day life we come across cuboids
such as rectangular room, rectangular box,
brick, rectangular fish tank, etc.
The following are the formulae for the surface area of cuboid :
FORMULAE
1.
2.
3.
Total surface area of a cuboid = 2 (lb + bh + lh)
Vertical surface area of a cuboid = 2 (l + b) × h
Volume of a cuboid = l × b × h
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
1.
Sol.
The dimensions of a cuboid in cm are 16 × 14 × 20. Find its total surface
area.
(2 marks)
Length of a cuboid (l) = 16 cm
its breadth (b) = 14 cm
its height (h) = 20 cm
Total surface area of a cuboid = 2 (lb + bh + lh)
= 2 (16 × 14 + 14 × 20 + 16 × 20)
= 2 (224 + 280 + 320)
= 2 × 824
= 1648 cm2
 Total surface area of a cuboid is 1648 cm2.
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
4.
Sol.
The cuboid water tank has length 2 m, breadth 1.6 m and height 1.8m.
Find the capacity of the tank in litres.
(2 marks)
Length of the cuboidal water tank (l) = 2 m
its breadth (b) = 1.6 m
and its height (h) = 1.8 m.
Volume of cuboidal water tank = l × b × h
= 2 × 1.6 × 1.8
= 5.76 m3
= 5.76 × 1000 litres [l m3 = 1000 litres]
 Volume of cuboidal water tank is 5760 litres.
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
6.
Sol.
A fish tank is in the form of a cuboid whose external measures are
80cm × 40cm × 30cm. The base, side faces and back faces are to be covered
(2 marks)
with a coloured paper. Find the area of the paper needed.
Length of cuboidal fish tank (l) = 80 cm
its breadth (b) = 40 cm
its height (h) = 30 cm
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Cuboid is make up of 6 rectangular faces
Area of the base of the fish tank = l × b
= 80 × 40
= 3200 cm2
Area of two side faces = 2 × b × h
= 2 × 40 × 30
= 2400 cm2
Area of back face = l × h
= 80 × 30
= 2400 cm2
Area of the paper needed = 3200 + 2400 + 2400
= 8000 cm2
 The area of the paper needed is 8000 cm2.
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
7.
Sol.
Find the total cost of white washing the 4 walls of a cuboidal room at
the rate Rs. 15 per m2. The internal measures of the cuboidal room are
(3 marks)
length 10 m, breadth 4 m and height 4 m.
Length of the cuboidal room (l) = 10m
Its breadth (b) = 14m
Its height (h) = 4m
Vertical Surface area of the room = 2 (l + b) × h
= 2 (10 + 4) × 4
= 2 × 14 × 4
= 112m 2

Area of white washing = 112m 2
Rate of white washing = Rs 15 per m2

Total cost = Area of white washing × rate of
white washing
= 112 × 15
= 1680
 Total cost of white washing is Rs. 1680.
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
9.
Sol.
A beam 4m long, 50cm wide and 20cm deep is made of wood which
(3 marks)
weighs 25kg per m3. Find the weight of the beam.
Length of the beam (l) = 4m
its breadth (b) = 50cm
=
50
m
100
5
m
10
= 20 cm
=
its height (h)
306
=
20
m
100
=
2
m
10
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Volume of the beam
= l×b×h
5
2
= 4×
×
10
10
40 3
=
m
100
Weight of the beam = 25kg per m3
40

Total weight of the beam = 25 ×
100
= 10 kg
 The weight of the beam is 10 kg.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
5.
Sol.
The length, breadth and height of a cuboid are in the ratio 5:4:2. If the
total surface area is 1216 cm2, find the dimensions of the solid. (3 marks)
Ratio of the length, breadth and height of a cuboid is 5 : 4 : 2
Let the common multiple be ‘x’

Length of a cuboid = 5x cm
its breadth = 4x cm
and its height = 2x cm
Total surface area of a cuboid = 1216 cm2
Total surface area of a cuboid = 2 (lb + bh + lh)

1216 = 2 [(5x) (4x) + (4x) (2x) + (5x) (2x)]
1216

= 20x2 + 8x2 + 10x2
2

608 = 38x2
608

= x2
38

x2 = 16

x = 4
[Taking square roots]
Length of a cuboid = 5x
= 5 (4)
= 20 cm
its Breadth = 4x
= 4 (4)
= 16cm
and its height = 2x
= 2 (4)
= 8 cm
 Dimensions of a cuboid are 20 cm, 16 cm and 8 cm.
CUBE
A cube is a cuboid bounded by six equal square faces.
Hence its length, breadth and height are equal.
 The edge of the cube = length = breadth = height
The edge of the cube is denoted as ‘l’
l
A dice is an example of cube.
The following are the formulae for the surface area of the cube :
FORMULAE
1. Total surface area of a cube = 6l2
2. Vertical surface area of a cube = 4l2
3. Volume of cube = l3
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EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
2.
Sol.
The side of a cube is 60 cm. Find the total surface area of the cube.
Side of a cube (l) = 60 cm
Total surface area of a cube = 6l2
= 6 (60)2
= 6 × 60 × 60
= 21600 cm2
 Total surface area of a cube is 21600 cm2.
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
3.
Sol.
The perimeter of one face of a cube
Find (i) the total area of the 6 faces
Perimeter of one face of a cube =
Perimeter of one face of a cube =

4l =

l =

l =
Total surface area of a cube =
=
=
=
Volume of the cube =
=
=
is 24 cm.
(ii) the volume of the cube.
24 cm
4l
24
24
4
6 cm.
6l2
6 (6)2
6×6×6
216 cm2
l3
63
216 cm3.
 Total area of the 6 faces is 216 cm2 and volume of the cube is 216 cm3.
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
5.
Sol.
The volume of a cube is 1000 cm3. Find its total surface area.
Volume of a cube = 1000 cm3
Volume of a cube = l3

l3 = 1000

l = 10 cm
[Taking cube roots]
Total surface area of a cube = 6l2
= 6 × 102
= 6 × 10 × 10
= 600 cm2
 Total surface area of a cube is 600 cm2.
EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
8.
Sol.
A solid cube is cut into two cuboids exactly at middle. Find the ratio of
the total surface area of the given cube and that of the cuboid.
Side of a cube = l

Total Surface of a cube = 6l2
l
Length of cuboid (l1) = side of a cube
l1 = l
l
2
Its height (h1) = l
Its Breadth (b1) =
308
l
l
l 2
2
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Total surface area of a cuboid = 2 (l1b1 + b1h1 + l1h1)
l
l


+ × l + l × l
= 2 l ×
2
2


 l2

l2
+
+ l2
= 2
2
 2

 2l 2

+ l2 
= 2 
 2

= 2 × 2l2
Total surface area of a cuboid = 4l2
T.S.A of a cube
Ratio of the total surface area of a cube and cuboid = T.S.A of a cuboid
=
6l 2
4l 2
3
2
= 3:2
=
 The ratio of the total surface area of the given cube and that
of the cuboid is 3 : 2.
RIGHT CIRCULAR CYLINDER
A right circular cylinder (Cylinder) is a solid figure
bounded by two flat circular surfaces and a curved
surface.
h
The perpendicular distance between the two base faces
is called height of the cylinder and is denoted by ‘h’.
The radius of the base of the cylinder is denoted by ‘r’.
r
The cylinders which we see regularly are drum,
pipe, road roller, coins, test tube, refill of a ball pen, syringe etc.
The following are the formulae for the surface area of a right circular
cylinder :
FORMULAE
1. Curved surface area of a right circular cylinder = 2rh
2. Total surface area of a right circular cylinder = 2r (r + h)
3. Volume of a right circular cylinder = r2h
EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173)
1.
The radius of the base of a right circular cylinder is 3 cm height is 7cm.
Find (i) curved surface area (ii) total surface area (iii) volume of the
22 


7 

Radius of a right circular cylinder = 3cm
its height (h)
= 7cm
(i) Curved surface area of a cylinder = 2 rh
22
= 2×
×3×7
7
 Curved surface area of a cylinder = 132 cm2
closed righ t circu lar cylin der.  Given  =
Sol.
S C H O O L S E C TI O N
(3 marks)
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(ii) Total Surface area of a cylinder = 2r (r + h)
22
× 3 (3 + 7)
= 2×
7
22
× 3 × 10
= 2×
7
1320
=
7
= 188.57 cm2
(iii)
Volume of the cylinder = r 2 h
22
×3×3×7
=
7
= 198 cm3
 Curved surface area is 132 cm2 Total surface area is 188.57cm2 and
volume of the cylinder is 198 cm3
EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173)
2.
Sol.
The volume of a cylinder is 504 cm3 and height is 14 cm. Find its curved
surface area and total surface area. Express answer in terms of .(3 marks)
Volume of a cylinder = 504  cm3
Its height (h) = 14 cm
Volume of a cylinder = r 2 h

504  =  × r2 × 14
504

= r2
14

r 2 = 36

r = 6 cm [Taking square roots]
Curved surface area of a cylinder = 2rh
= 2 ×  × 6 × 14
= 168  cm2
Total surface area of a cylinder = 2r (r + h)
= 2 ×  × 6 (6 + 14)
= 2 ×  × 6 × 20
= 240 cm2
 Curved surface area of a cylinder is 168 cm2
and Total surface area of a cylinder is 240 cm2
EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173)
3.
Sol.
310
The radius and height of a cylinder are n same ratio 3:7 and its volume
is 1584 cm3. Find its radius.
(3 marks)
The ratio of radius and height of a cylinder is 3 :7
Let the common multiple be ‘x’

Radius of cylinder (r) = ‘3x’ cm
and its height (h) = ‘7x’ cm
Volume of a cylinder = 1584 cm3
Volume of a cylinder = r 2 h
22

1584 =
× (3x) × (3x) × (7x)
7

1584 = 22 × 9 × x3
1584

= x3
22 × 9

x3 = 8

x = 2
[Taking cube roots]
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Radius (r) = 3x
= 3(2)
= 6cm
 Radius of the cylinder is 6 cm.
EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173)
4.
Sol.
Keeping the height same, how many times the rod of the given cylinder
should be made to get the cylinder of double the volume of given cylinder ?
(3 marks)
Let the radius and the volume of the given cylinder be r1 and v1 respectively.
The radius and the volume of the required cylinder be r2 and v2 respectively.
Let the heights of the cylinder be h
[ their heights are same]
From the given condition,
v2 = 2v1
2

r2 h = r12 h

r22 = 2r12

r 2 = 2 r1
[Taking square roots]
 The radius of the required cylinder should be
the given cylinder
2 times the radius
EXERCISE - 6.8 (TEXT BOOK PAGE NO. 184)
1.
Sol.
A cylindrical hole of diameter 30 cm is bored
through a cuboid wooden block with side 1
meter. Find the volume of the object so formed
( = 3.14)
(4 marks)
side of cubical wooden block = 1 m
= 100 cm
Volume of cubical wooden block = l3
= (100) 3
= 1000000 cm3
A cylindrical hole is bored through the cubical wooden block

Height of cylindrical hole (h) = 1m
= 100 cm
Diameter of cylindrical hole = 30 cm

Its radius (r)
=
=
Volume of cylindrical hole =
=
=
Volume of the object so formed =
=
=
30
2
15 cm
r 2 h
3.14 × 15 × 15 × 100
70650 cm3
Volume of cubical wooden block
– Volume of cylindrical hole
1000000 – 70650
929350 cm3
 Volume of the object so formed is 929350 cm3.
S C H O O L S E C TI O N
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4.
Sol.
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EXERCISE - 6.8 (TEXT BOOK PAGE NO. 185)
An ink container of cylindrical shape is filled with ink upto 91%. Ball
pen refills of length 12 cm and inner diameter 2 m are filled upto 84%.
If the height and radius of the ink container are 14 cm and 6 cm
respectively, find the number of refills that can be filled with this ink.
(4 marks)
Height of the cylindrical container (h) = 14cm
Its radius (r) = 6 cm
Volume of cylindrical container =  r2h
=  × 6 × 6 × 14
= 504 cm3
But, volume of ink filled
= 91% of 504 
in the cylindrical container
91
× 504 cm 3
=
100
Length of ball pen refill (h1) = 12m
its inner diameter = 2 mm

Its radius (r1) = 1 mm
1
=
cm
10
Volume of the refill = r 1 2 h 1
1
1
= ×
×
× 12
10 10
12 
=
cm3
100
12 

But, volume of ink filled = 84% of
100
84 12 
×
=
cm3
100 100
Number of refills that can be filled with ink
Volume of ink filled in the cylindrical container
=
Volume of ink filled in each refill
91 × 504 
100
84 ×12 
=
100 ×100
91 × 504  100 ×100
×
=
100
84 ×12 
= 4550
 Number of refills that can be filled with this ink is 4550.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
15.
Sol.
A building has 8 right cylindrical pillars whose cross sectional diameter
is 1 m and whose height is 4.2 m. Find the expenditure to paint those
pillars at the rate of Rs. 24 per m2.
(3 marks)
Diameter of a pillar = 1 m

312
 1
=   m
2
Its height (h) = 4.2 m
Curved surface area of a pillars = 2rh
22
1
= 2×
×
× 4.2
7
2
= 13.2 m2
Its radius (r)
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

Curved surface area of 8 pillars =
=
Rate of painting =
Total expenditure =
=
=
8 × 13.2
105.6 m2
Rs. 24 per m2
Area to be painted × Rate of painting
105.6 × 24
2534.40
Total expenditure to paint the pillars is Rs. 2534.40.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
16.
Sol.
A 10 m deep well of diameter 1.4 m us dug up in a field and the earth
from digging is spread evenly on the adjoining cuboid field. The length
and breadth of that filled are 55m and 14 m respectively. Find the
thickness of the earth layer spread.
(4 marks)
Diameter of well = 1.4 m
1.4

Its radius (r) =
2
= 0.7 m
Its depth (h) = 10 m
Volume of cylindrical well = r 2 h
22
 0.7  0.7  10
=
7
22
7
7


 10
=
7
10 10
154
=
10
= 15.4 m3
 Volume of earth dug is 15.4 m3
Now, Earth dug from the well is spread evenly on the adjoining cuboid field
Volume of cuboid = Volume of earth dug
= 15.4 m3
Length of a cuboid (l) = 55 m
Its breadth (b) = 14 m
Volume of cuboid = l × b × h

15.4 = 55 × 14 × h
154

10 × 55 × 14 = h
1

h =
m
50

h = 0.02 m

The thickness of the earth layer spread is 0.02 cm.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
18.
Sol.
A roller of diameter 0.9 m and length 1.8 m is used to press the ground.
Find the area of ground pressed by it in 500 revolutions. (Given  = 3.14)
(3 marks)
Diameter of the roller = 0.9 m
0.9

its radius (r) =
2
= 0.45 m
its length (h) = 1.8 m
S C H O O L S E C TI O N
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

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Curved surface area of the roller = 2rh
= 2 × 3.14 × 0.45 × 1.8
= 6.28 × 0.81
= 5.0868 m2
Area of the ground pressed by the roller in 1 revolution = curved surface
area of roller
Area of the ground pressed in one revolution = 5.0868 m2
Area of the ground pressed in 500 revolution = 500 × 5.0868
50868
= 500 
10000
= 2543.4 m2
Area of the ground pressed by the roller = 2543.4 m2.
O
RIGHT CIRCULAR CONE
An ice-cream cone, a clown’s hat, a funnel are examples
of cones. A cone has one circular flat surface and one
l
h
curved surface.
In the diagram alongside,
seg OA is the height of the cone denoted by ‘h’.
P
A r
seg AP is the radius of the base denoted by ‘r’.
seg OP is the slant height of the cone denoted by ‘l’.
The h, r and l of a cone represents the sides of a right angled triangle
where l is the hypotenuse.
 l2 = r2 + h2.
FORMULA
Volume of a right circular cone =
1
× r2h
3
EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178)
1.
Sol.
The curved surface area of a cone is 4070 cm 2 and its diameter is
70 cm. What is its slant height ?
(2 marks)
Diameter of a cone = 70 cm.
70

Its radius (r) =
2
= 35 cm.
Curved surface area of a cone = 4070 cm2
Curved surface area of a cone = rl
22
 35  l

4070 =
7
4070

22 × 5 = l

l = 37
 Slant height of a cone is 37 cm.
EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178)
2.
Sol.
314
The base radii of two right circular cones of the same height are in the
ratio 2:3. Find the ratio of their volumes.
(3 marks)
Let the radii of two right circular cone be r1 and r2 and their volumes
be v1 and v2 respectively
S C H O O L S E C TI O N
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
r1
r2
=

v1
v2
=

v1
v2
r12
= 2
r2

v1
v2
 r1 
=  
 r2 
v1
v2 =
v1

v2 =
 v1 : v2 =

2
3
.......(i)
(Given)
1 2
r1 h
3
1 2
r2 h
3
2
2
2
 
3
4
9
4:9
[From (i)]
 Ratio of volumes of two right circular cone is 4 : 9
EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178)
3.
Sol.
A cone of height 24 cm has a plane base of surface area 154 cm2. Find
its volume.
(2 marks)
Height of a cone (h) = 24 cm
Surface area of base = 154 cm2
1
Volume of a cone =
× Surface area of base × height
3
1
× r2 × h
=
3
1
× 154 × 24
=
3
= 1232 cm3
3
 Volume of the cone is 1232 cm .
EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178)
4.
Sol.
Curved surface area of a cone with
Find the height of the cone.
Curved surface area of a cone =
its radius (r) =
Curved surface area of a cone =

1640 =
base radius 40 cm is 1640 sq.cm.
(3 marks)
1640cm2
40 cm.
rl
 × 40 × l
1640
= l
40
l = 41 cm


Now,






r2 + h2
402 + h2
h2
h2
h2
h
Height of a cone is 9 cm.
S C H O O L S E C TI O N
=
=
=
=
=
=
l2
412
412 – 402
1681 – 1600
81
9 cm [Taking square roots]
315
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
17.
Sol.
The total surface area of cone is 71.28 cm2. Find the volume of this
cone if the diameter of the base is 5.6 cm.
(3 marks)
Diameter of the base = 5.6 cm
5.6

Its radius (r) =
2
= 2.8 cm
Total surface area of a cone = 71.28 cm2
Total surface area of cone = r (r + l)
22

71.28 =
× 2.8 (2.8 + l)
7
7128
22
28

=
×
(2.8 + l)
100
7
10
7128  10

100  22  4 = 2.8 + l
81

= 2.8 + l
10

8.1 – 2.8 = l

l = 5.3 cm
r2 + h2 = l2

(2.8)2 + h2 = (5.3)2

h 2 = (5.3)2 – (2.8)2

h 2 = (5.3 + 2.8) (5.3 – 2.8)

h 2 = 8.1 × 2.5
81 × 25

h 2 = 10 × 10
9×5

h =
[Taking square roots]
10
45

h =
10

h = 4.5 cm
1 2
r h
Volume of a cone =
3
1 22

 2.8  2.8  4.5
=
3
7
1 22 28 28 45




=
3
7
10 10 10
36960
=
1000

Volume of a cone is 36.96 cm3.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
19.
Sol.
316
The diameter of the base of metallic cone is 2 cm and height is 10 cm.
900 such cones are molten to form 1 right circular cylinder whose radius
is 10 cm. Find total surface area of the right circular cylinder so formed.
(Given  = 3.14)
(4 marks)
Diameter of the base of metallic cone = 2 cm
2
 Its radius (r) =
= 1 cm
2
Its height (h) = 10 cm
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Volume of a metallic cone





=
1 2
r h
3
=
1
× × 1 × 1 × 10
3
=
10
cm3
3
10
3
3000 cm3
right circular cylinder
3000 
10 cm and height be h2
r12h1
 × 10 × 10 h2
30 cm
2r1 (r1 + h1)
2 × 3.14 × 10 (10 + 30)
6.28 × 40
2512 cm2
Volume of 900 metallic cones = 900 
=
900 cones are melted to form a
Volume of a cylinder =
For a cylinder, Radius (r2) =
Volume of a cylinder =
3000 =
h1 =
Total surface area of cylinder =
=
=
=
Total surface area of the right circular cylinder is 2512 cm2.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
20.
Sol.
The volume of a cone of height 5 cm is 753.6 cm3. This cone and a cylinder
have equal radii and height. Find the total surface area of cylinder.
(Given  = 3.14)
(3 marks)
Height of a cone (h) = 5 cm
Volume of a cone = 753.6 cm3
Volume of a cone
=
1 2
r h
3

753.6
=
1
× 3.14 × r2 × 5
3

7536
10
=
1 314
×
× r2 × 5
3 100

7536 × 3 × 100
10 × 314 × 5




= r2
r 2 = 144
r = 12 cm
[Taking square roots]
Cone and cylinder have equal radii and height
Radius of a cylinder = 12 cm and its heights = 5 cm.
Total surface area of cylinder = 2r (r + h)
= 2 × 3.14 × 12 (12 + 5)
= 75.36 × 17
= 1281.12 cm2
Total surface area of a cylinder is 1281.12 cm2.
S C H O O L S E C TI O N
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FRUSTUM OF THE CONE :
If the cone is cut off by a plane parallel to the base not
passing through the vertex, two parts are formed as
(i) cone (a part towards the vertex)
(ii) frustum of cone (the part left over on the
other side i.e. towards base of the original cone)
r2
r1
FORMULAE
1.
2.
3.
4.
l
h
Slant height (l) of the frustum = h 2 + r1 – r2 
Curved surface area = p (r1 + r2) l
2
Total surface area of the frustum =  r1 + r2  l +  r12 +  r22
1
 r12 + r22 + r1 × r2 h
Volume of the frustum =
3


EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178)
5.
Sol.
The curved surface area of the frustum of a cone is 180 sq. cm and the
perimeters of its circular bases are 18 cm and 6 cm respectively. Find
the slant height of the frustum of a cone.
(3 marks)
Curved surface area of the frustum of a cone = 180 cm2
Perimeters of circular bases are 18 cm and 6 cm

2r 1 = 18
........(i)
2r 2 = 6
........(ii)
Adding (i) and (ii), we get
2r1 + 2r2 = 18 + 6
 2 (r1 + r2) = 24
24

 (r1 + r2) =
2

 (r1 + r2) = 12
.......(iii)
Curved surface area of the frustum of a cone = (r1 + r2) l

180 = (r1 + r2) l

180 = 12 × l [From (iii)]

l = 15 cm
 Slant height of the frustum of a cone is 15 cm.
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
21.
The height of a cone is 40 cm. A small cone is cut off at the top of a
1
of the volume of the given
plane parallel to its base. If its volume be
64
(5 marks)
cone, at what height above the base is the section cut ?
Sol.


318
Let the radius, height and volume of the
smaller cone be r1, h1 and v1 respectively.
The radius height and volume of the
given figure cone be r2, h2 and v2 respectively.
h2 = 40 m
[Given]
Consider points A, B, C, E and F as shown in the figure,
In AEF and ABC,
A  A
[Common angle]
AEF  ABC
[Each is 90º]
AEF ~ ABC
[By AA test of similarity]
A
E
F
r1
l
h
B
r2
C
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


AE
EF
=
AB
BC
h1
r1
=
h2
r2
1
v2
V1 =
64
V1
1
V2 = 64
1 r2 h
V1
3 1 1
=
1 r2 h
V2
3 2 2

 r1 
h1
=   ×
r
h
 2
2

1
64
 h1 
h1
=   ×
h2
 h2 
1
64
h1
h2
h1
40
 h1 
=  
h 






......(i)
[Given]
......(ii)
2
1
64

[c.s.s.t.]
2
[From (i)]
3
2
1
=
4
1
=
4
40
h1 =
4
h 1 = 10 cm
The height above the base in the section cut = h2 – h1
= 40 – 10
= 30 cm
The height above the base in the section cut is 30 cm.
SPHERE
The set of all points of space which are at a fixed
distance from a fixed point is called a sphere.
The fixed point is called the centre and the fixed
distance is called the Radius of the sphere.
A
Oh• r
In the adjoining figure, point O is the centre of the
sphere and seg OA is the radius of the sphere which is
denoted as ‘r’.
Since the entire surface of the sphere is curved, its
area is called as curved surface area or simply surface
area of the sphere.
Some common examples of a sphere are cricket ball, football, globe,
spherical soap bubble etc.
The following are the formulae for surface area of the sphere :
FORMULAE
Surface area (curved surface area) of a sphere = 4r2
4
Volume of a sphere =
× r3
3
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EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
1.
Sol.
22 

Find the volume and surface area of a sphere of radius 4.2 cm.   =

7 

(3 marks)
Radius of a sphere (r) = 4.2 cm
4
Volume of a sphere
=
r3
3
4
22
=
×
× 4.2 × 4.2 × 4.2
3
7
4
22
42
42
42
=
×
×
×
×
3
7
10
10
10
310464
=
1000
= 310.464
= 310.46 cm3
 Surface area of a sphere = 4r 2
22
= 4×
× 4.2 × 4.2
7
22
42
42
= 4×
×
×
7
10
10
22176
=
100
= 221.76 cm2
 Volume of sphere is 310.46 cm3 and surface area of a sphere is 221.76 cm2.
EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
2.
Sol.
320
The volumes of two spheres are in the ratio 27 : 64. Find their radii if
the sum of their radii is 28 cm.
(3 marks)
Let the radii of two spheres be r1 and r2 and their volumes be v1 and v2.
v1
27
........(i)
[Given]
v 2 = 64
4 3
r1
3
v1

v 2 = 4 r23
3
4 3
r1
3
27

= 4 3
64
r2
3
r13
27

=
3
r2
64
r1
3

[Taking cube roots]
r2 = 4
Let the common multiple be x
 r1 = 3x and r2 = 4x
[Given]
r1 + r2 = 28
 3x + 4x = 28

7x = 28
28

x =
7

x = 4
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 r 1 = 3x
 r 1 = 3 (4)
 r 1 = 12 cm

r 2 = 4x
r 2 = 4 (4)
r 2 = 16 cm
 Radii of two spheres are 12 cm and 16 cm.
EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
3.
Sol.
22 

The surface area of a sphere is 616cm2. What is its volume ?   =

7 

(3 marks)
Surface area of sphere = 616 cm2
Surface area of a sphere = 4r 2
22

616 = 4 ×
× r2
7
616  7

= r2
4  22

r 2 = 49

r = 7 cm
[Taking square roots]
4 3
Volume of a sphere =
r
3
4 22

777
=
3
7
4312
=
3
= 1437.33 cm3
 The volume is 1437.33 cm3.
EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
4.
Sol.
If the radius of a sphere is doubled, what will be the ratio of its surface
area and volume as to that of the first sphere.
Let the original radius be ‘r1’
 Original surface Area (A 1)= 4r 1 2
New radius (r2) = 2r 1

New surface Area (A 2) = 4r 2 2
= 4 ×  × (2r1)2
= 4 ×  × 4r12
= 16r 1 2
A2
Ratio of New surface area to the original surface area = A
1
16r12
=
4r12
4
=
1
 Ratio of New surface area to the original surface area is 4 : 1
4
r 3
Now, original volume (v1) =
3 1
4
New volume (v2) =
r 3
3 2
4
=
(2r 1) 3
3
4
=
×  × 8r13
3
32
v2 = 3 r 1 3
S C H O O L S E C TI O N
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v2
Ratio of New volume to the original volume = v
1
32 3
r1
3
= 4 3
r1
3
32
=
4
8
=
1
 Ratio of New volume to the original volume is 8 : 1
HEMISPHERE
Half of a sphere is called as hemisphere.
Any hemisphere is made up of a curved surface and a plane
circular surface.
The following are the formulae for the surface area of a hemisphere :
FORMULAE
1. Curved surface area of a hemisphere = 2r2
2. Total surface area of a hemisphere = 3r2
2
× r 3
3. Volume of a hemisphere =
3
EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
5.
The curved surface area of a hemisphere is 905
Sol.
1
cm2, what is its volume?
7
(3 marks)
Curved surface area of a hemisphere = 905





Curved surface area of a hemisphere
1
905
7
6336
7
6336  7
7  2  22
r2
r
1
cm2
7
= 2r 2
22
= 2×
× r2
7
22
= 2×
× r2
7
= r2
= 144
= 12 cm [Taking square roots]
2 3
r
Volume of a hemisphere =
3
2
22
=
×
× 12 × 12 × 12
3
7
25344
=
7
= 3620.57 cm3
 Volume of a hemisphere is 3620.57 cm3.
322
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
14.
Sol.
The following shapes are made up of cones and hemispheres. Find their
volume.
(3 marks)
(a) For a hemisphere,
Height = radius (r) = 3.5 cm
Volume of the figure = Volume of cone + Volume of hemisphere
=
1
2 3
r1h +
r
3
3
=
1 2
r [h + 2r]
3
=
1 22

 3.5  3.5  [12.3  2(3.5)]
3
7
=
1 22 35 35



(12.3  7)
3
7
10 10
12.3 cm
3.5 cm
1 385

 19.3
3
10
385  193
= 3  10  10
=
74305
= 3  100
24768.33
100
= 247.68 cm3
=
 Volume of the given figure is 247.68 cm3.
(b) Diameter of smaller hemisphere = 3 cm
3
 Its radius r1 =
= 1.5 cm
2
Diameter of bigger figure = 10 cm
10
 Its radius r2 =
= 5 cm
2
Volume of the figure = Volume of smaller hemisphere +
volume of bigger hemisphere
2 3 2 3
r1  r2
=
3
3
2
3
 r1  r23
=
3
2 22
(1.5)3  53 

=
3
7
2 22

 (3.375  125)
=
3
7
2 22

 128.375
=
3
7
5648.5
=
21
= 268.98 cm3

3 cm
10 cm

 Volume of the given figure is 268.98 cm3.
S C H O O L S E C TI O N
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EXERCISE - 6.8 (TEXT BOOK PAGE NO. 184)
2.
Sol.
A toy is a combination of a cylinder, hemisphere and a cone, each with radius
10cm. Height of the conical part is 10 cm and total height is 60cm.
Find the total surface area of the toy. ( = 3.14, 2 = 1.41) (5 marks)
10 cm
10 cm
10 cm






60 cm
A toy is a combination of cylinder,
radius 10 cm
r = 10 cm
Height of the conical part (h) =
Height of the hemispherical part =
Total height of the toy =
Height of the cylindrical part (h1) =
l2 =
l2 =
l2 =
l2 =
l =
l =
Slant height of the conical part (l) =
=
=
Total surface area of the toy =
=
=
=
=
=
=
hemisphere and cone, each with
10 cm
its radius = 10cm
60cm
60 – 10 – 10 = 60 – 20 = 40 cm
r2 + h2
102 + 102
100 + 100
200
[Taking square roots]
200
10 2 cm
10 2
10 × 1.41
14.1 cm
Curved surface area of the conical
part + Curved surface area of the
cylindrical part + Curved surface
area of the hemispherical part
rl + 2rh1 + 2r2
r (l + 2h1 + 2r)
3.14 × 10 (14.1 + 2 × 40 + 2 × 10)
31.4 (14.1 + 80 + 20)
31.4 × 114.1
3582.74 cm2
 Total surface area of the toy is 3582.74 cm2.
EXERCISE - 6.8 (TEXT BOOK PAGE NO. 184)
3.
Sol.
324
A test tube has diameter 20 mm and height is
15 cm. The lower portion is a hemisphere in
the adjoining figure. Find the capacity of the
test tube. ( = 3.14)
(5 marks)
Diameter of a test tube = 20 mm
15 cm
20

its radius (r) =
2
= 10 mm
= 1 cm
Its height (h) = 15 cm
Height of hemispherical part (h1) = radius of hemisphere
= 1 cm

Height of cylindrical part (h2) = h – h1
= 15 – 1
= 14 cm
S C H O O L S E C TI O N
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Volume of test tube = Volume of cylindrical part +
Volume of hemispherical part
2 3
= r2h2 +
r
3
2 

r
= r2 h2 +
3


2

= 3.14 (1) 14 + 
3

44
= 3.14 ×
3
138.16
=
3
Volume of test tube = 46.05 cm3
 Capacity of a test tube is 46.05 cm3
EXERCISE - 6.8 (TEXT BOOK PAGE NO. 185)
5.
Sol.
A cylinder of radius 12 cm contains water upto depth of 20 cm. A
spherical iron ball is dropped into the cylinder and thus water level is
(5 marks)
raised by 6.75 cm. what is the radius of the ball ?
Radius of the cylinder (r) = 12 cm
A spherical iron ball is dropped into the cylinder and the water level
rises by 6.75 cm
 Volume of water displaced = volume of the iron ball
Height of the raised water level (h) = 6.75 m
Volume of water displaced = r 2 h
=  × 12 × 12 × 6.75 cm3

Volume of iron ball =  × 12 × 12 × 6.75 cm3
4 3
r
But, Volume of iron ball =
3
4

 × 12 × 12 × 6.75 =
×  × r3
6.75 cm
3
12 × 12 × 6.75 × 3

= r3
4
20 cm

r 3 = 3 × 12 × 6.75 × 3

r 3 = 3 × 3 × 3 × 4 × 6.75

r 3 = 3 × 3 × 3 × 27

r = 3 3 × 3 × 3 × 3 × 3 × 3 [Taking cube roots]

r = 3×3

r = 9

 Radius of the iron ball is 9 cm.
22.
Sol.
A piece of cheese is cut in the
shape of the sector of a circle of
radius 6 cm. The thickness of the
cheese is 7 cm. Find
(i) The curved surface area of the cheese.
(ii) The volume of the cheese piece.
(4 marks)
For a sector,
Measure of arc () = 60º
Radius (r) = 6 cm
S C H O O L S E C TI O N
6 cm
60º
7 cm
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
325
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(i) Curved surface area of the cheese = Length of arc × height

 2r  h
=
360
60
22
2
67
=
360
7
= 44 cm2
 The curved surface area of the cheese is 44 cm2.
(ii)
Volume of the cheese piece
= A (sector) × height

 r 2  h
=
360
60
22

667
=
360
7
= 132 cm3
 The volume of the cheese piece 132 cm3.
EXERCISE - 6.3 (TEXT BOOK PAGE NO. 164)
1.
Which are polyhedrons from the following ?
Nail
(i)
Sol.
(i) No
Unsharpened
Pencil
(ii)
Tile
Diamond
Test tube
(iii)
(iv)
(v)
(ii) Yes
(iii) Yes
(iv) Yes
(v) No
EXERCISE - 6.3 (TEXT BOOK PAGE NO. 164)
2.
Sol.
Using Euler’s formula, find V, if E = 30, F = 12. If the solid figure is a
prism, how many sides the base polygon has.
(1 mark)
F+V = E+2
 12 + V = 30 + 2

V = 32 – 12

V = 20
1
 20 = 10
 Number of sides of base polygon =
2
EXERCISE - 6.3 (TEXT BOOK PAGE NO. 164)
3.
(i)
Sol.
Verify Euler’s formula for these solids :
(2 marks)




326
F = 8, V = 12, E = 18
L.H.S. = F + V
= 8 + 12
L.H.S. = 20
R.H.S. = E + 2
= 18 + 2
R.H.S. = 20
L.H.S. = R.H.S.
F+V = E+2
......(i)
......(ii)
[From (i) and (ii)]
S C H O O L S E C TI O N
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(2 marks)
(ii)
Sol.




F = 8, V = 6, E = 12
L.H.S. = F + V
= 8+6
L.H.S. = 14
R.H.S. = E + 2
= 12 + 2
R.H.S. = 14
L.H.S. = R.H.S.
F+V = E+2
......(i)
......(ii)
[From (i) and (ii)]
(2 marks)
(iii)
Sol.




F = 8, V = 12, E = 18
L.H.S. = F + V
= 8 + 12
L.H.S. = 20
R.H.S. = E + 2
= 18 + 2
R.H.S. = 20
L.H.S. = R.H.S.
F+V = E+2
......(i)
......(ii)
[From (i) and (ii)]
(2 marks)
(iv)
Sol.




F = 6, V = 6, E = 10
L.H.S. = F + V
= 6+6
L.H.S. = 12
R.H.S. = E + 2
= 10 + 2
R.H.S. = 12
L.H.S. = R.H.S.
F+V = E+2
......(i)
......(ii)
[From (i) and (ii)]
HOTS PROBLEM
(Problems for developing Higher Order Thinking Skill)
47.
A sphere and a cube have the same surface area. Show that the ratio of
the volume of the sphere to that of the cube is
6: .
(4 marks)
Proof :
Surface are of sphere = surface area of cube

4r 2 = 6l2
......(i)

r2
l2
=
6
4

r2
l2
=
3
2

r
l
=
Volume of sphere
Volume of cube
4
=
=
S C H O O L S E C TI O N
3
......(ii)
2× 
r3
3
l3
4 r 3
3l 3
327
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GEOMETRY
=
=
=
=
=
=

Volume of sphere
Volume of cube
=
4 r 2 × r
3l 2 × l
4 r 2 r
×
3l 2
l
6l 2 r
×
[From (i)]
3l 2 l
r
2×
l
3
2×
2× 
2× 2× 3
2× 
6

 Ratio of the volume of the sphere to that of the cube is
48.
Sol.
49.
Sol.
328
EDUCARE LTD.
6

Marbles of diameter 1.4 cm are dropped into a beaker containing some
water and are fully submerged. The diameter of the beaker is 7 cm. Find
how many marbles have been dropped in it if the water rises by 5.6 cm.
(5 marks)
Diameter of marble = 1.4 cm
1.4

its radius (r) =
2
= 0.7 cm
4 3
r
Volume of a marble =
3
4
7
7
7
××
×
×
=
cm3
3
10 10 10
 Marbles are submerged fully in the water, water level rises by 5.6 cm
 Height of water displaced (h) = 5.6 cm
Diameter of beaker = 7 cm
7

Its radius (r1) =
cm
2
Volume of water displaced = r12 h
7 7 56
cm3
= × × ×
2 2 10
Volume of water displaced
Number of marbles =
Volume of marble
7 7 56  4
7
7
7
  


= ×  

2 2 10  3
10 10 10 
7 7 56 3 1 10 10 10
× × ×
×
×
= × × ×
2 2 10 4 
7
7
7
= 150
 Number of marbles is 150.
Water flows at the rate of 10 m per minute through a cylindrical pipe
having its diameter is 20 mm. How much time will it take to fill a
conical vessel of base diameter 40 cm and depth 24 cm ?
(5 marks)
Diameter of conical vessel = 40 cm
40
 Its radius (r) =
= 20cm
2
S C H O O L S E C TI O N
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Its depth (h) = 24 cm
Volume of conical vessel
=
=
=
=
Diameter of cylindrical pipe= 20
 Its radius (r1)
=
=
1 2
r h
3
1
×  × 20 × 20 × 24
3
20 × 20 × 8 × 
3200 cm3
mm
20
2
10 mm
10
cm
[ 1 cm = 10 mm]
10
= 1 cm
Water flowing in 1 minute (h) = 10 m
= 10 × 100 cm [ 1 m = 100 cm]
= 1000 cm
Volume of water flowing in 1 minute through a cylindrical pipe
r12 h
 × 1 × 1 × 1000
1000 cm3
Volume of conical vessel
Time taken to fill conical vessel = Volume of water flowing in 1 minute
=
=
=
=
=
3200
1000
=
32
mins
10
32
× 60 secs [1 minute = 60 seconds]
10
= 192 seconds
= 3 minutes and 12 seconds
=
 The time taken to fill the conical vessel is 3 minutes and 12 seconds.
50.
Sol.
Find the length of 13.2 kg. copper wire of diameter 4 mm, when 1 cubic
cm of copper weighs 8.4 gm.
(4 marks)
3
Volume of 8.4 gm of copper = 1 cm
13200
8.4
13200 × 10
=
84
11000
cm3
=
7
Diameter of copper wire = 4 mm
Volume of 13.2 kg i.e. 13200 gm of copper =

4
2
= 2 mm
Its radius (r) =
=
S C H O O L S E C TI O N
2
cm
10
329
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Volume of copper wire = r 2 h




11000
7
11000 × 7 × 10 × 10
7 × 22 × 2 × 2
h
h
=
22
2
2
×
×
×h
7 10 10
= h
= 12500 cm
= 125 m
[ 1 metre = 100 cm]
 Length of wire is 125 m.
51.
Sol.
A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4
(4 marks)
cm thick iron. Find the volume of the iron.
Girth of garden roller = 440 cm
Girth of garden roller = 2 r

2r = 440





22
× r = 440
7
440 × 7
r = 2 × 22
r = 70 cm
Radius of outer cylinder (r) = 70 cm
Width of the roller (h) = 63 cm
Thickness of the roller = 4 cm
Radius of inner cylinder (r1) = 70 – 4
= 66 cm
Volume of iron = Volume of outer cylinder – Volume of inner cylinder
= r2h – r12 h
2×

= h r2 – r12
=

22
× 63 × 702 – 662 
7
= 198 (70  66) (70 – 66)
= 198 × 136 × 4
= 107712 cm3
 The volume of the iron is 107712 cm3.
52.
Sol.
A semi-circular sheet of metal of diameter 28 cm is bent into an open
conical cup. Find the depth and capacity of cup. ( 3 = 1.73) (5 marks)
Diameter of semicircle sheet = 28 cm
28

Its radius (r) =
2
= 14 cm
A semicircular sheet is bent to form a open cone

Slant height of a cone (l) = radius of a semicircular sheet

l = 14 cm
Circumference of a base of a cone = length of semicircle
= r
22
× 14
7
= 44 cm
=
330
S C H O O L S E C TI O N
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Let the radius of a cone be r1
Circumference of a base of a cone = 2r 1

44 = 2r 1
22
× r1

44 = 2 ×
7
44 × 7

2 × 22 = r 1

r 1 = 7 cm




l2
142
h2
h2
h2
=
=
=
=
=
r12  h2
72 + h2
142 – 72
196 – 49
147

h
=

h
=
147
49 × 3



h
h
h
= 7 3
= 7 × 1.73
= 12.11 cm
1 2
r1 h
=
3
1 22
×
× 7 × 7 × 12.11
=
3
7
1864.94
=
3
= 621.646
= 621.65 cm3
Volume of conical cup
 Depth of a conical cup is 12.11 cm and volume of conical cup is 621.65 cm3.
53.
Sol.
A cone and a hemisphere have equal bases and equal volumes. Find the
ratio of their heights.
(3 marks)
A cone and a hemisphere have equal bases
 their radii are equal
Height of hemisphere = radius of hemisphere
Volume of a cone = Volume of hemisphere
1 2
2 3
r h =
r
3
3

h = 2r
h
2

=
r
1
 Ratio of heights of a cone and a hemisphere is 2 : 1

54.
Sol.
A bucket is in the form of a frustum of a cone and holds 28.490 litres of
water. The radii of the top and bottom are 28 cm and 21 cm respectively.
Find the height of the bucket.
(4 marks)
Radii of circular ends are 25 cm and 21 cm
 r1 = 28 cm and r2 = 21 cm
Volume of bucket = 28.490 litres
[ 1 litre = 1000 cm3]
= 28.490 × 1000 cm3
3
= 28490 cm
S C H O O L S E C TI O N
331
MT
GEOMETRY
Volume of bucket =

28490 =

28490 =

28490 =


EDUCARE LTD.
1
 r12  r22  r1  r2 × h
3
1 22
×
282  212  28 × 21 × h
3
7
22
784  441  588 × h
21
22
× 1813 × h
21




28490 × 21
22 × 1813 = h
h = 15 cm
 The height of the bucket is 15 cm.
55.
Sol.
A oil funnel of tin sheet consists of a cylindrical portion 10 cm long
attached to a frustum of a cone. If diameter of the top and bottom of
the frustum is 18 cm and 8 cm respectively and the slant height of the
frustum of cone is 13 cm. Find the surface area of the tin required to
make the funnel. (Express your answer in terms of )
(4 marks)
Diameters of circular ends of frustum are 18 cm and 8 cm
18
8
 r1 =
= 9 cm and r2 =
= 4 cm
2
2
Slant height (l) = 13 cm
Curved surface area of frustum of frustum =  (r1 + r2) l
=  (9 + 4) × 13
=  × 13 × 13
= 169 cm2
Radius of a cylinder (r2) = 4 cm
Its height (h) = 10 cm
Curved surface area of a cylinder
= 2r 2 h
= 2 ×  × 4 × 10
= 80 cm2
Surface area of tin required to make the funnel
= Curved surface area of frustum + curved surface area of cylinder
= 169 + 80
= 249 cm2
 The surface area of the tin required to make the funnel is 249 cm2.
56.
Sol.
332
There are 3 stair-steps as shown in the figure. Each stair-step has width
25 cm, height 12 cm and length 50 cm. How many bricks have been
(5 marks)
used in it if each brick is 12.5 cm × 6.25 cm × 4 cm.
Length of a stair-step (l) = 50 cm
its breadth (b) = 25 cm
its height (h) = 12 cm
Volume of a stair-step = l × b × h
= 50 × 25 × 12
= 15000 cm3
 Volume of 3 stair-step = 6 × 15000
= 90000 cm3
Length of a brick (l1) = 12.5 cm
its breadth (b2) = 6.25 cm
its height (h1) = 4 cm
S C H O O L S E C TI O N
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Volume of a brick = l1 × b1 × h1
= 12.5 × 6.25 × 4
= 312.5 cm3
Volume of 3 stair-steps
Number of bricks required =
Volume of each brick
90000
=
312.5
6 × 50 × 25 × 12
= 12.5 × 6.25 × 4
= 288
 Number of bricks required is 288.
57.
If V is the volume of a cuboid of dimensions a × b × c and S its surface
area, then prove that
1 2  1 1 1
=  + + .
V S  a b c
(3 marks)
1
V
1
 L.H.S. =
.......(i)
abc
2 1 1 1
 
R.H.S. =
S  a b c 
2
 bc  ac  ab 
= 2 (ab  bc  ac) 

abc

1
(ab + bc + ac)
= ab + bc + ac ×
abc
1
R.H.S. =
......(ii)
abc
 L.H.S. = R.H.S.
[From (i) and (ii)]
Proof :

L.H.S. =
2 1 1 1 
1
+ +
=
S  a b c 
V
MCQ’s
1.
An arc of a circle having measure 45º has length 25 cm. What is the
circumference of the circle ?
(a) 200 cm
(b) 100 cm
(c) 50 cm
(d) 45 cm
2.
The measure of arc of circle is 90º. If the radius of the circle is 7 cm. What
is the area of sector ?
(b) 77 cm2
(a) 78.5 cm2
2
(c) 35.8 cm
(d) 38.5 cm2
3.
P is the centre of a circle, m (arc RYS) = 60º. If the radius of the circle is
4.2 cm, what is the perimeter of P-RYS ?
(a) 4.4 m
(b) 8.4 m
(c) 12.8 m
(d) 17.2 m
S C H O O L S E C TI O N
333
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GEOMETRY
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4.
The radius of a circle is 3.5 cm. What is the measure of arc whose length is
5.5 cm ?
(a) 30º
(b) 45º
(c) 60º
(d) 90º
5.
A cylinder with base radius 8 cm and height 2 cm is melted to form a cone
of height 6 cm. What is the radius of the cone ?
(a) 2 cm
(b) 4 cm
(c) 8 cm
(d) 16 cm
6.
Which of the following represents Euler’s formula ?
(a) F – V = E + 2
(b) F + V = E + 2
(c) F + E = V + 2
(d) F – V = E – 2
7.
What is the curved surface area of a cone of height 15 cm and base radius
8 cm ?
(a) 60 cm2
(b) 68 cm2
2
(c) 120 cm
(d) 136 cm2
8.
How many solid metallic spheres each of diameter 6 cm are required to be
melted to form a solid metallic cylinder of height 45 cm and diameter 4 cm.
(a) 3
(b) 4
(c) 15
(d) 6
9.
The radius and slant height of a cone are 5 cm and 10 cm respectively. What
is its curved surface area ?
(a) 314 cm2
(b) 157 cm2
2
(c) 78.5 cm
(d) 100 cm2
10.
The radii of the circular ends of a bucket which is 24 cm high are 14 cm
and 7 cm respectively. What will be its slant height ?
(a) 12 cm
(b) 21 cm
(c) 45 cm
(d) 25 cm
11.
The diameter of a sphere 6 cm is melted and drawn into a wire of diameter
2 mm. What will be the length of the wire ?
(a) 12 m
(b) 18 m
(c) 24 m
(d) 36 m
12.
Two cubes each with 12 m edge are joined end to end. What is the difference
in surface area of the resulting cuboid and the surface area of two cubes ?
(a) 288 m2
(b) 144 m2
2
(c) 1440 m
(d) 770 m2
13.
Area of a sector with central angle 60º will be ............ of area of a circle.
14.
334
(a)
2
3
(c)
1
2
rd
1
6
th
(b)
1
4
th
(d)
If area of semicircle is 77 cm2 its perimeter is ................. .
(a) 72 cm
(b) 120 cm
(c) 36 cm
(d) 40 cm
S C H O O L S E C TI O N
MT
15.
GEOMETRY
EDUCARE LTD.
Area of minor segment AXB is ................ cm2.
(a) 38.5 cm2
(b) 24.5 cm2
2
(c) 14.0 cm
(d) 154 cm2
O
7
B
x
16.
The capacity of a bowl is 144 cm . Find the radius.
(a) 8 cm
(b) 4 cm
(c) 7 cm
(d) 6 cm
17.
A solid metallic ball of radius 14 cm is melted and recasted into small balls
of radius 2 cm. Find how many such balls can be made ?
(a) 434
(b) 343
(c) 433
(d) 344
18.
Find the capacity of swimming pool of length 20 m breadth 5 m and depth 4 m ?
(a) 40000 l
(b) 400000 l
(c) 4000 l
(d) 4000000 l
19.
A cube of side 40 cm is divided into 8 equal cubes. Then its surface area
will increase .................. times.
(a) 4
(b) 8
(c) 2
(d) 5
20.
Find the number of coins 2.4 cm in diameter and 2 mm thick to be melted
to form a right circular cylinder of height 12 cm and diameter 6 cm ?
(a) 350
(b) 370
(c) 400
(d) 375
21.
The curved surface area of a right cone is double that of another right cone.
If the ratio of their slant heights is 1 : 2, find the ratio of their radii ?
(a) 1 : 4
(b) 2 : 3
(c) 3 : 2
(d) 4 : 1
22.
The area swept out by a horse tied in a rectangular grass field with a rope
8 m long is ............... .
(a) 16 cm2
(b) 64 cm2
2
(c) 48 cm
(d) 32 cm2
23.
The angle swept by the minute hand of a clock of length 9 cm in 15 mins is
................. .
(a) 90
(b) 45
(c) 30
(d) 60
24.
A (sector) =
25.
A cylinder and a cone have equal radii and equal heights. If the volume of
the cylinder is 300 cm3, then what is the volume of the cone ?
(a) 100 cm3
(b) 10 cm3
3
(c) 110 cm
(d) 300 cm3
3
A
1
A (circle). Hence, measure of the corresponding central
12
angle will be ................ .
(a) 30
(b) 45
(c) 60
(d) 90
S C H O O L S E C TI O N
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GEOMETRY
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: ANSWERS :
1.
3.
5.
7.
9.
11.
(a)
(c)
(c)
(d)
(b)
(d)
13.
(b)
15.
17.
19.
21.
23.
25.
(c)
(b)
(c)
(d)
(a)
(a)
200 cm
12.8 m
8 cm
136 cm2
157 cm2
36 m
2.
4.
6.
8.
10.
12.
(d)
(d)
(b)
(c)
(d)
(a)
38.5 cm2
90º
F+V=E+2
15
25 cm
288 m2
14.
(c) 36 cm
16.
18.
20.
22.
24.
(d)
(b)
(d)
(a)
(a)
th
1
6
14.0 cm2
343
2
4:1
90
100 cm2
6 cm
400000 l
375
16 cm2
30
1 Mark Sums
1.
Sol.
Find the length of the arc when the corresponding central angle is 270º
and circumference is 31.4 cm.
Measure of central angle () = 270º
Circumfernce (2r) = 31.4 cm
Length of the arc =
2.
Sol.

 2r
360

l =
270
× 31.4
360

l =
3  31.4
4

l =
94.2
4

l = 23.55 cm
If length of an arc is 7 cm, 2r = 36, then find the angle subtended at
the centre by the arc.
Length of the arc (l) = 7 cm
2r = 36


 2r
360

 36
7 =
360
7  360
 =
36
 = 7 × 10

 = 70º
l =



336
The angle subtended at the centre by the arc is 70º.
S C H O O L S E C TI O N
MT
3.
Sol.
Find the area of a circle with radius 7 cm.
Radius of circle (r) = 7 cm

Area of the circle = r 2
22
77
=
7
= 154 cm2

4.
Sol.
The area of a circle is 154 cm2.
Using Euler’s formula, write the value of V, if E = 30 and F = 12.
F+ V = E+2

12 + V = 30 + 2

12 + V = 32

V = 32 – 12

5.
Sol.
Sol.
Sol.
area of its minor sector is 31.4 cm2.
Area of circle – Area of minor sector
314 – 3.14
282.6 cm2
The area of the major sector is 282.6 cm2.
The length of the side of a cube is 6 cm.
A cubical tank has each side of length
in cubic metres.
Side of a cubical tank (l) =
Volume of cubical tank =
=
=

8.
= 20
Perimeter of one face of a cube is 24 cm. Find the length of its side.
Perimeter of one face of a cube = 24 cm
Perimeter of one face of a cube = 4l

24 = 4l
24

l =
4

l =6

7.
V
The area of a circle is 314 cm2 and the
Find the area of its major sector.
Area of major sector =
=
=

6.
Sol.
GEOMETRY
EDUCARE LTD.
2 m. Find the capacity of the tank
2m
l3
23
8 m3
Capacity of the cubical tank is 8 m3.
If the radius is 2 cm and length of corresponding arc is 3.14 cm, find the
area of a sector.
Radius (r) = 2 cm
Length of arc (l) = 3.14 cm
r
Area of sector = l 
2
2
= 3.14 ×
2
= 3.14 cm2

The area of a sector is 3.14 cm2.
S C H O O L S E C TI O N
337
MT
GEOMETRY
9.
Sol.
The dimensions of a cuboid are
Length of a cuboid (l)
Its breadth (b)
Its height (h)
Volume of a cuboid

10.
Sol.
Sol.
Sol.
side 5 cm ?
5 cm
l3
(5) 3
125 cm3
Volume of the cube is 125 cm3.
The value of R is 10.
The radius of the base of a cone
its slant height ?
Radius of base of cone (r)
Its height (h)
l2

l2

l2

l2

l

338
Length of the side of cube is 10 cm.
The area of a circle with radius R is equal to the sum of the areas of
circles with radii 6 cm and 8 cm. What is the value of R ?
According to given condition,
R 2 =  × (6)2 +  × (8)2

R 2 =  [62 + 82]

R 2 = 36 + 64

R 2 = 100

R = 10
[Taking square roots]

14.
The total surface area of cube is 6 cm2.
What is the volume of a cube with
Side of a cube (l) =
Volume of cube =
=
=

13.
Volume of cuboid is 60 cm3.
Volume of a cube is 1000 cm3, find the length of its side.
Volume of a cube = 1000 cm3
Volume of a cube = l3

l3 = 100

l = 10
[Taking cube roots]

12.
Sol.
5 cm, 4 cm and 3 cm. Find its volume.
= 5 cm
= 4 cm
= 3 cm
= l×b×h
= 5×4×3
= 60 cm3
Find the total surface area of a cube with side 1 cm.
Length of side of cube (i) = 1 cm
 Total surface area of a cube = 6l2
= 6 (l)2
= 6 (1)
= 6 cm2

11.
Sol.
EDUCARE LTD.
is 7 cm and its height is 24 cm. What is
=
=
=
=
=
=
=
7 cm
24 cm
r2 + h2
72 + 242
49 + 576
625
25
[Taking square roots]
Slant height of cone is 25 cm.
S C H O O L S E C TI O N
MT
15.
Sol.
EDUCARE LTD.
Using Euler’s formula, find F, if V = 6 and E = 12.
F+V=E+2
 F + 6 = 12 + 2
 F + 6 = 14
 F = 14 – 6

16.
Sol.
Sol.
Sol.
Sol.
20.
Sol.
Length of the arc is 22 cm.
Radius of a circle is 10 cm. The length of an arc of this circle is 25 cm.
What is the area of the sector ?
Radius of circle (r) = 10 cm
Length of arc (l) = 25 cm
r
Area of sector = l 
2
10
= 25 ×
2
= 25 × 2
= 125 cm2

19.
Area of the sector is 456 cm2.
The corresponding central angle of an arc is 90º. What is the length of
this arc, if the radius of the circle is 14 cm ?
Measure of central angle () = 90º
Radius (r) = 14 cm

 2r
Length of the arc (l) =
360
90
22
2
 14

l =
360
7

l = 22

18.
F=8
The area of a circle is 1368 cm2. What is the area of the sector of the
circle whose corresponding central angle is 120º ?
Area of circle = 1368 cm2
 Area of sector of the circle whose corresponding central is 120º
1
 1368
=
3
= 456 cm2

17.
GEOMETRY
The area of the sector is 125 cm2.
A cylinder and a cone have equal radii and equal heights ? If the volume
of the cylinder is 300 cm3, what is the volume of the cone ?
A cylinder and cone have equal height and equal radii
1
 volume of cylinder
 Volume of cone =
3
1
 300
=
3
= 100 cm3
 Volume of the cone is 100 cm3.
What is the corresponding angle of a sector whose area is one-forth of
the area of the circle ?
The corresponding angle of a sector whose area is one-forth of the area
of the circle is 90º.
S C H O O L S E C TI O N
339
MT
GEOMETRY
21.
Sol.
The area of a sector with corresponding angle 45º is 8cm2. What is the
area of the circle ? ( = 3.14)
Area of sector = 8cm 2
Measure of the arc ()= 45º

 r 2
Area of sector =
360
45
 r 2

8 =
360
360

r 2 = 8 ×
45
360
2

r = 8  3.14 
45

r 2 = 3.14 × 64

r 2 = 200.96

22.
Sol.
Sol.
Sol.
Sol.
r = 8
The radius of the base of a
its slant height ?
Radius of base of cone (r)
its height (h)
l2

l2

l2

l2

l

25.
x = 8
The area of a circle with radius 17 cm is equal to the sum of the areas of
circles with radii r cm and 15 cm respectively. What is the value of r ?
According to given condition,
 (17) 2 = r2 +  (15)2

 (17) 2 =  (r2 + 152)

172 = r2 + 152

r 2 = 172 – 152

r 2 = 289 – 225

r 2 = 64

24.
Area of the circle is 200.96 cm2.
The dimensions of a cuboid are 3 cm × 9 cm × x cm. The volume of this cuboid
is equal to the volume of a cube with side 6 cm. What is the value of x ?
Volume of cuboid = Volume of cube
[Given]
3

3 × 9 × x = (6)

3×9×x = 6×6×6
666

x =
39

23.
EDUCARE LTD.
[Taking square roots]
cone is 7 cm and its height is 24 cm. What is
=
=
=
=
=
=
=
7 cm
24 cm
r2 + h2
72 + 212
49 + 576
625
25
[Taking square roots]
Slant height of cone is 25 cm
What is the corresponding central angle of a sector whose area is onetenth the area of the circle ?
The corresponding central angle of a sector whose area is one tenth
the area of the circle is 36º.

340
S C H O O L S E C TI O N
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GEOMETRY
S.S.C.
Marks : 30
CHAPTER 6 : Mensuratiion
GEOMETRY
SET - A
Duration : 1 hr. 15 min.
Q.I. Solve the following :
(4)
(i) If the area of the minor sector is 392.5 sq. cm and the corresponding
central angle is 72º, find the radius. ( = 3.14)
(ii)
Find the length of the arc of a circle with radius 0.7 m and area of
the sector is 0.49 m2.
Q.II. Attempt the following :
(i) Calculate the area of the shaded
A
region in the adjoining figure where
ABCD is a square with side 8 cm each.
(9)
B
X
D
(ii)
8 cm
C
Two arcs of the same circle have their lengths in the ratio 4:5. Find
the ratio of the areas of the corresponding sectors.
(iii) In the adjoining figure,
What will be the area of the part
of the field in which the horse can
graze, if the pole was fixed on a side
exactly at the middle of the side?
10 m
10 m
30 m
Q.III. Solve the following :
(i) Adjoining figure depicts a racing track
whose left and right ends are
semicircular. The distance between
two inner parallel line segments is 70
m and they are each 105 m long. If
the track is 7 m wide, find the
difference in the lengths of the inner
edge and outer edge of the track.
20
(12)
7m
105 m
70 m
70 m
105 m
7m
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In the adjoining figure,
seg QR is a tangent to the circle
with centre O. Point Q is the point
of contact. Radius of the circle is 10 cm.
OR = 20 cm. Find the area of the
shaded region. ( = 3.14,
O
10 cm
(ii)
GEOMETRY
3 = 1.73 )
T
R
Q
(iii) In the adjoining figure,
PR and QS are two diameters of the circle.
P
S
120º
If PR = 28 cm and PS = 14 3 cm, find
(a) Area of triangle OPS
Q
(b) The total area of two shaded segments.
O
R
( 3 = 1.73)
Q.IV. Solve the following :
(i) In the adjoining figure,
PR = 6 units and PQ = 8 units.
Semicircles are draw taking
sides PR, RQ and PQ as diameters
as shown in the figure. Find out the
area of the shaded portion. ( = 3.14)
(5)
P
R
Q
Best of Luck
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GEOMETRY
S.S.C.
Marks : 30
CHAPTER 6 : Mensuratiion
GEOMETRY
SET - B
Duration : 1 hr. 15 min.
Q.I. Solve the following :
(4)
(i) The dimensions of a cuboid in cm are 16 × 14 × 20. Find its total
surface area.
(ii)
The volume of a cube is 1000 cm3. Find its total surface area.
Q.II. Attempt the following :
(9)
3
(i) The volume of a cylinder is 504 cm and height is 14 cm. Find
its curved surface area and total surface area. Express answer in
terms of .
(ii)
The total surface area of cone is 71.28 cm2. Find the volume of this
cone if the diameter of the base is 5.6 cm.
1
(iii) The curved surface area of a hemisphere is 905 cm2, what is its
7
volume?
Q.III. Solve the following :
(12)
(i) A cylinder of radius 12 cm contains water upto depth of 20 cm. A
spherical iron ball is dropped into the cylinder and thus water level
is raised by 6.75 cm. what is the radius of the ball ?
(ii)
The diameter of the base of metallic cone is 2 cm and height is 10
cm. 900 such cones are molten to form 1 right circular cylinder
whose radius is 10 cm. Find total surface area of the right circular
cylinder so formed. (Given  = 3.14)
(iii) A toy is a combination of a cylinder, hemisphere and a cone, each
10 cm
10 cm
10 cm
60 cm
with radius 10cm. Height of the conical part is 10 cm and total height
22
MAHESH TUTORIALS PVT. LTD.
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GEOMETRY
is 60cm. Find the total surface area of the toy. ( = 3.14, 2 = 1.41)
Q.IV. Solve the following :
(5)
(i) An ink container of cylindrical shape is filled with ink upto 91%.
Ball pen refills of length 12 cm and inner diameter 2 m are filled
upto 84%. If the height and radius of the ink container are 14 cm
and 6 cm respectively, find the number of refills that can be filled
with this ink.
Best of Luck
MAHESH TUTORIALS PVT. LTD.
23