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SCHOOL SECTION286

` Introduction :Mensuration is a special branch of mathematics that deals with themeasurement of geometric figures.In previous classes we have studied certain concepts related to areas ofplane figures (shapes) such as triangles, quadrilaterals, polygons and circles.

Now we will study how to find some measurements related to circle andthe surface area and the volume of solid figures.

` Circle : Arc, Sector, Segment : Area of sector :

Sector of a circle is the partof the circle enclosed by tworadii of the circle and theirintercepted arc. (i.e. arc betweenthe two ends of radii)

Area of the sector (A) = × r360

2

Length of an arc :Length of an arc of a circle (arc length) is the distance along the curvedline making up the arc.

Length of the arc (l) = × 2 r360

Relation between the area of the sector and the length of an arc :

Area of the sector =r

× length of arc2

EXERCISE - 6.1 (TEXT BOOK PAGE NO. 157)

1. The diameter of a circle is 10 cm. Find the length of the arc, when thecorresponding central angle is as given below : ( = 3.14)

(i) 144º (2 marks)Sol. Diameter of a circle = 10 cm

Its radius (r) =10

2= 5 cm

l =360

× 2r

l =144

360 × 2 × 3.14 × 5

l =144

360 × 3.14 × 10

l = 12.56 cm

The length of the arc is 12.56 cm.

Mensuration6.

Major arc

Central angle

Minor arc

rO

A

B

X

Y

MT EDUCARE LTD. GEOMETRY

SCHOOL SECTION 287

(ii) 45º (2 marks)Sol. Diameter of a circle = 10 cm

Its radius (r) =10

2= 5 cm

l =360

× 2r

l =45

360 × 2 × 3.14 × 5

l =45

360 × 3.14 × 10

l =5

4 ×

314

100

l =785

2 100

l =392.5

100 l = 3.925 l = 3.93 cm


(iii) 270º (2 marks)Sol. Diameter of a circle = 10 cm

It radius (r) =10

2= 5 cm

l =360

× 2r

l =270

360 × 2 × 3.14 × 5

l =3

4 × 3.14 × 10

l = 23.55 cm


(iv) 180º (2 marks)Sol. Diameter of a circle = 10 cm

Its radius (r) =10

2= 5 cm

l =360

× 2r

=180

360 × 2 × 3.14 × 5

= 15.70 cm


GEOMETRY MT EDUCARE LTD.

SCHOOL SECTION288


2. Find the angle subtended at the centre of a circle by an arc, given thefollowing :

(i) radius of circle = 5.5 m, length of arc = 6.05 m ( = 227

) (2 marks)

Sol. l =360

× 2r

6.05 =360

× 2 ×

22

7 × 5.5

605

100=

360

×

22

7 × 11

605 × 360 × 7

100 × 22 ×11 =

= 63º

Measure of the arc is 63º.

(ii) radius of circle = 20 m length of arc = 78.50 m ( = 3.14) (2 marks)

Sol. l =360

× 2r

78.50 =360

× 2 × 3.14 × 20

785

10=

360

×

314

100 × 40

785 × 9 ×100

10 × 314 =

= 225º

Measure of an arc is 225º.


3. The radius of the circle is 7 cm and m (arc RYS) = 60º, with the help ofthe figure, answer the following questions : (3 marks)

(i) Name the shaded portion.Sol. P-RYS

(ii) Find the area of the circle.Sol. Area of circle = r2

=22

7 × 7 × 7

= 154 cm2

Area of a circle is 154 cm2.

(iii) Find A (P-RYS)

Sol. Area of the sector =360

× r2

=60

360 ×

22

7 × 7 × 7

=77

3= 25.67

Area of the sector P -RYS is 25.67 cm2

S

Y

R

PX 60º


SCHOOL SECTION 289

(iv) Find A (P-RXS)Sol. 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.


4. The radius of a circle is 7 cm. Find area of the sector of this circle if theangle of the sector is :

(i) 30º (2 marks)


× r2

=30

360 ×

22

7 × 7 × 7

=77

6= 12.83

Area of the sector is 12.83 cm2.

(ii) 210º (2 marks)


× r2

=210

360 ×

22

7 × 7 × 7

=539

6= 89.83


(iii) 3 rt. angles (2 marks)


× r2

=270

360 ×

22

7 × 7 × 7

=231

2= 115.50



5. An arc of a circle having measure 36 has length 176 m. Find thecircumference of the circle. (2 marks)

Sol. Length of arc (l) = 176 m measure of arc () = 36º

l =360

× 2r

176 =36

360× 2r


SCHOOL SECTION290

176 =1

10 × 2r

176 × 10 = 2r 2r = 1760

But, circumference = 2r

Circumference of the circle is 1760 m.


6. 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)

Sol. Length of arc (l) = 4 cm measure of arc () = 40º

l =360

× 2r

4 =40

360 × 2 × × r

4 × 9

2= r

r = 18 cm.

Area of the sector =× r

2

l

=4 ×18

2

= 36cm2

Radius of the circle is 18 cm and Area of the sector is 36cm2.


7. If the area of the minor sector is 392.5 sq. cm and the correspondingcentral angle is 72º, find the radius. ( = 3.14) (2 marks)

Sol. Measure of arc () = 72ºArea of the sector = 392.5 cm2

Area of the sector =360

× r2

392.5 =72

360 × 3.14 × r2

3925

10=

72

360 ×

314

100 × r2

3925 × 360 ×100

10 × 72 × 314 = r2

r2 = 625 r = 25 [Taking square roots]

Radius of the circle is 25 cm.


8. Find the area of sector whose arc length and radius are 10 cm and 5 cmrespectively. (2 marks)

Sol. Length of arc (l) = 10 cmRadius of a circle (r) = 5 cm


SCHOOL SECTION 291


2 × l

=5

2 × 10

= 25 cm2

Area of the sector is 25 cm2.


9. 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)

Sol. Radius of a circle = 11.2 cmArea of the sector = 49.28 cm2

Area of the sector =360

× r2

49.28 =360

×

22

7 × 11.2 × 11.2

4928

100=

360

×

22

7 ×

112

10 ×

112

10

4928 × 360 × 7 ×10 ×10

100 × 22 ×112 ×112 =

= 45º

Measure of arc is 45º.


10. Find the length of the arc of a circle with radius 0.7 m and area of thesector is 0.49 m2. (2 marks)

Sol. Radius of a circle = 0.7 cmArea of the sector = 0.49 m2


2 × l

0.49 =0.7

2 × l

49

100=

7

20 × l

49 20

100 7

= l

l = 1.4

The length of the arc is 1.4 m.


11. Two arcs of the same circle have their lengths in the ratio 4:5. Find theratio of the areas of the corresponding sectors. (2 marks)

Sol. 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 respectivelyLet the lengths of two arcs be ‘l1’, and ‘l2’ and Areas of their correspondingsectors be A1 and A2.


SCHOOL SECTION292

l1 = (4x) units l2 = (5x) units

Both Arcs are of the same circle. Their radii are equal

Now,

A1 =r

2 × l1 .......(i)

A2 =r

2 × l2 ......(ii)

Dividing (i) and (ii) we get,

1

2

A

A =1

2

l

l

1

2

A

A =4x

5x A1 : A2 = 4 : 5

Ratio of the areas of sectors is 4 : 5.


12. Adjoining figure depicts a racing trackwhose left and right ends are semicircular.The distance between two inner parallelline segments is 70 m and they are each105 m long. If the track is 7 m wide, findthe difference in the lengths of the inneredge and outer edge of the track. (4 marks)

Sol. Diameter of inner circular edge (d1) = 70 mWidth 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 twosemicircles and parallel segments of length 105 m each

Length of outer edge =1

2d2 + 105 +

1

2d2 + 105

= d2 + 210

= (84 + 210) m

Length of inner edge =1

2d1 + 105 +

1

2d1 + 105

= d1 + 210= (70 + 210) m

Difference in the lengths of= (84 + 210) – (70 + 210)

inner and outer edge= 84 + 210 – 70 – 210= 14

= 14 × 22

7= 44 m

The difference in the lengths of inner edge and outer edge of thetrack is 44 m.

105 m

105 m

70 m70 m

7 m

7 m


SCHOOL SECTION 293


13. In the adjoining figure,A horse is tied to a pole fixedat one corner of a 30 m × 30 m squarefield of grass by means of a 10 m long rope.( = 3.14). (3 marks)

(i) Find the area of that part of thefield in which the horse can graze.

Sol. Side of a square = 30 m.Length of the rope = radius of the sector

Radius of the sector (r) = 10 mMeasure of arc () = 90º [Angle of a square]

Area of field that can be grazed = Area of sector

=360

× r2

=90

360 × 3.14 × 10 × 10

=1

4 × 314

Area of field that can be grazed = 78.5 m2

(ii) In the adjoining figure,What will be the area of the partof the field in which the horse cangraze, if the pole was fixed on a sideexactly at the middle of the side?

Sol. If the pole is fixed at the middle ofthe middle of the side of asquare, thenArea 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. 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)

Sol. Area of a circle = 314 cm2

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


3. Prove A = 12

C r, for a circle having radius, circumference and area r, C,

A respectively. (2 marks)Sol. L.H.S. = A

L.H.S. = r2 ........(i)

30 m

10 m

10 m

10 m 10 m

30 m


SCHOOL SECTION294

R.H.S. =1

Cr2

R.H.S =1

2 × 2r × r

R.H.S. = r2 ........(ii)

L.H.S. = R.H.S [From (i) and (ii)]

A =12

Cr


4. The radius of a circle is 3.5 cm and area of the sector is 3.85 cm2. Findthe length of the corresponding arc and the measure of arc. (3 marks)

Sol. Radius of a circle (r) = 3.5 cm

Area of the sector = 3.85 cm2

Area of sector =r

2 × l

3.85 =3.5

2 × l

3.85 × 2

3.5= l

3.85 × 2 ×10

100 × 35 = l

l =22

10 l = 2.2 cm

Area of sector = r360

2

3.85 =22

3.5 3.5360 7

3.85 =22 35 35

360 7 10 10

385

100=

3511

360 10

385 × 360 ×10

100 ×11 × 35 =

= 36º

Length of arc is 2.2 cm and measure of an arc is 36º.


6. Find the perimeter of each ofthese sectors. (Give your answersin terms of ) (3 marks)

Sol. Radius of the sector (r) = 8 cmMeasure of arc () = 40º

10 cm

120º40º8 cm


SCHOOL SECTION 295

Length of arc (l) = 2 r360

=40

360 × 2 × × 8

=16

9 cm

Perimeter of the sector = r + r + l

= 8 + 8 + 16

9

=16

169

= 16 1 +9

Perimeter of the sector =16 (9 + )

9

cm

(b) Radius of a sector (r) = 10 cmMeasure of arc () = 126º

Length of arc (l) = 2 r360

=126

360 × 2 × × 10

= 7 cmPerimeter of the sector = r + r + l

= 10 + 10 + 7 Perimeter of the sector = (20 + 7 ) cm.


7. Find the area of the shaded part. (Give your answers in terms of )

(3 marks)

Sol. (a) Area of shaded part = Area of sector I + Area of sector II

=30 90

3 3 4 4360 360

Area of shaded part =3

4 sq.units4

(b) Area of shaded part = Area of sector I + Area of sector II

=70 50

9 9 8 8360 360

Area of shaded part =63 80

sq. units4 9

3 cm

4 cm

30º

(a)

9 c

m

8 cm

70º

50º(b)


SCHOOL SECTION296


10. In the adjoining figure, seg QR isa 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 theshaded region. ( = 3.14, 3 = 1.73 ) (4 marks)

Sol. In OQR,m OQR = 90º [Radius is perpendicular to the tangent]

OQ2 + QR2 = OR2 [By Pythagoras theorem] 102 + QR2 = 202

QR2 = 400 – 100 QR2 = 300 QR = 300

QR = 100 × 3

QR = 10 3 QR = 10 (1.73) QR = 17.3 cm

Area of OQR =1

2 × Product of Perpendicular sides

=1

2 × OQ × QR

=1

10 17.32

= 86.5 cm2

In OQR, m OQR = 90ºOQ = 10 cmOR = 20 cm

OQ =1

2 OR

By converse of 30º - 60º - 90º triangle theorem. m ORQ = 30º m QOR = 60º [Remaining angle]

Now, For sector O-QXTMeasure of arc () = 60º

Radius (r) = 10 cm

Area of Sector O-QXT =360

× r2

=60

360 × 3.14 × 10 × 10

=157

3Area 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

Area of the shaded region is 34.17 cm2

Q R

T

O

10 c

m


SCHOOL SECTION 297


11. In the adjoining figure,PR = 6 units and PQ = 8 units.Semicircles are draw takingsides PR, RQ and PQ as diametersas shown in the figure. Find out the areaof the shaded portion. ( = 3.14) (5 marks)

Sol. Diameter PR = 6 units 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]

QR2 = PR2 + PQ2 [By Pythagoras theorem] QR2 = 62 + 82 QR2 = 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)]

A (PQR) =1

2× product of perpendicular sides

=1

2 × PR × PQ

=1

2 × 6 × 8

= 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

=1

2r1

2 + 1

2r2

2 + 24 – 1

2r3

2

=1 1 1

r r – r 242 2 2

2 2 2

1 2 3

=1

2 (r1

2 + r22 – r3

2) + 24

=1

2 × 3.14 (32 + 42 – 52) + 24

=1

2 × 3.14 × (9 + 16 – 25) + 24

=1

2 × 3.14 (0) + 24

= 0 + 24= 24 sq. units

Area of shaded portion = 24 sq.units

P

R Q


SCHOOL SECTION298

` Area of segment of a circle :A segment of a circle is the regionbounded by a chord and an arc.

Area of segment = sin

r –360 2

2


1. In the adjoining figure,A (O-AXB) = 75.36 cm2 andradius = 12 cm, find the areaof the segment AXB. ( = 3.14). (2 marks)

Sol. Radius of a circle (r) = 12 cmA (O-AXB) = 75.36 cm2

Area of OAB =1

2 r2 sin

=1

2 × 12 × 12 × sin 60º

= 72 × 3

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

Area of the segment AXB is 13.08 cm2.


2. Calculate the area of the shadedregion in the adjoining figure whereABCD is a square with side 8 cm each. (3 marks)

Sol. Mark point X as shown in the figureABCD is a square [Given]side = 8 cmRadius (r) = side of a square

r = 8 cmMeasure of arc () = 90º [Angle of a square]

Area of the segment AXC = r2sin

360 2

= 823.14 90 sin 90

360 2

= 64 1.57 1

2 2

= 64 1.57 1

2

A

O

P Q

R

A

X

B

O 60º

BA

D C8 cm

X


SCHOOL SECTION 299

=64 0.57

2

=36.48

2 cm2

Area of shaded region = 2 × Area of segment AXC

= 2 × 36.48

2= 36.48 cm2

Area of shaded region is 36.48 cm2.


3. In the adjoining figure,P is the centre of the circle withradius 18 cm. If the area of thePQR is 100 cm2 and area of thesegment QXR is 13.04 cm2.Find the central angle . ( = 3.14) (3 marks)

Sol. Radius of a circle (r) = 18 cmArea of PQR = 100 cm2

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 =360

× r2

113.04 =360

× 3.14 × 18 × 18

11304 =360

× 314 × 18 × 18

11304 × 360

314 ×18 ×18 =

= 40

Central angle is 40º.


4. In the adjoining figure, the centre of thecircle is A and ABCDEF is a regular hexagonof side 6 cm. Find the following :( 3 = 1.73, = 3.14)(i) Area of segment BPF(ii) Area of the shaded portion (5 marks)

Sol. 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 =360

× r2

=120

360 × 3.14 × 6 × 6

= 3.14 × 12= 37.68 cm2

P

Q R

X

18 c

m

A

P

B F

C E

D

M

P


SCHOOL SECTION300

In ABF,seg AB seg AF [Radii of the same circle]

ABF AFB ......(i) [Isosceles triangle theorem]mBAF + mABF + mAFB = 180º [Sum of measures of angles

of a triangle is 180º] 120 + mABF + mABF = 180 [Given, from (i)] 2mABF = 180 – 120 2mABF = 60

mABF = 60

2 mABF = 30º mABM = 30º .......(ii) [B - M - F]

In AMB,mAMB = 90º [By construction]mABM = 30º [From (ii)]

mBAM = 60º [Remaining angle] AMB is 30º - 60º - 90º triangle, By 30º - 60º - 90º triangle theorem,

AM =1

2 AB [Side opposite to 30º]

AM =1

2 × 6

AM = 3 cm

BM = 3

2 × AB [Side opposite to 60º]

BM = 3

2 × 6

BM = 3 3 cm

seg AM chord BF

BM = 1

2 BF

[The perpendicular drawn

from the centre of circle to

a chord, bisec ts the chord]

3 3 = 1

2 BF

BF = 6 3 BF = 6 (1.73) BF = 10.38 cm.

Area of ABF =1

2 × base × height

=1

2 × BF × AM

=1

2 × 10.38 × 3

= 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


SCHOOL SECTION 301

(ii) side = 6 cm

Area of regular hexagon ABCDEF =3 3

2 × (side)2

=3 3

2 × 6 × 6

= 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.


8. Find the area of the shaded region.

( = 3.14, 3 = 1.73) (3 marks)

Sol. Radius of the sector (r) = 12 cmMeasure of arc () = 60º

Area of segment = r2sin

–360 2

=3.14 × 60 sin 60

12 –360 2

2

=3.14 3 1

144 –6 2 2

=3.14 3

144 –6 4

=6.28 – 3(1.73)

14412

=6.28 – 5.19

14412

=144 ×1.09

12= 12 × 1.09

= 13.08 cm2

Area of shaded region is 13.08 cm2.

12 cm

60º


SCHOOL SECTION302


9. In the adjoining figure,m POQ = 30º and radius OP = 12 cm.Find the following (Given = 3.14)(i) Area of sector O-PRQ (ii) Area of OPQ(iii) Area of segment PRQ (3 marks)

Sol. Radius of the circle (r) = 12 cmMeasure of arc () = 30º

Area of sector O - PRQ =360

× r2

=30

360 × 3.14 × 12 × 12

= 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

PM =1

2 OP

=1

2 × 12

PM = 6 cm.OP = OQ = 12 cm [Radii of same circle]

Area of OPQ =1

2 × base × height

=1

2 × OQ × PM

=1

2 × 12 × 6

= 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


12. In the adjoining figure,PR and QS are two diameters of the circle.If PR = 28 cm and PS = 14 3 cm, find(i) Area of triangle OPS(ii) The total area of two shaded segments.

( 3 = 1.73) (4 marks)Sol. Draw seg OM side PS

OP = 1

PR2 [Radius is half of diameter]

OP = 1

282

OP = 14 cm

O

12 cm 30º M

Q

P R•

P S

Q R

O

120º


SCHOOL SECTION 303

seg OM chord PS [By construction]

PM =1

PS2 [The perpendicular drawn from the

centre of a circle to a chord bisec ts

the chord]

PM =1

14 32

PM = 7 3 cmIn OMP,OMP = 90º [By construction]OM2 + PM2 = OP2 [By Pythagoras theorem]

OM2 + 7 32

= 142

OM2 = 196 – 147 OM2 = 49 OM = 7 cm [Taking square roots]

Area of OPS =1

2 × base × height

Area of OPS =1

2 × PS × OM

=1

2 × 14 3 × 7

= 49 3= 49 (1.73)

Area of OPS = 84.77 cm2

Area of sector OPS =360

× r2

=120 22

14 14360 7

=616

3= 205.33 cm2

Area of segment PS = Area of sector OPS – Area of OPS= 205.33 – 84.77= 120.56 cm2

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 is241.12 cm2.


13. 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

(Given = 3.14, 3 = 1.73) (5 marks)

Sol. For a segment PMQ,radius (r) = 10 cmmeasure of arc () = 60º

P Q

M

Q•

O

10 c

m

10 cm

60º


SCHOOL SECTION304

Area of segment PMQ =sin

r –360 2

2

=3.14 × 60 sin 60

10 –360 2

2

=3.14 3 1

100 –6 2 2

=3.14 3

100 –6 4

=6.28 – 3(1.73)

10012

=6.28 – 5.19

10012

=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 ofangles of a triangle is 180º]

x + x + 60 = 180 2x = 180 – 60 2x = 120

x = 120

2 x = 60 m POQ = m OPQ = m OQP = 60º OPQ is an equilateral triangle [An equiangular triangle is an

equilateral triangle] OP = OQ = PQ = 10 cm [Sides of an equilateral triangle]

Diameter PQ = 10 cm

Radius (r) = 10

2= 5 cm

Area of semicircle =1

r2

2

=1

3.14 5 52

= 39.25 cm2

Area of the shaded portion = Area of semicircle – Area of segment PMQ= 39.25 – 9.08= 30.17 cm2

The area of shaded portion is 30.17 cm2.


SCHOOL SECTION 305

CUBOID [RECTANGULAR PARALLELOPIPED]

l

h

bA cuboid is a solid figure bounded by sixrectangular faces, where the opposite facesare equal.A cuboid has a length, breadth and heightdenoted as ‘l’, ‘b’ and ‘h’ respectively as shownin the figure,

In our day to day life we come across cuboidssuch as rectangular room, rectangular box,brick, rectangular fish tank, etc.The following are the formulae for the surface area of cuboid :

FORMULAE

1. Total surface area of a cuboid = 2 (lb + bh + lh)2. Vertical surface area of a cuboid = 2 (l + b) × h3. Volume of a cuboid = l × b × h


1. The dimensions of a cuboid in cm are 16 × 14 × 20. Find its total surfacearea. (2 marks)

Sol. Length of a cuboid (l) = 16 cmits breadth (b) = 14 cmits 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.


4. 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)

Sol. Length of the cuboidal water tank (l) = 2 mits 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.


6. A fish tank is in the form of a cuboid whose external measures are80cm × 40cm × 30cm. The base, side faces and back faces are to be coveredwith a coloured paper. Find the area of the paper needed. (2 marks)

Sol. Length of cuboidal fish tank (l) = 80 cmits breadth (b) = 40 cmits height (h) = 30 cm


SCHOOL SECTION306

Cuboid is make up of 6 rectangular facesArea 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.


7. Find the total cost of white washing the 4 walls of a cuboidal room atthe rate Rs. 15 per m2. The internal measures of the cuboidal room arelength 10 m, breadth 4 m and height 4 m. (3 marks)

Sol. Length of the cuboidal room (l) = 10mIts breadth (b) = 14mIts height (h) = 4m

Vertical Surface area of the room = 2 (l + b) × h= 2 (10 + 4) × 4= 2 × 14 × 4= 112m2

Area of white washing = 112m2

Rate of white washing = Rs 15 per m2

Total cost = Area of white washing × rate ofwhite washing

= 112 × 15= 1680

Total cost of white washing is Rs. 1680.


9. A beam 4m long, 50cm wide and 20cm deep is made of wood whichweighs 25kg per m3. Find the weight of the beam. (3 marks)

Sol. Length of the beam (l) = 4mits breadth (b) = 50cm

=50

100 m

=5

10 m

its height (h) = 20 cm

=20

100m

=2

10m


SCHOOL SECTION 307

CUBE

l

Volume of the beam = l × b × h

= 4 × 5

10 ×

2

10

=40

100m3

Weight of the beam = 25kg per m3

Total weight of the beam = 25 × 40

100= 10 kg

The weight of the beam is 10 kg.


5. The length, breadth and height of a cuboid are in the ratio 5:4:2. If thetotal surface area is 1216 cm2, find the dimensions of the solid. (3 marks)

Sol. Ratio of the length, breadth and height of a cuboid is 5 : 4 : 2Let the common multiple be ‘x’

Length of a cuboid = 5x cmits breadth = 4x cm

and its height = 2x cmTotal 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

2= 20x2 + 8x2 + 10x2

608 = 38x2

608

38= x2

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.

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 = heightThe edge of the cube is denoted as ‘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


SCHOOL SECTION308


2. The side of a cube is 60 cm. Find the total surface area of the cube.Sol. 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.


3. The perimeter of one face of a cube is 24 cm.Find (i) the total area of the 6 faces (ii) the volume of the cube.

Sol. Perimeter of one face of a cube = 24 cmPerimeter of one face of a cube = 4l

4l = 24

l =24

4 l = 6 cm.


= 6 (6)2

= 6 × 6 × 6= 216 cm2

Volume of the cube = l3

= 63

= 216 cm3.

Total area of the 6 faces is 216 cm2 and volume of the cube is 216 cm3.


5. The volume of a cube is 1000 cm3. Find its total surface area.Sol. Volume of a cube = 1000 cm3

Volume of a cube = l3

l3 = 1000 l = 10 cm [Taking cube roots]


= 6 × 102

= 6 × 10 × 10= 600 cm2

Total surface area of a cube is 600 cm2.


8. A solid cube is cut into two cuboids exactly at middle. Find the ratio ofthe total surface area of the given cube and that of the cuboid.

Sol. Side of a cube = l Total Surface of a cube = 6l2

Length of cuboid (l1) = side of a cubel1 = l

Its Breadth (b1) =2

l

Its height (h1) = l2

l 2

l

l

l


SCHOOL SECTION 309

RIGHT CIRCULAR CYLINDER

h

r

Total surface area of a cuboid= 2 (l1b1 + b1h1 + l1h1)

= 2 × + × + ×2 2

l ll l l l

= 2 + +2 2

2 22l l

l

= 2 2

2

2l

+ 2l

= 2 × 2l2

Total surface area of a cuboid= 4l2

Ratio of the total surface area of a cube and cuboid=T.S.A of a cube

T.S.A of a cuboid

=6

4

2

2

l

l

=3

2= 3 : 2

The ratio of the total surface area of the given cube and thatof the cuboid is 3 : 2.

A right circular cylinder (Cylinder) is a solid figurebounded by two flat circular surfaces and a curvedsurface.The perpendicular distance between the two base facesis called height of the cylinder and is denoted by ‘h’.The radius of the base of the cylinder is denoted by ‘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 circularcylinder :

FORMULAE 1. Curved surface area of a right circular cylinder = 2rh 2. Total surface area of a right circular cylinder = 2r (r + h) 3. Volume of a right circular cylinder = r2h


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

closed right circular cylinder.

22Given =

7 (3 marks)

Sol. Radius of a right circular cylinder = 3cmits height (h) = 7cm

(i) Curved surface area of a cylinder = 2 rh

= 2 × 22

7 × 3 × 7

Curved surface area of a cylinder = 132 cm2


SCHOOL SECTION310

(ii) Total Surface area of a cylinder = 2r (r + h)

= 2 × 22

7 × 3 (3 + 7)

= 2 × 22

7 × 3 × 10

=1320

7= 188.57 cm2

(iii) Volume of the cylinder = r2h

=22

7 × 3 × 3 × 7

= 198 cm3

Curved surface area is 132 cm2 Total surface area is 188.57cm2 andvolume of the cylinder is 198 cm3


2. The volume of a cylinder is 504 cm3 and height is 14 cm. Find its curvedsurface area and total surface area. Express answer in terms of .(3 marks)

Sol. Volume of a cylinder = 504 cm3

Its height (h) = 14 cmVolume of a cylinder = r2h

504 = × r2 × 14

504

14= r2

r2 = 36 r = 6 cm [Taking square roots]

Curved surface area of a cylinder = 2rh= 2 × × 6 × 14= 168 cm2

Total surface area of a cylinder = 2r (r + h)= 2 × × 6 (6 + 14)= 2 × × 6 × 20= 240 cm2

Curved surface area of a cylinder is 168cm2

and Total surface area of a cylinder is 240cm2


3. The radius and height of a cylinder are n same ratio 3:7 and its volumeis 1584 cm3. Find its radius. (3 marks)

Sol. The ratio of radius and height of a cylinder is 3 :7Let the common multiple be ‘x’

Radius of cylinder (r) = ‘3x’ cmand its height (h) = ‘7x’ cm

Volume of a cylinder = 1584 cm3

Volume of a cylinder = r2h

1584 =22

7 × (3x) × (3x) × (7x)

1584 = 22 × 9 × x3

1584

22 × 9 = x3

x3 = 8 x = 2 [Taking cube roots]


SCHOOL SECTION 311

Radius (r) = 3x= 3(2)= 6cm

Radius of the cylinder is 6 cm.


4. Keeping the height same, how many times the rod of the given cylindershould be made to get the cylinder of double the volume of given cylinder ?

(3 marks)Sol. 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

r h 22 = r h 2

1

r22 = 2r2

1

r2 = 2 r1 [Taking square roots]

The radius of the required cylinder should be 2 times the radiusthe given cylinder


1. A cylindrical hole of diameter 30 cm is boredthrough a cuboid wooden block with side 1meter. Find the volume of the object so formed( = 3.14) (4 marks)

Sol. 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) =30

2= 15 cm

Volume of cylindrical hole = r2h

= 3.14 × 15 × 15 × 100

= 70650 cm3

Volume of the object so formed = Volume of cubical wooden block

– Volume of cylindrical hole

= 1000000 – 70650

= 929350 cm3

Volume of the object so formed is 929350 cm3.


SCHOOL SECTION312

EXERCISE - 6.8 (TEXT BOOK PAGE NO. 185)4. 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 cmrespectively, find the number of refills that can be filled with this ink.

(4 marks)Sol. Height of the cylindrical container (h) = 14cm

Its radius (r) = 6 cmVolume of cylindrical container = r2h

= × 6 × 6 × 14= 504 cm3

But, volume of ink filledin the cylindrical container = 91% of 504

=91

× 504100

cm3

Length of ball pen refill (h1) = 12mits inner diameter = 2 mm

Its radius (r1) = 1 mm

=1

10cm

Volume of the refill = r12 h1

= × 1

10 ×

1

10 × 12

=12

100

cm3

But, volume of ink filled = 84% of 12

100

=84 12

×100 100

cm3

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

10084 ×12

100 ×100

=91× 504 100 ×100

×100 84 ×12

= 4550

Number of refills that can be filled with this ink is 4550.


15. A building has 8 right cylindrical pillars whose cross sectional diameteris 1 m and whose height is 4.2 m. Find the expenditure to paint thosepillars at the rate of Rs. 24 per m2. (3 marks)

Sol. Diameter of a pillar = 1 m

Its radius (r) =1

2 m

Its height (h) = 4.2 mCurved surface area of a pillars = 2rh

= 2 × 22

7 ×

1

2 × 4.2

= 13.2 m2


SCHOOL SECTION 313

Curved surface area of 8 pillars = 8 × 13.2= 105.6 m2

Rate of painting = Rs. 24 per m2

Total expenditure = Area to be painted × Rate of painting= 105.6 × 24= 2534.40

Total expenditure to paint the pillars is Rs. 2534.40.


16. A 10 m deep well of diameter 1.4 m us dug up in a field and the earthfrom digging is spread evenly on the adjoining cuboid field. The lengthand breadth of that filled are 55m and 14 m respectively. Find thethickness of the earth layer spread. (4 marks)

Sol. Diameter of well = 1.4 m

Its radius (r) =1.4

2= 0.7 m

Its depth (h) = 10 mVolume of cylindrical well = r2h

=22

0.7 0.7 107

=22 7 7

107 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 fieldVolume of cuboid = Volume of earth dug

= 15.4 m3

Length of a cuboid (l) = 55 mIts breadth (b) = 14 m

Volume of cuboid = l × b × h 15.4 = 55 × 14 × h

154

10 × 55 ×14 = h

h =1

50m

h = 0.02 m

The thickness of the earth layer spread is 0.02 cm.


18. 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)Sol. Diameter of the roller = 0.9 m

its radius (r) =0.9

2= 0.45 m

its length (h) = 1.8 m


SCHOOL SECTION314

RIGHT CIRCULAR CONE

hl

rA P

O

Curved surface area of the roller = 2rh= 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 surfacearea 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

50010000

= 2543.4 m2

Area of the ground pressed by the roller = 2543.4 m2.

An ice-cream cone, a clown’s hat, a funnel are examplesof cones. A cone has one circular flat surface and onecurved surface.In the diagram alongside,seg OA is the height of the cone denoted by ‘h’.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 trianglewhere l is the hypotenuse.

l2 = r2 + h2.FORMULA

Volume of a right circular cone = 13

× r2h


1. The curved surface area of a cone is 4070 cm2 and its diameter is70 cm. What is its slant height ? (2 marks)

Sol. Diameter of a cone = 70 cm.

Its radius (r) =70

2= 35 cm.

Curved surface area of a cone = 4070 cm2

Curved surface area of a cone = r l

4070 =22

357

l

4070

22 × 5 = l

l = 37

Slant height of a cone is 37 cm.


2. The base radii of two right circular cones of the same height are in theratio 2:3. Find the ratio of their volumes. (3 marks)

Sol. Let the radii of two right circular cone be r1 and r2 and their volumesbe v1 and v2 respectively


SCHOOL SECTION 315

r

r1

2=

2

3.......(i) (Given)

v

v1

2=

1r h

31

r h3

21

22

v

v1

2=

r

r

2122

v

v1

2=

r

r

2

1

2

v

v1

2=

2

3

2

[From (i)]

v

v1

2=

4

9 v1 : v2= 4 : 9

Ratio of volumes of two right circular cone is 4 : 9


3. A cone of height 24 cm has a plane base of surface area 154 cm2. Findits volume. (2 marks)

Sol. Height of a cone (h) = 24 cmSurface area of base = 154 cm2

Volume of a cone =1

3 × Surface area of base × height

=1

3 × r2 × h

=1

3 × 154 × 24

= 1232 cm3

Volume of the cone is 1232 cm3.


4. Curved surface area of a cone with base radius 40 cm is 1640 sq.cm.Find the height of the cone. (3 marks)

Sol. Curved surface area of a cone = 1640cm2

its radius (r) = 40 cm.Curved surface area of a cone = r l

1640 = × 40 × l

1640

40= l

l = 41 cmNow,

r2 + h2 = l2

402 + h2 = 412

h2 = 412 – 402

h2 = 1681 – 1600 h2 = 81 h = 9 cm [Taking square roots] Height of a cone is 9 cm.


SCHOOL SECTION316


17. The total surface area of cone is 71.28 cm2. Find the volume of thiscone if the diameter of the base is 5.6 cm. (3 marks)

Sol. Diameter of the base = 5.6 cm

Its radius (r) =5.6

2= 2.8 cm

Total surface area of a cone = 71.28 cm2

Total surface area of cone = r (r + l)

71.28 =22

7 × 2.8 (2.8 + l)

7128

100=

22

7 ×

28

10 (2.8 + l)

7128 10

100 22 4

= 2.8 + l

81

10= 2.8 + l

8.1 – 2.8 = l l = 5.3 cm

r2 + h2 = l2

(2.8)2 + h2 = (5.3)2

h2 = (5.3)2 – (2.8)2

h2 = (5.3 + 2.8) (5.3 – 2.8) h2 = 8.1 × 2.5

h2 =81 × 25

10 ×10

h = 9 × 5

10[Taking square roots]

h = 45

10 h = 4.5 cm

Volume of a cone =1

r h3

2

=1 22

2.8 2.8 4.53 7

=1 22 28 28 45

3 7 10 10 10

=36960

1000

Volume of a cone is 36.96 cm3.


19. 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 radiusis 10 cm. Find total surface area of the right circular cylinder so formed.(Given = 3.14) (4 marks)

Sol. Diameter of the base of metallic cone = 2 cm

Its radius (r) = 2

2 = 1 cm

Its height (h) = 10 cm


SCHOOL SECTION 317

Volume of a metallic cone =1

r h3

2

=1

3 × × 1 × 1 × 10

=10

3

cm3

Volume of 900 metallic cones =10

9003

= 3000 cm3

900 cones are melted to form a right circular cylinder Volume of a cylinder = 3000

For a cylinder, Radius (r2) = 10 cm and height be h2

Volume of a cylinder = r h21 1

3000 = × 10 × 10 h2

h1 = 30 cmTotal surface area of cylinder = 2r1 (r1 + h1)

= 2 × 3.14 × 10 (10 + 30)= 6.28 × 40= 2512 cm2

Total surface area of the right circular cylinder is 2512 cm2.


20. The volume of a cone of height 5 cm is 753.6 cm3. This cone and a cylinderhave equal radii and height. Find the total surface area of cylinder.(Given = 3.14) (3 marks)

Sol. Height of a cone (h) = 5 cmVolume of a cone = 753.6 cm3

Volume of a cone =1

r h3

2

753.6 =1

× 3.14 × r × 53

2

7536

10=

1 314× × r × 5

3 1002

7536 × 3 ×100

10 × 314 × 5 = r2

r2 = 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 = 2r (r + h)

= 2 × 3.14 × 12 (12 + 5)

= 75.36 × 17

= 1281.12 cm2

Total surface area of a cylinder is 1281.12 cm2.


SCHOOL SECTION318

FRUSTUM OF THE CONE :If the cone is cut off by a plane parallel to the base notpassing 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)

FORMULAE

1. Slant height (l) of the frustum = 221 2h + r – r

2. Curved surface area = p (r1 + r2) l

3. Total surface area of the frustum = 2 21 2 1 2r + r + r + r l

4. Volume of the frustum = 2 21 2 1 2

1r + r + r × r h

3


5. The curved surface area of the frustum of a cone is 180 sq. cm and theperimeters of its circular bases are 18 cm and 6 cm respectively. Findthe slant height of the frustum of a cone. (3 marks)

Sol. Curved surface area of the frustum of a cone = 180 cm2

Perimeters of circular bases are 18 cm and 6 cm 2r1 = 18 ........(i)

2r2 = 6 ........(ii)Adding (i) and (ii), we get2r1 + 2r2 = 18 + 6

2 (r1 + r2) = 24

(r1 + r2) =24

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.


21. The height of a cone is 40 cm. A small cone is cut off at the top of a

plane parallel to its base. If its volume be 164

of the volume of the given

cone, at what height above the base is the section cut ? (5 marks)

Sol. Let the radius, height and volume of thesmaller cone be r1, h1 and v1 respectively.The radius height and volume of thegiven 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]

hl

r1

r2

hl

r2

r1

A

F

B

E

C


SCHOOL SECTION 319

SPHERE

AE

AB=

EF

BC[c.s.s.t.]

h

h1

2=

r

r1

2......(i)

V1 =1

v64 2 [Given]

V

V1

2=

1

64......(ii)

V

V1

2=

1 r h31 r h3

21 1

22 2

1

64=

r h×

r h

2

1 1

2 2

1

64=

h h×

h h

2

1 1

2 2[From (i)]

1

64=

h

h

3

1

2

h

h1

2=

1

4

h

401 =

1

4

h1 =40

4 h1 = 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.

The set of all points of space which are at a fixeddistance from a fixed point is called a sphere.The fixed point is called the centre and the fixeddistance is called the Radius of the sphere.In the adjoining figure, point O is the centre of thesphere and seg OA is the radius of the sphere which isdenoted as ‘r’.Since the entire surface of the sphere is curved, itsarea is called as curved surface area or simply surfacearea 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 :

FORMULAESurface area (curved surface area) of a sphere = 4r2

Volume of a sphere = 43

× r3

h•O Ar


SCHOOL SECTION320


1. Find the volume and surface area of a sphere of radius 4.2 cm.

22=

7

(3 marks)Sol. Radius of a sphere (r) = 4.2 cm

Volume of a sphere =4

3r3

=4

3 ×

22

7 × 4.2 × 4.2 × 4.2

=4

3 ×

22

7 ×

42

10 ×

42

10×

42

10

=310464

1000= 310.464= 310.46 cm3

Surface area of a sphere = 4r2

= 4 × 22

7 × 4.2 × 4.2

= 4 × 22

7 ×

42

10 ×

42

10

=22176

100= 221.76 cm2

Volume of sphere is 310.46 cm3 and surface area of a sphere is 221.76 cm2.


2. The volumes of two spheres are in the ratio 27 : 64. Find their radii ifthe sum of their radii is 28 cm. (3 marks)

Sol. Let the radii of two spheres be r1 and r2 and their volumes be v1 and v2.v

v1

2=

27

64........(i) [Given]

v

v1

2=

4r

34

r3

31

32

27

64=

4r

34

r3

31

32

r

r

3132

=27

64

r

r1

2=

3

4[Taking cube roots]

Let the common multiple be x r1 = 3x and r2 = 4x

r1 + r2 = 28 [Given] 3x + 4x = 28 7x = 28

x =28

7 x = 4


SCHOOL SECTION 321

r1 = 3x r2 = 4x r1 = 3 (4) r2 = 4 (4) r1 = 12 cm r2 = 16 cm

Radii of two spheres are 12 cm and 16 cm.


3. The surface area of a sphere is 616cm2. What is its volume ?

22=

7

(3 marks)Sol. Surface area of sphere = 616 cm2

Surface area of a sphere = 4r2

616 = 4 × 22

7 × r2

616 7

4 22

= r2


Volume of a sphere =4

3r3

=4 22

7 7 73 7

=4312

3= 1437.33 cm3

The volume is 1437.33 cm3.


4. If the radius of a sphere is doubled, what will be the ratio of its surfacearea and volume as to that of the first sphere.

Sol. Let the original radius be ‘r1’ Original surface Area (A1)= 4r1

2

New radius (r2) = 2r1 New surface Area (A2) = 4r2

2

= 4 × × (2r1)2

= 4 × × 4r12

= 16r12

Ratio of New surface area to the original surface area =A

A2

1

=

16 r

4 r

2121

=4

1 Ratio of New surface area to the original surface area is 4 : 1

Now, original volume (v1) =4

3r1

3

New volume (v2) =4

3r2

3

=4

3(2r1)

3

=4

3 ×× 8r1

3

v2 =32

3 r13


SCHOOL SECTION322

HEMISPHERE

Ratio of New volume to the original volume =v

v2

1

=

32r

34

r3

31

31

=32

4

=8

1

Ratio of New volume to the original volume is8 : 1

Half of a sphere is called as hemisphere.Any hemisphere is made up of a curved surface and a planecircular surface.The following are the formulae for the surface area of a hemisphere :

FORMULAE 1. Curved surface area of a hemisphere = 2r2

2. Total surface area of a hemisphere = 3r2

3. Volume of a hemisphere = × r323


5. The curved surface area of a hemisphere is 90517

cm2, what is its volume?

(3 marks)

Sol. Curved surface area of a hemisphere = 9051

7cm2

Curved surface area of a hemisphere = 2r2

9051

7= 2 ×

22

7 × r2

6336

7= 2 ×

22

7 × r2

6336 7

7 2 22

= r2


Volume of a hemisphere =2

3r3

=2

3 ×

22

7 × 12 × 12 × 12

=25344

7= 3620.57 cm3

Volume of a hemisphere is 3620.57 cm3.


SCHOOL SECTION 323


14. The following shapes are made up of cones and hemispheres. Find theirvolume. (3 marks)

Sol. (a) For a hemisphere,Height = radius (r) = 3.5 cmVolume of the figure = Volume of cone + Volume of hemisphere

=1

3r1h +

2

3r3

=1

3r2 [h + 2r]

=1 22

3.5 3.5 [12.3 2(3.5)]3 7

=1 22 35 35

(12.3 7)3 7 10 10

=1 385

19.33 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

Its radius r1 = 3

2 = 1.5 cm

Diameter of bigger figure = 10 cm

Its radius r2 = 10

2 = 5 cm

Volume of the figure = Volume of smaller hemisphere +volume of bigger hemisphere

=2 2

r r3 3

3 31 2

= 2r r

33 3

1 2

=2 22

(1.5) 53 7

3 3

=2 22

(3.375 125)3 7

=2 22

128.3753 7

=5648.5

21= 268.98 cm3

Volume of the given figure is 268.98 cm3.

12.3 cm

3.5 cm

3 cm

10 cm


SCHOOL SECTION324


2. A toy is a combination of a cylinder, hemisphere and a cone, each with radius10cm. 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)

Sol.

A toy is a combination of cylinder, hemisphere and cone, each withradius 10 cm

r = 10 cm Height of the conical part (h) = 10 cm

Height of the hemispherical part = its radius = 10cmTotal height of the toy = 60cm

Height of the cylindrical part (h1) = 60 – 10 – 10 = 60 – 20 = 40 cml2 = r2 + h2

l2 = 102 + 102

l2 = 100 + 100l2 = 200

l = 200 [Taking square roots]

l = 10 2 cm

Slant height of the conical part (l) = 10 2= 10 × 1.41= 14.1 cm

Total surface area of the toy = Curved surface area of the conicalpart + Curved surface area of thecylindrical part + Curved surfacearea of the hemispherical part

= rl + 2rh1 + 2r2

= 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.


3. A test tube has diameter 20 mm and height is15 cm. The lower portion is a hemisphere inthe adjoining figure. Find the capacity of thetest tube. ( = 3.14) (5 marks)

Sol. Diameter of a test tube = 20 mm

its radius (r) =20

2= 10 mm= 1 cm

Its height (h) = 15 cmHeight of hemispherical part (h1) = radius of hemisphere

= 1 cm Height of cylindrical part (h2) = h – h1

= 15 – 1= 14 cm

60 cm

10 cm10 cm

10 c

m

15 cm


SCHOOL SECTION 325

Volume of test tube = Volume of cylindrical part +Volume of hemispherical part

= r2h2 + 2

3r3

= r22

h + r3

2

= 3.14 (1) 2

14 +3

= 3.14 × 44

3

=138.16

3Volume of test tube = 46.05 cm3

Capacity of a test tube is 46.05 cm3


5. A cylinder of radius 12 cm contains water upto depth of 20 cm. Aspherical iron ball is dropped into the cylinder and thus water level israised by 6.75 cm. what is the radius of the ball ? (5 marks)

Sol. Radius of the cylinder (r) = 12 cmA spherical iron ball is dropped into the cylinder and the water levelrises by 6.75 cm

Volume of water displaced = volume of the iron ballHeight of the raised water level (h) = 6.75 mVolume of water displaced = r2h

= × 12 × 12 × 6.75 cm3

Volume of iron ball = × 12 × 12 × 6.75 cm3

But, Volume of iron ball =4

r3

3

× 12 × 12 × 6.75 =4

3 × × r3

12 ×12 × 6.75 × 3

4= r3

r3 = 3 × 12 × 6.75 × 3 r3 = 3 × 3 × 3 × 4 × 6.75 r3 = 3 × 3 × 3 × 27 r = 3 × 3 × 3 × 3 × 3 × 33 [Taking cube roots] r = 3 × 3 r = 9

Radius of the iron ball is 9 cm.


22. A piece of cheese is cut in theshape of the sector of a circle ofradius 6 cm. The thickness of thecheese is 7 cm. Find(i) The curved surface area of the cheese.(ii) The volume of the cheese piece. (4 marks)

Sol. For a sector,Measure of arc () = 60ºRadius (r) = 6 cm

6.75 cm

20 cm

60º6 cm

7 c

m


SCHOOL SECTION326

(i) Curved surface area of the cheese = Length of arc × height

= 2 r h360

=60 22

2 6 7360 7

= 44 cm2

The curved surface area of the cheese is 44 cm2.

(ii) Volume of the cheese piece = A (sector) × height

= r h360

2

=60 22

6 6 7360 7

= 132 cm3

The volume of the cheese piece 132 cm3.


1. Which are polyhedrons from the following ?

Sol. (i) No (ii) Yes (iii) Yes (iv) Yes (v) No


2. Using Euler’s formula, find V, if E = 30, F = 12. If the solid figure is aprism, how many sides the base polygon has. (1 mark)

Sol. F + V = E + 2 12 + V= 30 + 2 V = 32 – 12 V = 20

Number of sides of base polygon = 1

202 = 10


3. Verify Euler’s formula for these solids :(i) (2 marks)Sol. F = 8, V = 12, E = 18

L.H.S. = F + V= 8 + 12

L.H.S. = 20 ......(i)R.H.S. = E + 2

= 18 + 2 R.H.S. = 20 ......(ii) L.H.S. = R.H.S. [From (i) and (ii)] F + V = E + 2

Diamond Test tubeTileUnsharpenedPencil

Nail

(i) (ii) (iii) (iv) (v)


SCHOOL SECTION 327

(ii) (2 marks)Sol. F = 8, V = 6, E = 12

L.H.S. = F + V= 8 + 6

L.H.S. = 14 ......(i)R.H.S. = E + 2


(iii) (2 marks)Sol. F = 8, V = 12, E = 18

L.H.S. = F + V= 8 + 12

L.H.S. = 20 ......(i)R.H.S. = E + 2


(iv) (2 marks)Sol. F = 6, V = 6, E = 10

L.H.S. = F + V= 6 + 6

L.H.S. = 12 ......(i)R.H.S. = E + 2


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 4r2 = 6l2 ......(i)

r2

2l=

6

4

r2

2l=

3

2

r

l=

3

2 × ......(ii)

Volume of sphere

Volume of cube =4 r3 3

3l

=4 r

3

3

3

l


SCHOOL SECTION328

=4 r × r

3 ×

2

2l l

=4 r r

×3

2

2l l

=6 r

×3

2

2

l

l l[From (i)]

=r

2 ×l

=3

2 ×2 ×

=2 × 2 × 3

2 ×

Volume of sphere

Volume of cube =6

Ratio of the volume of the sphere to that of the cube is 6

48. Marbles of diameter 1.4 cm are dropped into a beaker containing somewater and are fully submerged. The diameter of the beaker is 7 cm. Findhow many marbles have been dropped in it if the water rises by 5.6 cm.

(5 marks)Sol. Diameter of marble = 1.4 cm

its radius (r) =1.4

2= 0.7 cm

Volume of a marble =4

r3 3

=4 7 7 7

× × × ×3 10 10 10

cm3

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

Its radius (r1) =7

2cm

Volume of water displaced = r h 21

=7 7 56

× × × cm2 2 10

3

Number of marbles =Volume of water displaced

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.

49. Water flows at the rate of 10 m per minute through a cylindrical pipehaving its diameter is 20 mm. How much time will it take to fill aconical vessel of base diameter 40 cm and depth 24 cm ? (5 marks)

Sol. Diameter of conical vessel = 40 cm

Its radius (r) = 40

2 = 20cm


SCHOOL SECTION 329

Its depth (h) = 24 cm

Volume of conical vessel =1

r h3 2

=1

× × 20 × 20 × 243

= 20 × 20 × 8 × = 3200 cm3

Diameter of cylindrical pipe= 20 mm

Its radius (r1) =20

2= 10 mm

=10

10 cm [ 1 cm = 10 mm]

= 1 cmWater 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= r h 2

1

= × 1 × 1 × 1000= 1000 cm3

Time taken to fill conical vessel =Volume of conical vessel

Volume of water flowing in 1 minute

=3200

1000

=32

10mins

=32

× 6010

secs [1 minute = 60 seconds]

= 192 seconds= 3 minutes and 12 seconds

The time taken to fill the conical vessel is 3 minutes and 12 seconds.

50. Find the length of 13.2 kg. copper wire of diameter 4 mm, when 1 cubiccm of copper weighs 8.4 gm. (4 marks)

Sol. Volume of 8.4 gm of copper = 1 cm3

Volume of 13.2 kg i.e. 13200 gm of copper = 13200

8.4

=13200 ×10

84

=11000

cm7

3

Diameter of copper wire = 4 mm

Its radius (r) =4

2= 2 mm

=2

10 cm


SCHOOL SECTION330

Volume of copper wire = r2h

11000

7=

22 2 2× × × h

7 10 10

11000 × 7 ×10 ×10

7 × 22 × 2 × 2 = h

h = 12500 cm h = 125 m [ 1 metre = 100 cm]

Length of wire is 125 m.

51. A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4cm thick iron. Find the volume of the iron. (4 marks)

Sol. Girth of garden roller = 440 cmGirth of garden roller = 2 r

2r = 440

22

2 × × r7

= 440

r =440 × 7

2 × 22 r = 70 cm Radius of outer cylinder (r) = 70 cm

Width of the roller (h) = 63 cmThickness 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 – r h 21

= h r – r 2 21

=22

× 63 × 70 – 667

2 2

= 198 (70 66) (70 – 66)= 198 × 136 × 4= 107712 cm3

The volume of the iron is 107712 cm3.

52. 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)

Sol. Diameter of semicircle sheet = 28 cm

Its radius (r) =28

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

×147

= 44 cm


SCHOOL SECTION 331

Let the radius of a cone be r1

Circumference of a base of a cone = 2r1

44 = 2r1

44 =22

2 × × r7 1

44 × 7

2 × 22 = r1

r1 = 7 cm

l2 = r h2 21

142 = 72 + h2

h2 = 142 – 72

h2 = 196 – 49 h2 = 147

h = 147

h = 49 × 3

h = 7 3 h = 7 × 1.73 h = 12.11 cm

Volume of conical cup =1

r h3 2

1

=1 22

× × 7 × 7 ×12.113 7

=1864.94

3= 621.646= 621.65 cm3

Depth of a conical cup is 12.11 cm and volume of conical cup is 621.65 cm3.

53. A cone and a hemisphere have equal bases and equal volumes. Find theratio of their heights. (3 marks)

Sol. A cone and a hemisphere have equal bases their radii are equal

Height of hemisphere = radius of hemisphereVolume of a cone = Volume of hemisphere

1

r h3 2

=2

r3 3

h = 2r

h

r=

2

1 Ratio of heights of a cone and a hemisphere is 2 : 1

54. A bucket is in the form of a frustum of a cone and holds 28.490 litres ofwater. The radii of the top and bottom are 28 cm and 21 cm respectively.Find the height of the bucket. (4 marks)

Sol. Radii of circular ends are 25 cm and 21 cm r1 = 28 cm and r2 = 21 cm

Volume of bucket = 28.490 litres= 28.490 × 1000 cm3 [ 1 litre = 1000 cm3]= 28490 cm3


SCHOOL SECTION332

Volume of bucket = 1r r r r × h

3 2 2

1 2 1 2

28490 = 1 22× 28 21 28 × 21 × h

3 7 2 2

28490 = 22784 441 588 × h

21

28490 =22

×1813 × h21

28490 × 21

22 ×1813 = h

h = 15 cm

The height of the bucket is 15 cm.

55. A oil funnel of tin sheet consists of a cylindrical portion 10 cm longattached to a frustum of a cone. If diameter of the top and bottom ofthe frustum is 18 cm and 8 cm respectively and the slant height of thefrustum of cone is 13 cm. Find the surface area of the tin required tomake the funnel. (Express your answer in terms of ) (4 marks)

Sol. Diameters of circular ends of frustum are 18 cm and 8 cm

r1 = 18

2 = 9 cm and r2 =

8

2 = 4 cm

Slant height (l) = 13 cmCurved surface area of frustum of frustum = (r1 + r2) l

= (9 + 4) × 13= × 13 × 13= 169 cm2

Radius of a cylinder (r2) = 4 cmIts height (h) = 10 cm

Curved surface area of a cylinder = 2r2h= 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. There are 3 stair-steps as shown in the figure. Each stair-step has width25 cm, height 12 cm and length 50 cm. How many bricks have beenused in it if each brick is 12.5 cm × 6.25 cm × 4 cm. (5 marks)

Sol. Length of a stair-step (l) = 50 cmits breadth (b) = 25 cmits 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 cmits breadth (b2) = 6.25 cmits height (h1) = 4 cm


SCHOOL SECTION 333

Volume of a brick = l1 × b1 × h1

= 12.5 × 6.25 × 4

= 312.5 cm3

Number of bricks required =Volume of 3 stair-steps

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)

Proof : L.H.S. =1

V

L.H.S. =1

abc.......(i)

R.H.S. =2 1 1 1

S a b c

=2 bc ac ab

2 (ab bc ac) abc

=1 (ab + bc + ac)

×ab + bc + ac abc

R.H.S. =1

abc......(ii)

L.H.S. = R.H.S. [From (i) and (ii)]

1V

=

2 1 1 1+ +

S a b c

MCQ’s1. 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. Whatis the area of sector ?(a) 78.5 cm2 (b) 77 cm2

(c) 35.8 cm2 (d) 38.5 cm2

3. P is the centre of a circle, m (arc RYS) = 60º. If the radius of the circle is4.2 cm, what is the perimeter of P-RYS ?(a) 4.4 m (b) 8.4 m(c) 12.8 m (d) 17.2 m


SCHOOL SECTION334

4. The radius of a circle is 3.5 cm. What is the measure of arc whose length is5.5 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 coneof 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 radius8 cm ?(a) 60 cm2 (b) 68 cm2

(c) 120 cm2 (d) 136 cm2

8. How many solid metallic spheres each of diameter 6 cm are required to bemelted 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. Whatis its curved surface area ?(a) 314 cm2 (b) 157 cm2

(c) 78.5 cm2 (d) 100 cm2

10. The radii of the circular ends of a bucket which is 24 cm high are 14 cmand 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 diameter2 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 differencein surface area of the resulting cuboid and the surface area of two cubes ?(a) 288 m2 (b) 144 m2

(c) 1440 m2 (d) 770 m2

13. Area of a sector with central angle 60º will be ............ of area of a circle.

(a)2

3

rd

(b)1

6

th

(c)1

2(d)

1

4

th

14. If area of semicircle is 77 cm2 its perimeter is ................. .(a) 72 cm (b) 120 cm(c) 36 cm (d) 40 cm


SCHOOL SECTION 335

15. Area of minor segment AXB is ................ cm2.(a) 38.5 cm2 (b) 24.5 cm2

(c) 14.0 cm2 (d) 154 cm2

16. The capacity of a bowl is 144cm3. 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 ballsof 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 areawill 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 meltedto 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 rope8 m long is ............... .(a) 16 cm2 (b) 64 cm2

(c) 48 cm2 (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) = 1

12 A (circle). Hence, measure of the corresponding central

angle will be ................ .(a) 30 (b) 45(c) 60 (d) 90

25. A cylinder and a cone have equal radii and equal heights. If the volume ofthe cylinder is 300 cm3, then what is the volume of the cone ?(a) 100 cm3 (b) 10 cm3

(c) 110 cm3 (d) 300 cm3

O 7 B

x

A


SCHOOL SECTION336

: ANSWERS :1. (a) 200 cm 2. (d) 38.5 cm2

3. (c) 12.8 m 4. (d) 90º

5. (c) 8 cm 6. (b) F + V = E + 2

7. (d) 136 cm2 8. (c) 15

9. (b) 157 cm2 10. (d) 25 cm

11. (d) 36 m 12. (a) 288 m2

13. (b)1

6

th

14. (c) 36 cm

15. (c) 14.0 cm2 16. (d) 6 cm

17. (b) 343 18. (b) 400000 l

19. (c) 2 20. (d) 375

21. (d) 4 : 1 22. (a) 16 cm2

23. (a) 90 24. (a) 30

25. (a) 100 cm2

1 Mark Sums1. Find the length of the arc when the corresponding central angle is 270º

and circumference is 31.4 cm.Sol. Measure of central angle () = 270º

Circumfernce (2r) = 31.4 cm

Length of the arc = 2 r360

l =270

× 31.4360

l =3 31.4

4

l =94.2

4

l = 23.55 cm

2. If length of an arc is 7 cm, 2r = 36, then find the angle subtended atthe centre by the arc.

Sol. Length of the arc (l) = 7 cm2r = 36

l = 2 r360

7 = 36360

=7 360

36

= 7 × 10

= 70º

The angle subtended at the centre by the arc is 70º.


SCHOOL SECTION 337

3. Find the area of a circle with radius 7 cm.Sol. Radius of circle (r) = 7 cm

Area of the circle = r2

=22

7 77

= 154 cm2

The area of a circle is 154 cm2.

4. Using Euler’s formula, write the value of V, if E = 30 and F = 12.Sol. F + V = E + 2

12 + V = 30 + 2 12 + V = 32 V = 32 – 12

V = 20

5. The area of a circle is 314 cm2 and the area of its minor sector is 31.4 cm2.Find the area of its major sector.

Sol. Area of major sector = Area of circle – Area of minor sector= 314 – 3.14= 282.6 cm2

The area of the major sector is 282.6 cm2.

6. Perimeter of one face of a cube is 24 cm. Find the length of its side.Sol. Perimeter of one face of a cube = 24 cm

Perimeter of one face of a cube = 4l 24 = 4l

l = 24

4 l = 6

The length of the side of a cube is 6 cm.

7. A cubical tank has each side of length 2 m. Find the capacity of the tankin cubic metres.

Sol. Side of a cubical tank (l) = 2 m

Volume of cubical tank = l3

= 23

= 8 m3

Capacity of the cubical tank is 8 m3.

8. If the radius is 2 cm and length of corresponding arc is 3.14 cm, find thearea of a sector.

Sol. Radius (r) = 2 cmLength of arc (l) = 3.14 cm

Area of sector = r

2l

= 3.14 × 2

2= 3.14 cm2

The area of a sector is 3.14 cm2.


SCHOOL SECTION338

9. The dimensions of a cuboid are 5 cm, 4 cm and 3 cm. Find its volume.Sol. Length of a cuboid (l) = 5 cm

Its breadth (b) = 4 cmIts height (h) = 3 cm

Volume of a cuboid = l × b × h= 5 × 4 × 3= 60 cm3

Volume of cuboid is 60 cm3.

10. Find the total surface area of a cube with side 1 cm.Sol. Length of side of cube (i) = 1 cm


= 6 (l)2

= 6 (1)= 6 cm2

The total surface area of cube is 6 cm2.

11. Volume of a cube is 1000 cm3, find the length of its side.Sol. Volume of a cube = 1000 cm3

Volume of a cube = l3

l3 = 100 l = 10 [Taking cube roots]

Length of the side of cube is 10 cm.

12. What is the volume of a cube with side 5 cm ?Sol. Side of a cube (l) = 5 cm

Volume of cube = l3

= (5)3

= 125 cm3

Volume of the cube is 125 cm3.

13. The area of a circle with radius R is equal to the sum of the areas ofcircles with radii 6 cm and 8 cm. What is the value of R ?

Sol. According to given condition,R2 = × (6)2 + × (8)2

R2 = [62 + 82] R2 = 36 + 64 R2 = 100 R = 10 [Taking square roots]

The value of R is 10.

14. The radius of the base of a cone is 7 cm and its height is 24 cm. What isits slant height ?

Sol. Radius of base of cone (r) = 7 cmIts height (h) = 24 cm

l2 = r2 + h2

l2 = 72 + 242

l2 = 49 + 576 l2 = 625 l = 25 [Taking square roots]

Slant height of cone is 25 cm.


SCHOOL SECTION 339

15. Using Euler’s formula, find F, if V = 6 and E = 12.Sol. F + V = E + 2

F + 6 = 12 + 2 F + 6 = 14 F = 14 – 6

F = 8

16. The area of a circle is 1368 cm2. What is the area of the sector of thecircle whose corresponding central angle is 120º ?

Sol. Area of circle = 1368 cm2

Area of sector of the circle whose corresponding central is 120º

=1

13683

= 456 cm2

Area of the sector is 456 cm2.

17. The corresponding central angle of an arc is 90º. What is the length ofthis arc, if the radius of the circle is 14 cm ?

Sol. Measure of central angle () = 90ºRadius (r) = 14 cm

Length of the arc (l) = 2 r360

l =90 22

2 14360 7

l = 22

Length of the arc is 22 cm.

18. 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 ?

Sol. Radius of circle (r) = 10 cmLength of arc (l) = 25 cm

Area of sector =r

2l

= 25 × 10

2= 25 × 2= 125 cm2

The area of the sector is 125 cm2.

19. A cylinder and a cone have equal radii and equal heights ? If the volumeof the cylinder is 300 cm3, what is the volume of the cone ?

Sol. A cylinder and cone have equal height and equal radii

Volume of cone =1

volume of cylinder3

=1

3003

= 100 cm3

Volume of the cone is 100 cm3.

20. What is the corresponding angle of a sector whose area is one-forth ofthe area of the circle ?

Sol. The corresponding angle of a sector whose area is one-forth of the areaof the circle is 90º.


SCHOOL SECTION340

21. The area of a sector with corresponding angle 45º is 8cm2. What is thearea of the circle ? ( = 3.14)

Sol. Area of sector = 8cm2

Measure of the arc ()= 45º

Area of sector = r360

2

8 = 45

r360

2

r2 = 8 × 360

45

r2 = 360

8 3.1445

r2 = 3.14 × 64 r2 = 200.96

Area of the circle is 200.96 cm2.

22. The dimensions of a cuboid are 3 cm × 9 cm × x cm. The volume of this cuboidis equal to the volume of a cube with side 6 cm. What is the value of x ?

Sol. Volume of cuboid = Volume of cube [Given] 3 × 9 × x = (6)3

3 × 9 × x = 6 × 6 × 6

x =6 6 6

3 9

x = 8

23. The area of a circle with radius 17 cm is equal to the sum of the areas ofcircles with radii r cm and 15 cm respectively. What is the value of r ?

Sol. According to given condition, (17)2 = r2 + (15)2

(17)2 = (r2 + 152) 172 = r2 + 152

r2 = 172 – 152

r2 = 289 – 225 r2 = 64

r = 8 [Taking square roots]

24. The radius of the base of a cone is 7 cm and its height is 24 cm. What isits slant height ?

Sol. Radius of base of cone (r) = 7 cmits height (h) = 24 cm

l2 = r2 + h2

l2 = 72 + 212

l2 = 49 + 576 l2 = 625 l = 25 [Taking square roots]

Slant height of cone is 25 cm

25. What is the corresponding central angle of a sector whose area is one-tenth the area of the circle ?

Sol. The corresponding central angle of a sector whose area is one tenththe area of the circle is 36º.

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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 ofthe sector is 0.49 m2.

Q.II. Attempt the following : (9)(i) Calculate the area of the shaded

region in the adjoining figure whereABCD is a square with side 8 cm each.

(ii) Two arcs of the same circle have their lengths in the ratio 4:5. Findthe ratio of the areas of the corresponding sectors.

(iii) In the adjoining figure,What will be the area of the partof the field in which the horse cangraze, if the pole was fixed on a sideexactly at the middle of the side?

Q.III. Solve the following : (12)(i) Adjoining figure depicts a racing track

whose left and right ends aresemicircular. The distance betweentwo inner parallel line segments is 70m and they are each 105 m long. Ifthe track is 7 m wide, find thedifference in the lengths of the inneredge and outer edge of the track.

CHAPTER 6 : Mensuratiion

SET - A

S.S.C.

GEOMETRY

Marks : 30

Duration : 1 hr. 15 min.

BA

D C8 cm

X

10 m 10 m

30 m

105 m

105 m

70 m70 m

7 m

7 m

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(ii) In the adjoining figure,seg QR is a tangent to the circlewith centre O. Point Q is the pointof contact. Radius of the circle is 10 cm.OR = 20 cm. Find the area of the

shaded region. ( = 3.14, 3 = 1.73 )

(iii) In the adjoining figure,PR and QS are two diameters of the circle.

If PR = 28 cm and PS = 14 3 cm, find

(a) Area of triangle OPS(b) The total area of two shaded segments.

( 3 = 1.73)

Q.IV. Solve the following : (5)(i) In the adjoining figure,

PR = 6 units and PQ = 8 units.Semicircles are draw takingsides PR, RQ and PQ as diametersas shown in the figure. Find out thearea of the shaded portion. ( = 3.14)

Q R

T

O

10 c

m

P S

Q R

O

120º

P

R Q

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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)(i) The volume of a cylinder is 504 cm3 and height is 14 cm. Find

its curved surface area and total surface area. Express answer interms of .

(ii) The total surface area of cone is 71.28 cm2. Find the volume of thiscone if the diameter of the base is 5.6 cm.

(iii) The curved surface area of a hemisphere is 9051

7cm2, what is its

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 levelis 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 10cm. 900 such cones are molten to form 1 right circular cylinderwhose radius is 10 cm. Find total surface area of the right circularcylinder so formed. (Given = 3.14)

(iii) 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

CHAPTER 6 : Mensuratiion

SET - B

S.S.C.

GEOMETRY

Marks : 30

Duration : 1 hr. 15 min.

60 cm

10 cm10 cm

10 c

m

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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 filledupto 84%. If the height and radius of the ink container are 14 cmand 6 cm respectively, find the number of refills that can be filledwith this ink.

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