mathamatics xam idea

64. With the vertices A, B and C of a triangle ABC as centres, arcs are drawn
with radii 6 cm each in Fig. 12.80. If AB = 20 cm, BC = 48 cm and
CA = 52 cm, then find the area of the shaded region. (Use p = 3.14)

Fig. 12.80

65. In Fig. 12.81, PQRS is a square lawn with side PQ = 42 metres. Two

circular flower beds are there on the sides PS and QR with centre

at O, the intersection of its diagonals. Find the total area of the two

flower beds (shaded parts). [CBSE (AI) 2015]

Fig. 12.81

Answers

1. (i) (b) (ii) (b) (iii) (b) (iv) (a) (v) (b)

2. (i) θ × 2πr + 2r (ii) quadrant (iii) 4p sq. units (iv) i rr (v) 4 : p
360 180

3. 44 cm 4. 50 cm 5. 140 cm2 6. 9 p cm2 7. 88 cm 8. 25 2 cm
r

9. 3 : 4 10. 3025 cm2 11. x2 cm2 12. pd cm 13. 9 : 16 14. 100°
2

15. p : 4 16. 3.5 cm 17. 4 p cm 18. 19.2p cm2 19. 8a cm 20. 144°

21. 7546 cm2 22. 7 cm 23. 88 cm 24. 30.96 cm2 25. 608 p cm 26. 25 p cm2

27. 30.5 cm2 28. 271.3 cm2 (Approx) 29. 128.25 cm2 30. 4.56 cm2 31. 924 cm2

32. 17.90 cm2 (Approx) 33. 3465 cm2 34. 154.88 cm2 35. 66.5 cm2 36. 145.33 cm2

37. 7.77 cm2 38. 86.6 cm2 39. ™517 40. 3.5 cm 41. 27104 m2
44. (32 + 2p)m2
42. (704 + 64 p) cm2 43. (248 – 4p)m2
52. 7.86 cm2
45. 4.05 cm2 46. 200 cm2 47. 80 cm2 49. 59.5 cm2
7 60. (180 – 8p) cm2 61. 1568 cm2
65. 504 m2
50. (i) 12.375 cm2 (ii) ™ 3.10 51. 15.84 km/h

53. ™ 68640 54. 560 55. 13.08 cm2 56. 150.92 cm2

57. d 25r + 25 n cm2 58. 168 cm2 59. 73 1 cm
4 2 3

62. 65 63. 1 : 8 : 17 64. 423.48 cm2

Areas Related to Circles 343

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) If θ is the angle (in degrees) of a sector of a circle of radius r, then area of the sector is

(a) rr2 i (b) rr2 i (c) 2rri (d) 2rri
360° 180° 360° 180°

(ii) An arc of a circle is of length 5π cm and the sector it bounds has an area of 20π cm2. The
radius of circle is

(a) 1 cm (b) 5 cm (c) 8 cm (d) 10 cm

(iii) A wire can be bent in the form of a circle of radius 35 cm. If it is bent in the form of a square,
then its area will be

(a) 3025 cm2 (b) 3025 cm2 (c) 1225 cm2 (d) 2450 cm2
2

(iv) If radius of circle is increased by 100% then its area will increased by

(a) 400% (b) 200% (c) 300% (d) 250%

2. Fill in the blanks. (3 × 1 = 3)

(i) If two circles touch internally, then the distance between their centres is equal to the
_____________ of their radii.

(ii) Distance moved by a rotating wheel in one revolution is equal to the _____________ of the
wheel.

(iii) The sector of a circle of angle 90° is also known as _____________ .

3. Solve the following questions. (3 × 1 = 3)

(i) If the radius of the circle is increased by 1 cm, then what is the ratio of the new circumference
to its new diameter?

(ii) If circumferences of two circles are equal, then what is the ratio between their areas?

(iii) Find the diameter of a circle whose area is equal to the sum of area of two circles with radii
24 cm and 7 cm.

Section B

QQ Solve the following questions. (3 × 2 = 6)
Fig. 12.82
4. In a circle of radius 21 cm, an arc subtends an angle of 60° at the
centre. Find the area of sector formed by the arc.

5. In Fig. 12.82, ABC is a triangle right angled at A. Semicircles are
drawn on AB and AC as diameters. Find the area of the shaded
region.

6. Find the area of the minor segment of a circle of radius 28 cm,
when the angle of the corresponding sector is 45°.

QQ Solve the following questions. (3 × 3 = 9)

7. A square of diagonal 18 cm is inscribed in a circle. Find the area included in the circle but not in
the square.

344 Xam idea Mathematics–X

8. In Fig. 12.83, ABCD is a square of side 14 cm. Semi-circles are drawn

with each side of square as diameter. Find the area of the shaded region.

 Use π = 22 
 7 

9. In Fig. 12.84, find the area of the shaded region, enclosed between two

concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°.  Use π = 22  Fig. 12.83
 7 

Fig. 12.84

QQ Solve the following questions. (3 × 5 = 15)

10. In Fig. 12.85, a chord AB of a circle, with centre O and radius 10 cm, that
subtends a right angle at the centre of the circle. Find the area of the minor
segment AQBP. Hence find the area of major segment ALBQA.

(Use p = 3.14)

11. A park is of the shape of a circle of diameter 7 m. It is surrounded by a path

of width of 0·7 m. Find the expenditure of cementing the path, if its cost is

™110 per sq. m.

12. In Fig. 12.86, ABCD is a trapezium with AB < DC, AB = 18 cm, DC = 32 cm Fig. 12.85

and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D

have been drawn, then find the area of the shaded region of the figure.

Fig. 12.86

Answers

1. (i) (a) (ii) (c) (iii) (a) (iv) (c)

2. (i) difference (ii) circumference (iii) quadrant
(iii) 50 cm
3. (i) π :1 (ii) 1 : 1
7. 92.57 cm2
4. 231 cm2 5. 39.285 cm2 6. (308 – 196 2 ) cm2
11. ™1863.40
8. 84 cm2 9. 410.67 cm2 1 0. 28.5 cm2, 285.5 cm2 12. 196 cm2
z z z

Areas Related to Circles 345

13 Surface Areas
and Volumes
BASIC CONCEPTS – A FLOW CHART

346 Xam idea Mathematics–X

cb I

(if hemisphere is solid)
b

Surface Areas and Volumes 347

MORE POINTS TO REMEMBER

 Surface area and volume of combinations of solids: To find the surface area and volume
of solids which are combinations of the basic solids like a cone, a cylinder, a sphere,
a hemisphere, a cube or a cuboid etc., following points should be kept in mind.

(i) To find the volume of combination of two or more basic solids, we
add their volumes.

(ii) To find the surface area of combination of two or more basic solids,
it is not necessary to add their surface areas directly.

For example, to find the surface area of solid given alongside, which is Fig. 13.1
combination of a cube and a hemisphere, the addition of surface area
of cube and hemisphere leads to an incorrect result.

 Frustum, its volume and surface area: If a right
circular cone is cut off by a plane parallel to its base,
then the portion of the cone between this plane and
the base is called the frustum of the cone.

Let h be the height, l the slant height and r1, r2 be the Fig. 13.2
radii of the circular bases of frustum ABCD, where
r1 > r2.

Let frustum be a portion of cone OAB of height h1,
slant height l1. (see Fig. 13.3)

Now OD = OC = l1 – l, OL = h1 – h
 DOMB ~ DOLC

⇒ OM = OB = MB ⇒ h1 h = l1 l = r1
OL OC LC h1 – l1 – r2

⇒ h1 = r1 and l1 = r1 ⇒ h1 – h = r2 = l1 – l = r2
h1 – r2 l1 – r2 h1 r1 l1 r1
h l

⇒ 1– h = r2 and 1 – l = r2 ⇒ h = r1 – r2 and l = r1 – r2 Fig. 13.3
h1 r1 l1 r1 h1 r1 l1 r1

⇒ h1 = hr1 and l1 = lr1
r1 – r2 r1 – r2

Now, volume of frustum ABCD (V) = Volume of cone OAB – Volume of cone ODC

⇒ V =13 rr12 h1 – 1 rr12 (h1 – h)= r #r12 h1 – r22 (h1 – h)-
3 3

⇒ = r * hr13 – r22 e hr1 – h o4 = r * hr13 – r22 e hr1 – hr1 + hr2 o4
3 r1 – r2 r1 – r2 3 r1 – r2 r1 – r2

= r * hr13 – hr23 4= r h_r13 – r23i
3 r1 – r2 r1 – r2 3 r1 – r2

= r h _r12 + r1 r2 + r22i [ a3 – b3 = (a – b)(a2 + ab + b2)]
3

348 Xam idea Mathematics–X

Again, curved surface area of frustum ABCD(S)
= curved surface area of cone OAB – curved surface area of cone ODC
⇒ S = pr1l1 – pr2(l1 – l)

= rr1 e lr1 o– rr2 ) r1 lr1 r2 – l 3 = r * lr12 – lr1 r2 – lr1 r2 + lr22 4
r1 – r2 – r1 – r2 r1 – r2

= r.l * r1 r12 – r22 4 = rl * r12 – r22 4
– r2 r1 – r2 r1 – r2

S = pl(r1 + r2) [ (a2 – b2) = (a + b)(a – b)]

\ Total surface area of frustum ABCD = rl (r1 + r2)+ rr12 + rr22 = r#l (r1 + r2)+ r12 + r22-

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. A funnel figure is the combination of [NCERT Exemplar]
(a) a cone and a cylinder Fig. 13.4
(b) frustum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) a hemisphere and a cone

2. If a marble of radius 2.1 cm is put into a cylindrical cup full of water of radius 5 cm and height

6 cm, then how much water flows out of the cylindricall cup? [NCERT Exemplar]

(a) 38.8 cm3 (b) 55.4 cm3 (c) 19.4 cm3 (d) 471.4 cm3

3. A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling

ice cream cones of radius 2 cm and height 7 cm upto its brim. How many children will get the

ice cream cones? [NCERT Exemplar]

(a) 163 (b) 263 (c) 363 (d) 463

4. The radii of the ends of a frustum of a cone of height h cm are r1 cm and r2 cm. The volume in
cm3 of the frustum of the cone is [NCERT Exemplar]
1 1
(a) 3 rh [r12 + r22 + r1 r2] (b) 3 rh [r12 + r22 – r1 r2]

(c) 1 rh [r12 – r22 + r1 r2] (d) 1 rh [r12 – r22 – r1 r2]
3 3

5. The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is

[NCERT Exemplar]

(a) 9.7 cm3 (b) 77.6 cm3 (c) 58.2 cm3 (d) 19.4 cm3

6. The shape of a gilli, in the gilli-danda game figure is a combination of [NCERT Exemplar]



(a) two cylinders Fig. 13.5
(c) two cones and a cylinder
(b) a cone and a cylinder
(d) two cylinders and a cone

Surface Areas and Volumes 349

7. A shuttle cock used for playing badminton has the shape of the combination of

[NCERT Exemplar]

(a) a cylinder and a sphere (b) a cylinder and a hemisphere

(c) a sphere and a cone (d) frustum of a cone and a hemisphere

8. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and
1
it is assumed that 8 space of the cube remains unfilled. Then the number of marbles that the

cube can accommodate is [NCERT Exemplar]

(a) 142296 (b) 142396 (c) 142496 (d) 142596

9. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded

to form a solid sphere. The radius of the sphere is [NCERT Exemplar]

(a) 21 cm (b) 23 cm (c) 25 cm (d) 19 cm

10. A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of
1
size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that 8 space is covered by the mortar.

Then the number of bricks used to construct the wall is [NCERT Exemplar]

(a) 11100 (b) 11200 (c) 11000 (d) 11300

11. A cylinder, a cone and a hemisphere are of the same base and of the same height. The ratio of

their volumes is

(a) 1 : 2 : 3 (b) 2 : 1 : 3 (c) 3 : 1 : 2 (d) 3 : 2 : 1

12. The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm respectively.
The curved surface area of the bucket is

(a) 4955 cm2 (b) 4952 cm2 (c) 4951 cm2 (d) 4950 cm2

13. If the height of right circular cylinder is equal to the diameter of the base, then its whole

surface area is equal to 2 3
4 3 2
(a) 3 rh2 (b) rh2 (c) rh2 (d) 4πh2

14. The radii of two cylinders are in the ratio 5 : 7 and their heights are in the ratio 3 : 5. The ratio

of their curved surface area is

(a) 3 : 7 (b) 7 : 3 (c) 5 : 7 (d) 3 : 5

15. The total surface area of a solid hemisphere of radius r is
(a) pr2 (b) 2pr2 (c) 3pr2 (d) 4pr2

16. The length of the diagonal of a cube is 8 3 cm. Its total surface area is
(a) 321 cm2 (b) 350 cm2 (c) 384 cm2 (d) 256 cm2

17. The length of the longest rod that can be placed in a 12m × 9m × 8m room is

(a) 15 m (b) 17 m (c) 21 m (d) 25 m

18. 1 m3 is equal to how many litres ?

(a) 100 (b) 1000 (c) 1 (d) 10

350 Xam idea Mathematics–X

19. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base
diameter 2 cm and height 16 cm. The diameter of each sphere is

(a) 4 cm (b) 3 cm (c) 2 cm (d) 6 cm

20. During conversion of a solid from one shape to another, the volume of the new shape will

(a) increase (b) decrease (c) remains unaltered (d) be doubled

21. Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is

(a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9

Answers

1. (b) 2. (a) 3. (c) 4. (a) 5. (d) 6. (c)
7. (d) 8. (a) 9. (a) 10. (b) 11. (c) 12. (d)
13. (c) 14. (a) 15. (c) 16. (c) 17. (b) 18. (b)
19. (c) 20. (c) 21. (d)

Fill in the Blanks [1 mark]

C omplete the following statements with appropriate word(s) in the blank space(s).

1. The volume of a hemisphere is _____________ the volume of a cylinder if its height and radius
is same as that of the cylinder.

2. A sharpened pencil is a combination of _____________ and _____________ shapes.
3. If we cut a cone by a plane parallel to its base, we obtain a _____________ and _____________ .
4. Solid figures are _____________ while plane figures are _____________ .
5. If the radius of a sphere is halved, its volume becomes _____________ time the volume of original

sphere.
6. Surahi is the combination of _____________ and _____________ .
7. The volume and surface area of a sphere are numerically equal, then the radius of sphere is

_____________ units.
8. In a right circular cone, the cross-section made by a plane parallel to the base is a

_____________ .
9. Volume of the frustum of cone is _____________ .
10. Total curved surface area of the frustum is _____________ .
11. The TSA, CSA stand for _____________ and _____________ respectively.
12. A shuttle cock used for playing badminton has the shape of the combination of _____________ of

cone and hemisphere.
13. _____________ is measured in square units.
14. In the gilli-danda game, the shape of a gilli is a combination of two cones and _____________ .
15. _____________ is measured in cubic units.

Answers

1. two-third 2. cylinder, cone 3. cone, frustum of a cone

4. 3-dimensional. 2-dimensional or cube, cuboid, etc. circle, square etc. 5. one-eighth

6. sphere, cylinder 7. 3 8. circle 9. 1 rh (r12 + r22 + r1 r2)
3

10. r (r1 + r2) l + rr12 + rr22 11. Total surface area, Curved surface area 12. Frustum

13. area 14. a cylinder 15. volume

Surface Areas and Volumes 351

Very Short Answer Questions [1 mark]

1. Two identical solid hemispheres of equal base radius r cm are struck together along their
bases. What will be the total surface area of the combination?

Sol. The resultant solid will be a sphere of radius r whose total surface area is 4pr2.

2. A solid ball is exactly fitted inside the cubical box of side a. What is the volume of the ball?

Sol. Diameter of the solid ball = edge of the cube = a

\    Volume of the ball = 4 ra a 3 = 4 # 1 ra3 = 1 ra3
3 2 3 8 6
k

3. A solid piece of iron in the form of a cuboid of dimension 49 cm × 33 cm × 24 cm is melted
to form a solid sphere. Find the radius of sphere.

Sol. Volume of iron piece = volume of the sphere formed

⇒ 49 × 33 × 24 = 4 pr3
3

⇒ r3 = 49 # 33 # 24 # 3 # 7 ⇒ r = 21 cm
4 # 22

4. A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of
1
size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that 8 space is covered by the mortar.
Find the number of bricks used to construct the wall.

Sol. Space occupied with bricks = 7 × volume of the wall
8

= 7 # 270 # 300 # 350
\ Number of bricks = 8

22.5 #11.25 # 8.75 7 # 270 # 300 # 350
8
space occupied with bricks = = 11,200

volume of one brick 22.5 #11.25 # 8.75

5. The radii of the ends of a frustum of a cone 40 cm high are 20 cm and 11 cm. Find its slant

height. [CBSE Delhi 2014]

Sol. l = h2 +(r1 – r2)2

= 402 +(20 – 11)2 = 1600 +81 = 1681 = 41 cm

6. Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of

hemisphere ? [CBSE Delhi 2017]

Sol. As per question

Volume of hemisphere = Surface area of hemisphere

2 rr3 = 3rr2 & r = 9 units
3 2

∴ d = 9 units

Short Answer Questions-I [2 marks]

1. What is the ratio of the volume of a cube to that of a sphere which will fit inside it?

Sol. Let edge of the cube be ‘a’.

Then, diameter of the sphere that will fit inside the given cube = a

\ Volume of the cube : Volume of the sphere

= a3 : 4 ra a 3 = a3 : 4 # 1 ra3 = a3 : 1 ra3 = 6 : r
3 2 3 8 6
k

352 Xam idea Mathematics–X

2. The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two
circular ends is 4 cm, find the height of the frustum.

Sol. Let r and R be radii of the circular ends of the frustum of the cone.

Then, R – r = 4, l=5

We know, l2 = (R – r)2 + h2

⇒ 52 = 42 + h2 or h2 = 25 – 16 = 9 ⇒ h = 3 cm

3. If the slant height of the frustum of a cone is 10 cm and the perimeters of its circular base are
18 cm and 28 cm respectively, what is the curved surface area of the frustum?

Sol. Let r and R be the radii of the two circular ends of the frustum of the cone.

Then, 2pr = 18 and 2pR = 28
and
⇒ r = 18 R = 28
2r 2r

⇒ r = 9 and R = 14
r r

Now, curved surface area of the frustum = p(r + R)l = ra 9 + 14 k × 10 = 23 × 10 = 230 cm2
r r

4. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular
ends are 18 cm and 6 cm. Find the curved surface area of the frustum. [NCERT, CBSE (AI) 2017]

Sol. We have, slant height, l = 4 cm

Let R and r be the radii of two circular ends respectively. Therefore, we have

2pR = 18 ⇒ pR = 9

2pr = 6 ⇒ pr = 3

\ Curved surface area of the frustum = (pR + pr)l

= (9 + 3) × 4 = 12 × 4 = 48 cm2 Fig. 13.6

5. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of

the hemisphere is 14 cm and the total height of the vessel is 13 cm.

Find the inner surface area of the vessel. [NCERT]

Sol. Here, radius of hemisphere = radius of cylinder = r cm = 7 cm

and height of cylinder, h = (13 – 7) cm = 6 cm

Now, inner surface area of the vessel Fig. 13.7

   = curved surface area of the cylindrical part + curved surface area of hemispherical part

  = (2prh + 2pr2) = 2pr (h + r)

   =2× 22 × 7 × (6 + 7) = 2 × 22 × 13 = 572 cm2
7
6. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to

1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of p.

[NCERT]

Sol. We have,

Height of cone is equal to its radius

i.e., h = r = 1 cm (Given)

Also, radius of hemisphere = r = 1 cm

Now, volume of the solid

= volume of the cone + volume of the hemisphere

Fig. 13.8

Surface Areas and Volumes 353

= 1 pr2h + 2 pr3
3 3

= 1 pr2 × r + 2 pr3 [ h = r]
3 3

= rr3 + 2 rr3 = r (1) 3 = r cm3
3 3

7. If the total surface area of a solid hemisphere is 462 cm2, find its volume. ;Take r = 22 E
7

[CBSE (AI) 2014]
Sol. Given, total surface area of solid hemisphere = 462 cm2

⇒ 3pr2 = 462 cm2

3× 22 × r2 = 462
7
r2 = 49
⇒ r = 7 cm

Volume of solid hemisphere = 2 pr3 = 2 × 22 × 7 × 7 × 7 = 718.67 cm3
3 3 7

Short Answer Questions-II [3 marks]

1. Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting

cuboid. [NCERT]

Sol. Let the length of each edge of the cube of volume 64 cm3 be x cm.

Then, Volume = 64 cm3

⇒ x3 = 64

⇒ x3 = 43 ⇒ x = 4 cm Fig. 13.9
The dimensions of cuboid so formed are

l = Length = (4 + 4) cm = 8 cm

b = Breadth = 4 cm and h = Height = 4 cm

\ Surface area of the cuboid = 2 (lb + bh + lh)

= 2 ( 8 × 4 + 4 × 4 + 8 × 4)

= 2 (32 + 16 + 32) = 160 cm2

2. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the

hemisphere can have? Find the surface area of the solid. [NCERT]

Sol. The greatest diameter that a hemisphere can have = 7 cm = l

Radius of the hemisphere (R) = 7 cm
2

\ Surface area of the solid after surmounting hemisphere

= 6l2 – pR2 + 2pR2 = 6l2 + pR2

= 6(7)2 + 22 #c 7 2
7 2
m

= 6× 49+ 22 # 7 # 7
7 2 2

= 294 + 38.5 = 332.5 cm2

354 Xam idea Mathematics–X

3. The dimensions of a solid iron cuboid are 4.4 m × 2.6 m × 1.0 m. It is melted and

recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm.

Find the length of the pipe. [CBSE (AI) 2017]

Sol. Let the length of pipe by h m.

Volume of cuboid = 4.4 × 2.6 × 1 m3

Inner and outer radii of cylindrical pipe are 30 cm, (30 + 5) cm = 35 cm

∴ Volume of material used = r (352 – 302) × h m3
1002
r
= 1002 × 65× 5h [Using a2 – b2 = (a + b)(a – b)]

Now r × 65 × 5h = 4.4 × 2.6
1002
7× 4.4 ×2.6×100×100
⇒ h= 22 × 65 × 5

⇒ h = 112 m

4. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The

total height of the toy is 15.5 cm. Find the total surface area of the toy. [NCERT]

OR

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its
circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

[CBSE (AI) 2017]

Sol. We have, CD = 15.5 cm and OB = OD = 3.5 cm

Let r be the radius of the base of cone and h be the height of conical part of the
toy.

Then, r = OB = 3.5 cm

h = OC = CD – OD = (15.5 – 3.5) cm = 12 cm

l = r2 + h2 = 3.52 +122 = 12.25+144 = 156.25 =12.5 cm

Also, radius of the hemisphere, r = 3.5 cm

\ Total surface area of the toy Fig. 13.10

= surface area of cone + surface area of hemisphere

= prl + 2pr2 = pr(l + 2r) = 22 × 3.5 (12.5 + 2 × 3.5)
7

= 22 × 3.5 × 19.5 = 214.5 cm2
7

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the

diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of

the remaining solid. [NCERT]

Sol. Here, we have

Edge of the cube = l = Diameter of the hemisphere

Therefore, radius of the hemisphere = l
2

\ Surface area of the remaining solid after cutting out the hemispherical

Depression = 6l2 – rc l 2 + 2rc l 2
2 2
m m

= = 6l2 + r× l2 = l2 (24 + r)
4 4

Surface Areas and Volumes 355

6. A tent is in the shape of a cylinder surmounted by a conical top. If

the height and diameter of the cylindrical part are 2.1 m and 4 m

respectively, and the slant height of the top is 2.8 m, find the area of

the canvas used for making the tent. Also, find the cost of the canvas
of the tent at the rate of ™ 500 per m2. (Note that the base of the tent

will not be covered with canvas). [NCERT]

Sol. We have, 4
2
Radius of cylindrical base = =2m

Height of cylindrical portion = 2.1 m Fig. 13.11

\ Curved surface area of cylindrical portion = 2prh

     =2× 22 × 2 × 2.1 = 26.4 m2
Radius of conical base = 2 m 7

Slant height of conical portion = 2.8 m 22
7
\ Curved surface area of conical portion = prl = × 2 × 2.8 = 17.6 m2

Now, total area of the canvas = (26.4 + 17.6)m2 = 44 m2

\ Total cost of the canvas used = ™ 500 × 44 = ™ 22,000

7. A medicine capsule is in the shape of a cylinder with two

hemispheres stuck to each of its ends (Fig. 13.12). The length

of the entire capsule is 14 mm and the diameter of the capsule

is 5 mm. Find its surface area. [NCERT]

Sol. Let the radius and height of the cylinder be r cm and h cm Fig. 13.12

respectively. Then,

r= 5 mm = 2.5 mm
2

and h = c14 – 2 # 5 m mm = 9 mm
2

Also, radius of hemisphere r = 5 mm
2
Fig. 13.13
Now, Surface area of the capsule

= curved surface of cylinder + surface area of two hemispheres
= 2prh + 2 × 2pr2 = 2pr (h + 2r)

= 2 # 22 # 5 # c9+2 # 5 m =2 # 22 # 5 #14 = 220 mm2
7 2 2 7 2

8. A wooden article was made by scooping out a hemisphere from each end of a
solid cylinder, as shown in Fig. 13.14. If the height of the cylinder is 10 cm, and
its base is of radius 3.5 cm, find the total surface area of the article. [NCERT]

Sol. We have, r = 3.5 cm and h = 10 cm

Total surface area of the article Fig. 13.14

= curved surface area of cylinder + 2 × curved surface area of hemisphere
= 2prh + 2 × 2pr2 = 2pr (h + 2r)

=2× 22 × 3.5 × (10 + 2 × 3.5) = 2 × 22 × 3.5 × 17 = 374 cm2
7 7

356 Xam idea Mathematics–X

9. Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical
depression at one end (Fig. 13.15). The height of the cylinder is 1.45 m and its radius is 30 cm.

Find the total surface area of the bird-bath. ccoTmamkeorn=ra2d72iums of the cylinder and
Sol. Let h be height of the cylinder, and r be the
hemisphere.

Then, the total surface area of the bird-bath

= curved surface area of cylinder + curved surface area of hemisphere

= 2prh + 2pr2 = 2pr (h + r)

=2× 22 × 30 (145 + 30) cm2
7

= 33,000 cm2 = 3.3 m2 Fig. 13.15

10. A juice seller was serving his customers using glasses as shown in Fig. 13.16. The inner

diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical

raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find

the apparent capacity of the glass and its actual capacity. (Use p = 3.14). [CBSE 2019 (30/5/1)]

Sol. Since, the inner diameter of the glass = 5 cm and height =10 cm,

the apparent capacity of the glass = pr2h

= (3.14 × 2.5 × 2.5 × 10) cm3 = 196.25 cm3

But the actual capacity of the glass is less by the volume of the hemisphere at the
base of the glass.

i.e., it is less by 2 pr3 = 2 × 3.14 × 2.5 × 2.5 × 2.5 cm3
3 3

= 32.71 cm3

So, the actual capacity of the glass

= apparent capacity of glass – volume of the hemisphere Fig. 13.16

= (196.25 – 32.71) cm3 = 163.54 cm3

11. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the

spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume
to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and

p = 3.14. [NCERT]

Sol. We have,

Radius of cylindrical neck = 1 cm and height of cylindrical neck = 8 cm

Radius of spherical part = 4.25 cm

Now, Volume of spherical vessel = pr2h + 4 pr3
3

= p(12) × 8 + 4 × p × (4.25)3
3

= 3.14 × ;8 + 4 # (4.25)3E
3

= 3.14 × [8 + 102.354]

= 3.14 × 110.354 = 346.51 cm3 Fig. 13.17

\   The answer found by the child is incorrect.

Hence, the correct answer is 346.51 cm3.

12. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to

form a platform 22 m by 14 m. Find the height of the platform. [NCERT]

Surface Areas and Volumes 357

Sol. Here, radius of cylindrical well = 7 m
2
Depth of cylindrical well = 20 m

Let H metre be the required height of the platform.

Now, the volume of the platform = volume of the earth dugout from the cylindrical well

i.e., 22 × 14 × H = pr2h

= 22 #c 7 2 # 20 = 22 # 7 # 7 # 20 =770 m3
7 2 7 2 2
m

⇒ H = 770 = 5 = 2.5 m
22 #14 2

\ Height of the platform = 2.5 m

13. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a

cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm? [NCERT]

Sol. We have, 1.75 = 0.875 cm
Radius of coin = 2

and thickness i.e., height = 2 mm = 2 cm = 0.2 cm
10

The shape of a coin will be like the shape of cylinder

\ Volume of the coin = pr2h = 22 × 0.875 × 0.875 × 0.2
7

Now, volume of the cuboid = 5.5 × 10 × 3.5 = Volume of the cuboid
\ Number of coins required to form a cuboid Volume of the coin

= 22 5.5 #10 # 3.5 = 400
7 # 0.875 # 0.875 # 0.2

14. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its

two circular ends are 4 cm and 2 cm. Find the capacity of the glass. [NCERT]

Sol. We have, R = 2 cm, r = 1 cm, h = 14 cm

\ Capacity of the glass = volume of the frustum

= 1 ph (R2 + r2 + Rr)
3

= 1 # 22 × 14 × [(2)2 + (1)2 + (2 × 1)]
3 7

44 44 308 2 Fig. 13.18
3 3 3 3
= × (4 + 1 + 2) = ×7= = 102 cm3

15. A fez, the cap used by the Turks, is shaped like the frustum of a cone
(Fig. 13.19). If its radius on the open side is 10 cm, radius at the upper base is
4 cm and its slant height is 15 cm, find the area of material used for making it.
[NCERT]

Sol. We have, R = 10 cm, r = 4 cm, l = 15 cm

\ Area of the material used for making the fez

= Surface area of frustum + Area of top circular section

= p (R + r) l + pr2 Fig. 13.19

358 Xam idea Mathematics–X

= 22 (10 + 4) × 15 + 22 ×4×4
7 7

= 22 × 14 × 15 + 22 × 16
7 7

= 22 (210 + 16) = 4972 = 710 2 cm2
7 7 7

Fig. 13.20

16. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm.

The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.

[CBSE (F) 2015]

Sol. Volume of smaller sphere = 4 pr3 = 4 p(3)3 = 4 p(27) = 36p
3 3 3

Volume of smaller sphere × density = mass

\ 36π (density of metal) = 1

Density of metal = 1
36r

\ Volume of bigger sphere × density = mass

4 p (R)3 # 1 =7
3 36r

       R3 = 7 # 36 # 3 = 7 × 9 × 3 ...(i)
4

Volume of new sphere = volume of smaller sphere + volume of bigger sphere

     4 p(R')3 = 4 pr3 + 4 pR3 (where R' is the radius of new sphere)
3 3 3

= 4 p(3)3 + 4 p(7 × 9 × 3) [using (i)]
3 3

   4 p(R')3 = 4 p [33 + 7 × 9 × 3]
3 3

(R')3 = [33 + 7 × 33]

     (R')3 = 33(1 + 7)

     (R')3 = 33 × 8

    (R')3 = 33 × 23

       R' = 3 × 2

       R' = 6 cm

\ Diameter of new sphere = 12 cm.

17. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which

is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km per

hour, in how much time will the tank be filled completely? [CBSE Delhi 2014]

Sol. Given, Diameter of tank = 10 m

Depth of tank (H) = 2 m

Internal diameter of pipe = 20 cm = 2 m
10

Rate of flow of water, v = 4 km / h = 4,000 m/h

Internal radius of pipe, r = 1 m
10

Surface Areas and Volumes 359

Let ‘t’ be the time taken to fill the tank.

\ Water flowing through pipe in t hours = Volume of tank

pr2 × v × t = pR2H

1 × 1 × 4,000 × t = 5 × 5 × 2
10 10

=t ü=ü4×ü,0×00 4 = 1 1 hours
4

18. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72

cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted

in this transfer. 36 [CBSE (AI) 2015]
2
Sol. Radius of hemispherical bowl, R = = 18 cm

Radius of cylindrical bottle, r = = 3 cm

Let height of cylindrical bottle = h

Since, 10% liquid is wasted, therefore only 90% liquid is filled into 72 cylindrical bottles.

\ Volume of 72 cylindrical bottles = 90% of the volume in bowl

⇒ 72 × pr2h = 90% of 2 pR3
3
2
72 × p × 3 × 3 × h = 90 # 3 × p × 18 × 18 × 18
100

h = 90 # 2 # r # #18 #18 #18
100 # 3 # r # 72 # 3 #3

h = 27 = 5.4 cm
5

19. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is

100 cm and the diameter of the hemispherical ends is 28 cm. Find the cost of polishing the

surface of the solid at the rate of 5 paise per sq. cm.

Sol. We have 28
2
r = radius of cylinder = radius of hemispherical ends = cm

h = height of the cylinder = 100 – 2 × 14 = 100 – 28 = 72 cm.

Total surface area

= curved surface area of cylinder + 2 × surface

area of hemispherical ends

= 2πrh + 2 × (2πr2)

= 2πrh + 4πr2 = 2πr (h + 2r)

=2× 22 × 14(72 + 28)
7
22
=2× 7 × 14 × 100 = 8800 cm2 Fig. 13.21

Rate of polishing = 5 paise per sq. cm.

Cost of polishing the surface = 8800 # 5 = ` 440
100
20. A hemispherical tank, of diameter 3 m, is full of water. It is being emptied by a pipe at the rate of

3 4 litre per second. How much time will it take to make the tank half empty? cUse r = 22 m
7 7

[CBSE (F) 2016]

360 Xam idea Mathematics–X

Sol. Radius of hemispherical tank = 3 m = 150 cm
2

Volume of water in the hemispherical tank = 2 pr3
3

= 2 # 22 × 150 × 150 × 150 cm3
3 7

Volume of water to be emptied = 1 # 2 # 22 # 150 #150 #150 L
2 3 7 1000

Time taken to empty the tank = 22 # 5 #15 #15 # 7 min
7 60 # 25

= 33 min = 16 1 min
2 2

21. The 3 th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The
4
water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water

in cylindrical vessel. [CBSE Delhi 2017]

Sol. Let the height of cylindrical vessel be h cm

According to question,

3 × volume of conical vessel = volume of cylindrical vessel
4

⇒ 3 × 1 × r ×5×5×24 = r ×10×10× h
4 3

⇒ h = 3 cm or 1.5 cm
2

22. A cylindrical tub, whose diameter is 12 cm and height 15 cm is full of ice cream. The whole

ice-cream is to be divided into 10 children in equal ice-cream cones, with conical base

surmounted by hemispherical top. If the height of conical portion is twice the diameter of

base, find the diameter of conical part of ice-cream cone. [CBSE (F) 2016]

Sol. Volume of ice-cream in the cylinder = pr2h = p(6)2 × 15 cm3

Volume of ice-cream in one ice-cream cone = 1 pr2 h + 2 pr3
3 3

= 1 pr2 (4r) + 2 pr3 = 2pr3 [ a h = 2d = 4r]
3 3

\ Volume of ice-cream in 10 such cones = 10 × 2pr3 = 20pr3

                20pr3 = p × 36 × 15

                     r3 = 36 #15 = 27 ⇒ r = 3 cm
20

    Diameter of conical ice-cream cup = 6 cm

23. The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If

the total surface area of the solid cylinder is 1628 sq. cm find the volume of the cylinder.

cUse r = 22 m [CBSE Delhi 2016]
7

Sol. Here r + h = 37 and 2pr(r + h) = 1628

⇒ 2pr = 1628 ⇒ r = 1628 # 7
37 37 # 2 # 22

⇒ r = 7 cm and h = 30 cm

Hence volume of cylinder = pr2h

= 22 × 7 × 7 × 30 = 4620 cm3
7

Surface Areas and Volumes 361

24. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with
water. If the sphere is completely submerged in water, the water level in the cylindrical vessel

rises by 3 5 cm. Find the diameter of the cylindrical vessel. [CBSE (AI) 2016]
9 p(6)3 cm3
4
Sol. Volume of sphere = 3

Volume of water rise in cylinder = pr2 c 32 m cm3
9

\ pr2 32 = 4 p(6)3 ⇒ r2 = 4 # 2 # 36 # 9 = 81
⇒ 9 3 32

r = 9 cm

25. Water is flowing at the rate of 5 km/hour through a pipe of diameter 14 cm into a rectangular

tank of dimensions 50 m × 44 m. Find the time in which the level of water in the tank will rise

by 7 cm. [CBSE Delhi 2017 (C)]

Sol. Let the time taken by pipe be t hours.

 Speed = 5 km/h
∴ Length in t hours = 5000 t m.
According to question,

Volume of water flown through pipe = Volume of water in tank

πr2h = l × b × h

⇒ 22 ×c 7 2 t = 50× 44 × 7 ;a height = 7 cm = 7 mE
7 100 100 100
m ×5000

⇒ t = 50× 44 ×7×7×10000
100 × 5000 × 22 × 7 × 7

⇒ t = 2

Hence required time is 2 hours.

26. The radius and height of a solid right circular cone are in the ratio of 5 : 12. If its volume is

314 cm3, find its total surface area. [ Take π = 3·14 ] [CBSE (F) 2017]

Sol. Given r : h = 5 : 12

Let r = 5x and h = 12x

Volume of cone = 1 rr2 h
3

314 = 1 × 3.14 (5x)2 × 12x
3

⇒ x3 = 314 ×3 ⇒ x3 = 1
3.14 ×25×12

⇒ x=1

So, the value of r = 5 cm and h = 12 cm

Now, l = (12)2 + (5)2 = 13 cm

TSA of cone = πr(l + r) = 3.14 × 5 (13 + 5)

= 3.14 × 90 = 282.6 cm2

27. A wire of diameter 3 mm is wound about a cylinder whose height is 12 cm and radius 5 cm so
as to cover the curved surface of the cylinder completely. Find the length of the wire.
[CBSE (F) 2017]

362 Xam idea Mathematics–X

Sol. CSA of cylinder = 2π(5) × 12 = 120 πcm2 3
20
Let length of wire = h cm and radius of wire = cm

According to question

CSA of wire = CSA of cylinder

2rc 3 mh = 120r
20
120 × 20
⇒ h= 2×3 ⇒ h = 400 cm

28. Mohan and Sohan are two farmers in a village. Mohan is the owner of a rectangular plot of land

of length 25m and breadth 20 m, whereas Sohan has a rectangular plot of length and breadth

30 m and 15m respectively. There are two ponds in the village. One pond of dimensions

10 m × 10 m × 5 m and other, whose dimension is not known. In previous year Mohan

irrigated his field by using the whole water of a pond, whose dimension is known but Sohan

irrigated his field by using the whole water of other pond 10 times up to 10cm water level

whose dimension is not known. In summer Mohan offers Sohan to choose any one of the pond

for irrigation purpose. Sohan was unable to decide that which pond can be used more times

for irrigation of his field for a water level of 10 cm. Finally, Sohan meets his son Naman who is

studying in class X in Jawahar Navodaya Vidyalaya Giridih and asks him to complete the task.

(i) Which pond should be selected by Naman so that Sohan’s field can be irrigated more
times?

(ii) Write the mathematical process adopted by Naman to reach on conclusion.

(iii) How many times Mohan’s field can be irrigated up to the height of 10cm water level by
using the water of unknown dimensional pond?

Sol.

(i) The first pond, whose dimension is known, is selected by Naman so that Sohan’s field can
be irrigated more times.

(ii) Naman makes following mathematical calculation:
Area of Mohan’s plot = 25 m × 20 m = 500 m2
Area of Sohan’s plot = 30 m × 15 m = 450 m2

Total volume of water used by Sohan to irrigate his field 10 times having height 10 cm

= 10× 450 × 10  10cm = 10 m
= 450 m3 100 100

⇒ Total capacity of pond, whose dimension is not known = 450 m3

⇒ Total capacity of pond, whose dimension is known = 10 m × 10 m × 5 m

= 500 m3

⇒ The capacity of first pond, whose dimension is known is more than the capacity
of pond, whose dimension is not known.

So, Naman selected first pond, whose dimension is known.

(iii) Volume of water required to irrigate Mohan’s field one time up to the height of 10 cm
10
= 500 m2 × 100 m
= 50 m3

No. of times Mohan’s field can be irrigated up to the height of 10cm water level by using

the water of unknown dimensional pond = 450 m3 = 9 times.
50 m3

Surface Areas and Volumes 363

29. In a medical laboratory a lab technician composed a capsule in the shape of a cylinder with
two hemispheres stuck to each of its ends. The length of entire capsule is 15 mm and the
diameter of the cylindrical portion is 7 mm.

After reading above passage answer the following questions:

(i) Find surface area of the capsule.

(ii) Volume of the capsule.

Sol. Let r cm be the radius of cylinder and hemisphere, h cm be the height of the cylinder.

∴ r = 7 mm , h = 15 – 7 = 8 mm
2

(i) Surface area of the capsule

= Curved surface area of cylinder + surface area of two hemisphere

= 2prh + 2 × 2pr2 = 2pr (h + 2r)

= 2× 22 × 7 (8 + 7) = 22 ×15 = 330 mm2
7 2

(ii) Volume of the capsule = Volume of cylinder + 2 × volume of hemisphere

= pr2h + 2× 2 r r3
3
4 22 4 7
= r r2 c h + 3 r m = 7 × 7 × 7 c8 + 3 × 2 m
2 2

= 11× 7 × 38 = 2926 = 487.67mm3
2 3 6

Long Answer Questions [5 marks]

1. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same

height and same diameter is hollowed out. Find the total surface area of the remaining solid
to the nearest cm2.
[NCERT]

OR

From a solid right circular cylinder of height 2·4 cm and radius 0·7 cm, a right circular cone

of same height and same radius is cut out. Find the total surface area of the remaining solid.

[CBSE (AI) 2017]

Sol. We have,

Radius of the cylinder = 1.4 = 0.7 cm
2

Height of the cylinder = 2.4 cm

Also, radius of the cone = 0.7 cm

and height of the cone = 2.4 cm

Now, slant height of the cone = (0.7)2 +(2.4)2

⇒ l = 0.49 + 5.76 = 6.25 = 2.5 cm Fig. 13.22

\ Total surface area of the remaining solid

= curved surface area of cylinder + curved surface area of the cone

+ area of upper circular base of cylinder

364 Xam idea Mathematics–X

= 2prh + prl + pr2 = pr(2h + l + r)

= 22 × 0.7 × [2 × 2.4 + 2.5 + 0.7] = 22 × 0.1 × (4.8 + 2.5 + 0.7)
7

= 2.2 × 8.0 = 17.6 cm2 . 18 cm2

2. The decorative block shown in figure is made of two solids—a cube and a hemisphere. The

base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter

of 4.2 cm. Find the total surface area of the block. cUse r = 22 m
7

Sol. The total surface area of the cube = 6 × (edge)2

= 6 × 5 × 5 cm2 = 150 cm2

\ Total surface area of the block

= total surface area of cube – base area of hemisphere

+ curved surface area of hemisphere

= 150 – pr2 + 2pr2 = (150 + pr2) cm2

= c150 + 22 4.2 4.2 cm2 = (150 + 13.86) cm2 = 163.86 cm2 Fig. 13.23
7 2 2
# # m

3. Rasheed got a playing top (lattu) as his birthday present, which surprisingly
had no colour on it. He wanted to colour it with his crayons. The top is

shaped like a cone surmounted by a hemisphere (Fig. 13.24). The entire
top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he

has to colour. cTake r = 22 m
7

Sol. Total surface area of the top

= Curved surface area of hemisphere + Curved surface area of cone. Fig. 13.24

Now, the curved surface area of hemisphere = 2pr2

= c2 # 22 # 3.5 # 3.5 m cm2
7 2 2

Also, the height of the cone = Height of the top – Height (radius) of the hemispherical part

= c5 – 3.5 m cm = 3.25 cm
2

So, the slant height of the cone (l) = r2 + h2 = c 3.5 2 + (3.25) 2 cm = 3.7 cm (approx)
2
m

Therefore, curved surface area of cone = prl = c 22 # 3.5 # 3.7mcm2
7 2

Thus, the surface area of the top = c2 # 22 # 3.5 # 3.5 m cm2 + c 22 # 3.5 # 3.7mcm2
7 2 2 7 2

= 272 # 32.5 (3.5 + 3.7) cm2 = 22 # 3.5 × 7.2 cm2
7 2
= 39.6 cm2 (approx).

4. A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The

raddii of the top and bottom of circular ends of the bucket are 20 cm and 12 cm respectively.

Find the height of the bucket and also the area of the metal sheet used in making it.

(Use p = 3.14) [CBSE 2019 (30/1/2)]

Sol. r = 12 cm, R = 20 cm, V = 12308.8 cm3

Volume of frustum = 1 rh (r2 + R2 + rR)
3

123088 = 1 × 314 ^144 + 400 + 240h × h
10 3 100

Surface Areas and Volumes 365

h = 123088 × 3 ×10 A OB
314 ×784
h = 15 cm h

l = h2 + ]R – rg2 = 225+ 64 = 289 = 17 cm O′ D
2 cm
Area of metal sheet used = p(R + r)l + pr2 Fig. 13.25

= p(20 + 12) × 17 + p × 144 C 2 cm = h2

= p(544 + 144) = 314 ×688 = 2160.32 cm2
100
Hence, height of bucket is 15 cm and area of metal sheet used is 2160.32 cm2.

5. Rachel, an engineering student, was asked to make a model shaped like a

cylinder with two cones attached at its two ends by using a thin aluminium

sheet. The diameter of the model is 3 cm and its length is 12 cm. If each

cone has a height of 2 cm, find the volume of air contained in the model that

Rachel made. (Assume the outer and inner dimensions of the model to be 8 cm = h1

nearly the same.) [NCERT] 3 cm

Sol. Here, radius of cylindrical portion = 3 cm
2
Height of each cone = 2 cm

Height of cylindrical portion = 12 – 2 – 2 = 8 cm 2 cm = h2
Fig. 13.26
Volume of the air contained in the model

= volume of the cylindrical part of the model

+ volume of two conical ends.

= pr2h1 + 2 × 1 pr2 h2 = pr2 c h1 + 2 h2 m
3 3

= p × c 3 2 # c8 + 2 #2m= 22 # 9 # 28 = 66 cm3
2 3 7 4 3
m

6. A gulab jamun, contains sugar syrup about 30% of its volume. Find approximately

how much syrup would be found in 45 gulab jamuns, each shaped like a

cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm

(Fig. 13.27). [NCERT]

Sol. We have,

Radius of cylindrical portion and hemispherical portion of a gulab jamun

= 2.8 = 1.4 cm Fig. 13.27
2

Length of cylindrical portion = 5 – 1.4 – 1.4 = 2.2 cm

Volume of one gulab jamun

= volume of the cylindrical part + volume of the hemispherical ends

= pr2h + 2 × 2 pr3 = pr2h + 4 pr3
3 3

= pr2 ch+ 4 rm= 22 # (1.4)2 c2.2+ 4 # 1.4 m
3 7 3

= 22 #1.4 #1.4 # c 6.6 + 5.6 m= 22 #1.96 # 12.2
7 3 7 3

22 12.2 Fig. 13.28
7 3
\ Volume of 45 gulab jamuns = 45 × #1.96 #

366 Xam idea Mathematics–X

\ Quantity of syrup in 45 gulab jamuns = 30% of their volume

= 30 # 45 # 22 #1.96 # 12.2
100 7 3

= 338.184 cm3 = 338 cm3 (approx.)

7. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the
cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right
circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the
toy. (Take p = 3.14)

Sol. Let BPC be the hemisphere and ABC be the cone standing on the base of the

hemisphere (see Fig. 13.29). 1
2
The radius BO of the hemisphere (as well as of the cone) = × 4 cm = 2 cm

So, Volume of the toy = 2 rr3 + 1 rr2 h
3 3

= ; 32 # 3.14 # (2) 3 + 1 # 3.14 # (2)2 # 2Ecm3 = 25.12 cm3 Fig. 13.29
3

Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of
the right circular cylinder = HP = BO = 2 cm and its height is

EH = AO + OP = (2 + 2) cm = 4 cm

So, the required volume = volume of the right circular cylinder – volume of the toy
= (3.14 × 22 × 4 – 25.12) cm3 = 25.12 cm3
Hence, the required difference of the two volumes = 25.12 cm3.

8. A pen stand made of wood is in the shape of a cuboid with

four conical depressions to hold pens. The dimensions

of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius

of each of the depressions is 0.5 cm and the depth is

1.4 cm. Find the volume of wood in the entire stand

(Fig. 13.30). [NCERT]

Sol. We have,

Length of cuboid = l = 15 cm

Breadth of cuboid = b = 10 cm Fig. 13.30
Height of cuboid = h = 3.5 cm

And radius of conical depression = 0.5 cm

Depth of conical depression = 1.4 cm

Now, Volume of wood in the entire pen stand

= volume of cuboid – 4 × volume of a conical depression

= lbh – 4 × 1 pr2h
3

= 15 × 10 × 3.5 – 4 × 1 # 22 × 0.5 × 0.5 × 1.4
3 7

= (525 – 1.47) cm3 = 523.53 cm3

9. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is

surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole,
given that 1 cm3 of iron has approximately 8 g mass. (Use p = 3.14).
[NCERT]

Surface Areas and Volumes 367

Sol. Let r1 and h1 be the radius and height of longer cylinder, respectively, and r2, h2
be the respective radius and height of smaller cylinder mounted on the longer
cylinder. Then we have,

r1 = 12 cm, h1 = 220 cm

r2 = 8 cm, h2 = 60 cm

Now, volume of solid iron pole

= volume of the longer cylinder + volume of smaller cylinder
=
= rr12 h1 + rr22 h2
3.14 × (12)2 × 220 + 3.14 × (8)2 × 60

= 3.14 × 144 × 220 + 3.14 × 64 × 60
= 99475.2 + 12057.6 = 111532.8 cm3

Hence, the mass of the pole = (111532.8 × 8) grams

= 111513020.08 # 8 kg = 892.2624 kg

Fig. 13.31
10. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a

hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such

that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the

cylinder is 60 cm and its height is 180 cm. [NCERT]

Sol. We have,

Radius of cylinder = radius of cone = radius of hemisphere = 60 cm

Height of cone = 120 cm

\ Height of cylindrical vessel = 120 + 60 = 180 cm

Volume of cylinder = pr2h = 22 × (60)2 × 180
7
Now, volume of the solid

= volume of cone + volume of hemisphere

= 1 # 22 # (60) 2 # 120 + 2 # 22 # (60) 3
3 7 3 7

Volume of water left in the cylinder

= volume of the cylinder – volume of the solid Fig. 13.32

= 22 × (60)2 × 180 – 1 # 22 × (60)2 × 120 – 2 # 22 × (60)3
7 3 7 3 7

= 22 (60)2 ;180 – 1 # 120 – 2 # 60E
7 3 3

= 22 × 3600[180 – 40 – 40] = 22 # 3600 # 100
7 7

= 7920000 cm3 = 792 m3 = 1.131 m3 (approx.)
7 700

11. A container shaped like a right circular cylinder having

diameter 12 cm and height 15 cm is full of ice cream. The

ice cream is to be filled into cones of height 12 cm and

diameter 6 cm, having a hemispherical shape on the top.

Find the number of such cones which can be filled with ice

cream. [NCERT] Fig. 13.33

368 Xam idea Mathematics–X

Sol. Volume of the cylinder = pR2H

= 22 × (6)2 × 15
7

= 22 × 36 × 15 cm3
Now, Volume of cone 7
1 2
having hemispherical shape on top = 3 rr2 h + 3 rr3

= 1 rr2 (h + 2r)
3

= 1 × 22 × (3)2 (12 + 2 × 3)
3 7

= 1 # 22 # 9 # 18 = 22 # 54 cm3
3 7 7

\ The number of cones that can be filled with ice cream

= Volume of cylinder
Volume of cone having hemispherical shape on top

22 # 36 # 15
7
= 22 = 10
7
# 54

12. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket

is emptied on the ground and a conical heap of sand is formed. If the height of the conical

heap is 24 cm, find the radius and slant height of the heap. [NCERT]

OR

A girl empties a cylindrical bucket, full of sand, of base radius 18 cm and height 32 cm, on the

floor to form a conical heap of sand . If the height of this conical heap is 24 cm, then find its

slant height correct up to one place of decimal. [CBSE (F) 2014, 2019 (30/5/1)]

Sol. We have,

Radius of cylindrical bucket = 18 cm

Height of cylindrical bucket = 32 cm

And height of conical heap = 24 cm

Let the radius of conical heap be r cm

Volume of the sand = volume of the cylindrical bucket

= pr2h = p × (18)2 × 32

Now, volume of conical heap = 1 pr2h = 1 pr2 × 24 = 8pr2
3 3

Here, volume of the conical heap will be equal to the volume of sand.

\ 8pr2 = p × (18)2 × 32

⇒ r2 = 18 × 18 × 4 = (18)2 × (2)2

⇒ r2 = (36)2 or r = 36 cm

\ S lant height, l = r2 + h2

= (36)2 + (24)2 = 1296 + 576

= 1872 = 12 13 cm

Surface Areas and Volumes 369

13. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of

a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively.

Find the cost of the milk which can completely fill the container, at the rate of ™ 20 per
litre. Also, find the cost of metal sheet used to make the container, if it costs ™ 8 per 100 cm2

(Take p = 3.14) [NCERT]

Sol. We have, R = 20 cm, r = 8 cm, h = 16 cm

  Slant height, l = h2 + (R – r)2 = (16)2 + (20 – 8)2

= 256 + 144 = 400 = 20 cm

  Capacity of the container = Volume of the frustum

= 1 ph (R2 + r2 + Rr)
3
Fig. 13.34

= 1 × 3.14 × 16 × [(20)2 + 82 + (20 × 8)]
3

= 50.24 × 624
3

= 50.24 × 208 cm3 = 10449.92 cm3

= 10449.92 litres ( 1 Litre = 1000 cm3)
1000

= 10.44992 litres

Now, cost of milk to fill the container completely at the rate of ™ 20 per litre

= ™ 20 × 10.44992 = ™ 208.9984 ≈ ™ 209

Also, Surface area = pl(R + r) + pr2

= 3.14 × 20 × (20 + 8) + 3.14 × 8 × 8
= 3.14 [560 + 64] = 3.14 ×624 = 1959.36 cm2

\  Total cost of metal sheet used to make the container at the rate of ™ 8 per 100 cm2

= ™ 8 × 1959.36 = ™ 156.75.
100

14. A metallic right circular cone 20 cm high whose vertical angle

is 60° which is cut into two parts at the middle of its height

by a plane parallel to its base. If the frustum so obtained be

drawn into a wire of diameter 1 cm, find the length of the
16
wire. [NCERT, CBSE (F) 2017]

Sol. Let VAB be the metallic right circular cone of height 20 cm.

Suppose this cone is cut by a plane parallel to its base at a point
O' such that VO' = O' O i.e., O' is the mid point of VO.

Let r1 and r2 be the radii of circular ends of the frustrum Fig. 13.35
ABB' A'.

Now, in DVOA and VO' A', we have

tan 30° = OA and tan 30° = OlAl
VO VOl

370 Xam idea Mathematics–X

⇒ 1 = r1 and 1 = r2
⇒ 3 20 3 10

r1 = 20 and r2 = 10
3 3

\ Volume of the frustum = 1 ph _r12 + r22 + r1 r2i
3

= 1 r # 10=d 20 2 + d 10 2 + 20 # 10 G
3 3 3 3 3
n n

= 10r ; 400 + 100 + 200 E
3 3 3 3

= 10r # 700 = 7000r cm3
3 3 9

Let the length of wire of diameter 1 cm be l cm. Then,
16

Volume of metal used in wire = p × c 1 2 # l = rl cm3
32 1024
m

Since, the frustum is recast into a wire of length l cm and diameter 1 cm.
16

\ Volume of the metal used in wire = volume of the frustum

⇒ rl = 7000r
1024 9

⇒ l = 7000 # 1024 = 796444.4 cm = 7964.444 m
9

15. In Fig. 13.36, a cone of radius 10 cm is divided into two parts by drawing a plane through the

mid-point of its axis, parallel to its base. Compare the volumes of the two parts.

[CBSE Delhi 2017]

Sol. Let BC = r cm, DE = 10 cm

Since, B is the mid-point of AD and BC is parallel to DE, therefore C is the mid-point of AE.

\ AC = CE

Also, DABC ~ DADE

\ AB = BC = AC = 1
AD DE AE 2

i.e., BC = 1 DE = 1 × 10 = 5 cm or r = 5 cm
2 2

Now, Volume of cone = 1 rr2 (AB)
Volume of the frustum 3

1 r (BD) [R2 + r2 + Rr]
3
1 h
3 r (5) 2 c 2 m 25 25 1
100 + 25 + 50 175 7
= = = Fig. 13.36

1 rc h m[(10)2 + (5)2 + 10 # 5]
3 2

\ The required ratio = 1 : 7.

Surface Areas and Volumes 371

16. An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical

base made of the same metallic sheet (Fig. 13.37). The diameters of the two circular ends of

the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40

cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet

used to make the bucket, where we do not take into account the handle of the

bucket. Also, find the volume of water the bucket can hold. cTake r = 22 m
OR 7

An open metallic bucket is in the shape of a frustum of a cone. If the diameters

of the two circular ends of a bucket are 45 cm and 25 cm and the vertical

height of the bucket is 24 cm, find the area of the metallic sheet used to make Fig. 13.37
the bucket. Also find the volume of the water it can hold. (Use 22 2019(30/5/1)]
r= 7 ) [CBSE

Sol. The total height of the bucket = 40 cm, which includes the height of the base. So, the height of
the frustum of the cone = (40 – 6) cm = 34 cm.

Therefore, the slant height of the frustum,

l = h2 + (r1 – r2)2

where,  r1 = 45 cm = 22.5 cm,
2

r2 = 25 cm = 12.5 cm and h = 34 cm
2

So,   l = 342 + (22.5 – 12.5)2 = 342 + 102 = 35.44 cm
Area of metallic sheet used

Curved surface area of frustum of cone + Area of circular base + Curved surface area of cylinder
= [p × 35.44 (22.5 + 12.5) + p × (12.5)2 + 2p × 12.5 × 6] cm2

= 22 (1240.4 + 156.25 + 150) cm2
7

= 4860.9 cm2

Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket)

= 1 rh # (r12 + r22 + r1 r2)
3

= 1 # 22 × 34 × [(22.5)2 + (12.5)2 + 22.5 × 12.5] cm3
3 7

= 1 # 22 × 34 × 943.75 = 33615.48 cm3
3 7

= 33.62 litres (approx.)

OR

Solution is similar as above only height of bucket has changed from (40 – 6 = 34) cm to 24 cm.
Area = 3351.07 cm2 (TSA = curved surface area of frustum of cone + area of circular base)
= πl(r1 + r2) + πr2

Volume 23728.57 cm3= 23.73 litres (Approx)

372 Xam idea Mathematics–X

17. 150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter
7 cm containing some water, which are completely immersed in water. Find the rise in the
level of water in the vessel.

Sol. Diameter of spherical balls = 1.4 cm
Radius of spherical balls, r = 0.7 cm
Diameter of cylinder = 7 cm
Radius of cylinder = 3.5 cm
No. of spherical balls = 150

Let the rise in water be h cm.

Now, 150 × volume of a spherical ball = Volume of cylinder with height h.

i.e., 150 × ; 4 rr3E = pR2h
3

150 × 4 × p × 0.7 × 0.7 × 0.7 = p × 3.5 × 3.5 × h
3

h = 150 × 4 # 0.7 # 0.7 # 0.7 = 28 = 5.6 cm
3 3.5 # 3.5 5

18. Two design of cans made of aluminium are shown below

Diameter

Diameter

Height

Specification of cans are given in the following table

Specifications Normal can Slim can

Volume 350 mL 220 mL

Total surface area 276.95 sq cm 220.05 sq cm

Diameter 7 cm 4.8 cm

Height 9.1 cm 12.2 cm

(a) The cost of production of a normal can is ™3. Find the cost of production of a slim can
giving your answer to the nearest ten paise.

(b) A normal can of drink is sold at ™35 while a slim can of the same drink is sold at ™ 25.
Determine which type provides better value for the customer.

(c) Based on your answers to question (a) and question (b), give two possible reasons why the
manufacturer has an incentive to market the slim can.

Sol. (a) The cost of production of a normal can = ™ 3

= 300 paise

The total surface area of normal can = 276.95 sq cm
⇒ The cost of production per sq cm 300
= 276.95 paise

⇒ The cost of production per sq cm = 1.083 paise

Now, The total surface area of slim can = 220.05 sq cm

Surface Areas and Volumes 373

⇒ The cost of production of slim can = 220.05 × 1.083 paise

= 238.31 paise

= 240 paise (nearest ten paise)

= ™ 2 and 40 paise

(b) Volume of normal can = pr2h

= 22 (3.5)2 × 9.1
7
22
= 7 ×12.25× 9.1

= 350.35 cm3

Volume of slim can = pr2h

= 22 (2.4) 2 ×12.2
7
22
= 7 × 5.76 ×12.2

= 220.85 cm3

Cost of drink in normal can = ™ 35 ÷ 350.35 cm3

= 10 paise per cm3

Cost of drink in slim can = ™ 25 ÷ 220.85 cm3
= 11 paise per cm3

Hence, normal can of drink provides better value for the customer.

(c) The manufacturer has an incentive to market the slim can because,

(i) The cost of production of slim can is less than the cost of production of normal can.

(ii) The profit to sell slim can of drink is more than that of normal can.

19. A sweet corner sells different types of sweets made of milk. Each sweet are circular in shape
and of same thickness. Following are the price of sweets and their diameter.

Sweets Thickness Diameter Price per piece

Sweet - 1 4 cm 2 cm ™ 10

Sweet - 2 4 cm 3 cm ™ 20

Sweet - 3 4 cm 4 cm ™ 30

Sweet - 4 4 cm 5 cm ™ 40

(a) Which sweet has better value for customers?
(b) Which sweet is better value for sweet corner?
(c) What strategy you have adopted to answers question (a) and question (b)?

Sol. Volume of sweet-1 = pr2h

= 22 (1) 2 × 4
7
= 22
7 × 4
=
= 88
Volume of sweet-2 =
7
= 12.57 cm3

374 Xam idea Mathematics–X pr2h

22 c 3 2 × 4
7 2
m

= 22 × 9 × 4
7 4
198
Volume of sweet-3 = 7

= 28.28 cm3

= pr2h

= 22 (2) 2 × 4
7
22
= 7 × 4× 4

= 352
7

= 50.28 cm3

Volume of sweet-4 = pr2h

= 22 c 5 2 × 4
7 2
m

= 22 × 25 × 4
7 4
550
= 7

= 78.57 cm3

Volume of sweet-1 for ` 1= 12.57 = 1.25 cm3
10

Volume of sweet-2 for ` 1 = 28.28 = 1.41 cm3
20

Volume of sweet-3 for ` 1 = 50.28 = 1.67 cm3
30

Volume of sweet-4 for ` 1 = 78.57 = 1.96 cm3
40
(a) Sweet-4 has better value for customers.

(b) Sweet-1 has better value for sweet corner.

(c) To answers question (a) and question (b), at first volume of each sweet is calculated then
volume of each sweet per one ™ is find out.

HOTS [Higher Order Thinking Skills]

1. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all

around it in the shape of a circular ring of width 4 m to form an embankment. Find the height

of the embankment.

Sol. Here, radius of the well = 3 = 1.5 m , depth of the well = 14 m
2

width of the embankment = 4 m and

\ Radius of the embankment = 1.5 + 4 = 5.5 m

Let h be the height of the embankment.

\ Volume of the embankment = volume of the well (cylinder)

⇒ p [(5.5)2 – (1.5)2] × h = p (1.5) 2 × 14

Fig. 13.38

Surface Areas and Volumes 375

⇒ (30.25 – 2.25) × h = 2.25 × 14

⇒ h= 2.25 # 14 = 1.125 m
28

2. A hollow cone is cut by a plane parallel to the base at some height and the upper portion is

removed. If the curved surface area of the remainder is 8 of the curved surface of the whole
9

cone, find the ratio of the two parts into which the cone’s altitude is divided. [CBSE (F) 2017]

Sol. In Fig. 13.39, the smaller cone APQ has been cut off through the
plane PQ BC. Let r and R be the radii of the smaller and larger cone
and l and L their slant heights respectively.

Here, in the adjoining figure OQ = r, MC = R, AQ = l, AC = L.

Now, DAOQ ~ DAMC

⇒ OQ = AQ ⇒ r = l ⇒ r= Rl … (i)
MC AC R L L
8
Since, curved surface area of the remainder = 9 of the curved
surface area of the whole cone
Fig. 13.39

⇒ CSA of smaller cone = 1 of the CSA of the whole cone
9

\ prl = 1 pRL ⇒ rc Rl ml = 1 pRL [using (i)]
9 L 9

⇒ l2 = L2 ⇒ l = 1
9 L 3

Now again in similar triangles, AOQ and AMC,we have

AO = AQ ⇒ AO = l = 1 ⇒ AO = AM
AM AC AM L 3 3

⇒ OM = AM – OA = AM – AM = 2 AM
\ AM 3 3

AO = 3 = 1
OM 2AM 2

3

Hence, the required ratio of their heights = 1 : 2

3. The height of the cone is 30 cm. A small cone is cut off at the top by a plane parallel to the
1
base. If its volume be 27 of the given cone, at what height above the base is the section made?

[CBSE (F) 2016]

OR

The height of a cone is 30 cm. From its topside a small cone is cut by a plane parallel to its
1
base. If volume of smaller cone is 27 of the given cone, then at what height it is cut from its
base ? [CBSE Delhi 2017]

Sol. Let the small cone APQ be cut off at the top by the plane PQ < BC

Let r and h be the radius and height of the smaller cone, respectively and also let the radius of
larger cone = R

Now, DAOQ ~ DAMC

376 Xam idea Mathematics–X

⇒ AO = OQ ⇒ h = r \ r = hR …(i)
 AM MC 30 R 30
\ 1
Volume of the smaller cone = 27 (Volume of larger cone)

1 rr2 h = 1 c 1 rR2 # 30m
3 27 3

⇒ r c hR 2 = 1 # 1 rR2 # 30 [From (i)]
⇒ 3 30 27 3
\ mh

h3 = 30 # 30 # 30 = 1000
27
Fig. 13.40
h = 10 cm

Hence, the smaller cone has been cut off at a height of (30 – 10) cm = 20 cm from the base.

4. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area
will it irrigate in 30 minutes, if 8 cm of standing water is needed?[NCERT; CBSE 2019(30/1/2)]

Sol. We have, width of the canal = 6 m

Depth of the canal = 1.5 m

Now, length of water flowing per hour = 10 km

\ Length of water flowing in half hour = 5 km = 5000 m
\ Volume of water flow in 30 minutes = 1.5 × 6 × 5000 = 45000 m3

Here, standing water needed is 8 cm = 0.08 m

\ Area irrigated in 30 minutes = volume = 45, 000
height 0.08

= 562500 m2 [1 hectare = 10000 m2]

= 56.25 hectares

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which
is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a
sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the
number of lead shots dropped in the vessel.

Sol. We have,

Height of conical vessel = h = 8 cm

and its radius = r = 5 cm

Now, volume of cone = volume of water in the cone

= 1 rr2 h = 1 # 22 # 5 # 5 # 8 = 4400 cm3
3 3 7 21
Fig. 13.41

Now, volume of water flows out = volume of lead shots

= 1 × volume of water in the cone = 1 # 4400 = 1100 cm3
4 4 21 21

Now, radius of the lead shots = 0.5 cm = 5 cm = 1 cm
10 2

Volume of one spherical lead shot = 4 rr3 = 4 # 22 # 1 # 1 # 1 = 11 cm3
3 3 7 2 2 2 21

\ Number of lead shots dropped in the vessel = volume of water flows out
volume of one lead shot

1100

= 21 = 1100 # 21 = 100
11 21 11
21

Surface Areas and Volumes 377

6. A right triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume
of double cone thus generated. (Use p = 3.14).

Sol. In the given Fig. 13.42, DPQR is a right triangle, where PQ = 3 cm, PR = 4 cm and QR = 5 cm.

Let OQ = x ⇒ OR = 5 – x and OP = y

Now in right angled-triangle POQ, we have

PQ2 = OQ2 + OP2 (by Pythagoras Theorem)

⇒ (3)2 = x2 + y2 ⇒ y2 = 9 – x2 …(i)

Also from right angled triangle POR, we have

OP2 + OR2 = PR2

⇒ y2 + (5 – x)2 = (4)2

⇒ y2 = 16 – (5 – x)2 …(ii) Fig. 13.42

From (i) and (ii), we get

9 – x2 = 16 – (5 – x)2

⇒ 9 – x2 = 16 – (25 + x2 – 10x)

or 9 – x2 = – 9 – x2 + 10x

⇒ 10x = 18 ⇒ x = 9
5
9 16
\ OR = 5 – x = 5 – 5 = 5

Now putting x = 9 in (i), we get
5

y2 = 9 – c 9 2 = 9 – 81 = 144 ⇒ y = 12
5 25 25 5
m

\ OP = y = 12
5
12
Now for the cone PQM, radius OP = 5 cm,

height OQ = 9 cm
\ rr2 5
Volume = 1 = 1 rc 12 2 # 9 = 432r cm3
3 h 3 5 5 125
m

Also for the cone PRM, radius OP = 12 cm,
5
16
height OR = 5 cm
\
Volume = 1 rc 12 2 # 16 cm3 = 768r cm3
3 5 5 125
m

Hence total volume, i.e., volume of the double cone

= c 432r + 768r m = 1200r
125 125 125

= 9.6 × 3.14 cm3 = 30.144 cm3

7. A right circular cone is divided into three parts by trisecting its height by two planes drawn

parallel to the base. Show that the volumes of the three portions starting from the top are in

the ratio 1 : 7 : 19. [CBSE (F) 2017]

Sol. From figure it is clear that

∆ACO′′ ~ ∆AEO′ [By AA similarity] and

378 Xam idea Mathematics–X

∆ACO′′ ~ ∆AGO [By AA similarity]

∴ r1 = h & 2r1 = r2 ...(i)
and r2 2h & 3r1 = r3 ...(ii)
...(iii)
r1 = h
r3 3h

Volume of cone ABC = 1 rr12 h
3

Volume of frustum BCED = volume of cone ADE – volume of cone ABC

= 1 rr22 (2h) – 1 rr12 h
3 3

= 1 r (2r1)2 × (2h) – 1 rr12 h [From equation (i)]
3 3

= 7 rr12 h ...(iv)
3
Volume of frustum DEGF = volume of cone AFG – volume of cone ADE

= 1 rr32 (3h) – 1 rr22 × 2h
3 3

= 1 r (3r1)2 × (3h) – 1 r (2r1)2 × 2h [From equation (i) and (ii)]
3 3

= 1 rr12 h (27 – 8)
3

= 19 rr12 h ...(v)
3

Now, from equation (iii), (iv) and (v), we get

Required Ratio = 1 rr12 h : 7 rr12 h : 19 rr12 h = 1 : 7 : 19
3 3 3

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) A cylinder and a cone are of the same base radius and same height. Find the ratio of the
volumes of the cylinder to that of the cone.

(a) 1 : 3 (b) 1 : 2 (c) 3 : 1 (d) 2 : 1

(ii) Total surface area of a hemispherical solid having radius 7 cm is
(a) 462 cm2 (b) 294 cm2 (c) 588 cm2 (d) 154 cm2

(iii) A piece of paper is in the shape of a semi circular region of radius 10 cm. It is rolled to form

a right circular cone. The slant height is

(a) 5 cm (b) 10 cm (c) 15 cm (d) 20 cm

(iv) The radius of the largest right circular cone that can be cut out from a cube of edge 4.2 cm
is

(a) 2.1 cm (b) 4.2 cm (c) 3.1 cm (d) 2.2 cm

Surface Areas and Volumes 379

(v) A rectangular block 6 cm × 12 cm × 15 cm is cut into exact number of equal cubes of edge
3 cm. The least possible number of cubes will be

(a) 6 (b) 11 cm (c) 33 cm (d) 40 cm

2. Fill in the blanks.
(i) A solid formed on revolving a right angled triangle about its height is a ___________.

(ii) The total surface area of the given solid (Fig. 13.43) is ___________.

l

r
h

Fig. 13.43
(iii) The total surface of a solid hemisphere having radius r is ___________.

(iv) If the radius of a sphere is halved its volume becomes ___________ times the volume of the
original sphere.

(v) Volume of a frustum of a cone is ______________ .

QQ Very Short Answer Questions: [1 mark each]

3. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas.

4. What geometrical shapes is the plumb line (Fig. 13.44) a combination of? [NCERT Exemplar]

Fig. 13.44

5. A cone is cut through a plane parallel to its base and then the cone that is formed on one side of

that plane is removed. What is the new part that is left over on the other side of the plane called?

[NCERT Exemplar]

6. A solid sphere of radius is melted and cast into the shape of a solid cone of height equal to the

diameter of the sphere. What will be the radius of the cone?

7. A solid piece of iron of dimensions 66 cm × 49 cm × 12 cm is moulded into a sphere. What is the

radius of the sphere?

8. A solid metallic cuboid of dimensions 9 m × 8 m × 2 m is melted and recast into solid cubes of

edge 2 m. Find the number of cubes so formed. [CBSE (F) 2017]

QQ Short Answer Questions–I: [2 marks each]

9. A right circular cylinder and a cone have equal bases and equal heights. If their curved surface

areas are in the ratio 8 : 5, show that the ratio between radius of their bases to their height is

3:4 [CBSE 2019 (C) (30/1/1)]

380 Xam idea Mathematics–X

State true or false for the following and justify your answer. (Q. 10 – Q. 11)

10. The volume of the frustum of a cone is 1 rh (r12 + r22 – r1 r2) where h is vertical height of the
3

frustum and r1, r2 are the radii of the ends. [NCERT Exemplar]

11. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius.

The total surface area of the shape so formed is 4prh + 2pr2. [NCERT Exemplar]
12. The surface area of a sphere is 616 cm2. Find its radius. [NCERT Exemplar]

13. The radii of two cylinders are in the ratio 2 : 5 and their heights are in the ratio 3 : 4. What is the

ratio of their curved surface areas?
14. A hemisphere and a cone have equal bases. If their heights are also equal, then what is the ratio

of their curved surfaces?

15. A sphere of maximum volume is cut-out from a solid hemisphere of radius 6 cm. What is the
volume of the cut-out sphere?

16. If the slant height of the frustum of a cone is 9 cm and the perimeters of its circular bases are

18 cm and 16 cm respectively. What is the curved surface area of the frustum?
17. If the areas of circular bases of a frustum of a cone are 16 cm2 and 36 cm2 respectively and the

height of the frustum is 15 cm. What is the volume of the frustum?

QQ Short Answer Questions–II: [3 marks each]

18. A wooden article was made by scooping out a hemisphere from each end of a solid

cylinder, as shown in Fig. 13.45. If the height of the cylinder is 10 cm and its base is

of radius 3.5 cm. Find the total surface area of the article. [CBSE 2018 (30/1)]

19. A solid is in the form of a cylinder with hemispherical ends. The total height of the

solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the Fig. 13.45
22
solid. (Use r= 7 ) [CBSE 2019 (30/2/1)]

20. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The

two spheres are melted to form a single big sphere. Find the diameter of the new sphere.

[CBSE 2019 (30/2/3)]

21. A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes

it in the form of a sphere. Find the radius of the sphere and hence find the surface area of this

sphere. [CBSE 2019 (30/4/2)]

22. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his

field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/

hr, in how much time will the tank be filled? [CBSE 2019 (30/4/2)]

23. A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones,

each of diameter 4 2 cm and height 3 cm. Find the number of cones so formed.
3

24. From a solid cylinder whose height is 24 cm and diameter 20 cm, a conical cavity of same height

and same diameter is hollowed out. Find the total surface area of the remaining solid.
[CBSE Delhi 2017 (C)]

25. Two identical cubes of volume 343 cm3 are joined together end to end. What is the surface area

of the resulting cuboid?

Surface Areas and Volumes 381

26. Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a

single cube. Find the surface area of the cube so formed. [NCERT Exemplar]

27. Two cones with same base diameter 14 cm and height 24 cm are joined together along their

bases. Find the surface area of the shape so formed.

28. A cone of maximum size is carved out from a cube of edge 20 cm. Find the surface area of the

solid left out after the cone is carved out.

29. Find the number of shots, each having diameter 6 cm that can be made from a cuboidal lead

which is solid of dimensions 24 cm × 22 cm × 12 cm.

30. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing

some water. Find the number of marbles that should be dropped into the beaker, so that the

water level rises by 5.6 cm.

31. A solid metallic sphere of diameter 16 cm is melted and recast into a number of smaller cones,

each of radius 4 cm and height 8 cm. Find the number of cones so formed. [CBSE Delhi 2017]

32. A wall 24 m long, 0.4 m thick and 6 m high which is constructed with the bricks, each of dimensions

25 cm × 16 cm × 10 cm. If the mortar occupies 1 th of the volume of the wall, then find the
10
number of bricks used in constructing the wall.

33. A cone of radius 7 cm is divided into two parts by drawing a plane through the mid-point of its

axis and parallel to its base. Compare the volumes of the two parts.

34. A bucket is in the form of frustum of a cone and holds 18.018 litres of water. The radii of the top

and bottom are 18 cm and 15 cm, respectively. Find the height of the bucket.

35. Two solid cones A and B are placed in a cylindrical tube in Fig. 13.46. The ratio of their capacities

are 2 : 1. Find the heights and capacities of cones. Also, find the volume of the remaining portion

of the cylinder.

AB

Fig. 13.46

36. An ice-cream cone is full of ice-cream having radius 5 cm and height 10 cm (in Fig. 13.47).

Calculate the volume of ice-cream, provided that its 1 part is left unfilled with ice-cream.
6
NCERT Exemplar]

Fig. 13.47

382 Xam idea Mathematics–X

37. A metal container, open from the top, is in the shape of a frustum of a cone of height 21 cm with

radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk

which can completely fill the container at the rate of ™35 per litre. (Take r = 22 )
7
[CBSE (F) 2016]

38. A metallic solid sphere of radius 10.5 cm is melted and recasted into smaller solid cones, each of

radius 3.5 cm and height 3 cm. How many cones will be made? [CBSE Delhi 2017]

39. A solid sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with

water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in

water, by how much will the level of water rise in the cylindrical vessel? [CBSE (F) 2017]

40. A hemispherical tank full of water is emptied by a pipe at the rate of 3 4 litres per second. How
7
22
much time will it take to empty the tank, if it is 3 m in diameter? (Take r = 7 )

[CBSE 2019 (C) (30/1/1)]

41. Two cones with same base diameter 16 cm and height 15 cm are joined together along their

bases. Find the surface area of the shape so formed. [CBSE 2019 (C) (30/1/1)]

QQ Long Answer Questions: [5 marks each]

42. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are

10 cm and 30 cm respectively. If its height is 24 cm, find: [CBSE 2018 (30/1)]

(i) The area of the metal sheet used to make the bucket.

(ii) Why we should avoid the bucket made by ordinary plastic? [Use π = 3.14]

43. A man donates 10 aluminum buckets to an orphanage. A bucket made of aluminum is of height

20 cm and has its upper and lowest ends of radius 36 cm and 21 cm respectively. Find the cost of

preparing 10 buckets if the cost of aluminum sheet is ™42 per 100 cm2. Write your comments on

the act of the man. [CBSE 2018 (C) (30/1)]

44. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is

surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole,

given that 1 cm3 of iron has approximately 8 gm mass. [Use π = 3.14] [CBSE 2019 (30/2/1)]

45. A right cylindrical container of radius 6 cm and height 15 cm is full of ice-cream, which has to be

distributed to 10 children in equal cones having hemispherical shape on the top. If the height of

the conical portion is four times its base radius, find the radius of the ice-cream cone.

[CBSE 2019 (30/2/3)]

46. In Fig. 13.48, a decorative block is shown which is made of two solids, a cube and a hemisphere.

The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a

diameter of 4.2 cm. Find

(a) the total surface area of the block. 22
7
(b) the volume of the block formed. (Take r= ) [CBSE 2019 (30/3/1)]

Surface Areas and Volumes 383

4.2 cm

6 cm

6 cm

6 cm

Fig. 13.48

47. A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The

radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of

the bucket and the area of metal sheet used in making the bucket. [Use π = 3.14]
[CBSE 2019 (30/3/1)]

48. A container opened at the top and made up of a metal sheet, is in the form of a frustum of a cone

of height 15 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the

cost of milk which can completely fill the container, at the rate of ™ 50 per litre. Also find the cost

of metal sheet used to make the container, if it costs ™ 10 per 100 cm2 . [Use π = 3.14]
[CBSE 2019 (30/4/1)]

49. The radii of the circular ends of the frustum of a cone of height 45 cm, are 28 cm and 7 cm. Find

its volume and curved surface area. [CBSE 2019 (C) (30/1/1)]

50. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all

around it in the shape of a circular ring of width 4 m to form an embankment. Find the height

of the embankment. [CBSE 2019 (C) (30/1/3)]

51. A factory manufactures 18000 cylindrical boxes daily. The length of each box is 32 cm and

circumference of base is 6 cm. Find the cost of colouring the curved surfaces of the boxes

manufactured in one day at ™ 0.08 per dm2.

52. Five hundred persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad.

What is the rise of water level in the pond, if the average displacement of the water by a person

is 0.04 m3?

53. How many cubic centimeter of iron is required to construct an open box whose external

dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm? If one

cubic cm of iron weighs 7.5 g, find the weight of the box. [NCERT Exemplar]

54. A well of diameter 3 m is dug 14 m deep. The soil taken out of it is spread evenly all around it to

a width of 5 m to form an embankment. Find the height of the embankment. [CBSE (F) 2017]

55. A milk container is made of metal sheet in the shape of frustum of a cone whose volume is

10459 3 cm3. The radii of its lower and upper circular ends are 8 cm and 20 cm, respectively,
7

find the cost of metal sheet used in making the container at the rate of ™ 1.40 per square cm.

56. A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-

cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape

on the top. Find the number of such cones which can be filled with ice-cream.

384 Xam idea Mathematics–X

57. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of

the cone is 6 cm and the diameter of the base is 7 cm. Determine the volume of the toy. If a cube

circumscribes the toy, find the difference of the volumes of cube and the toy.

58. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the

other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of

the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface

area of the toy, if the total height of the toy is 30 cm.

59. A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius

as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2 of
3

the volume of the hemisphere, calculate the height of the cone and the surface area of the toy.

60. The interior of a building is in the form of a cylinder of diameter 4 m and height 3.5 m,

surmounted by a cone of the same base with vertical angle as a right angle. Find the curved

surface area and volume of the interior of the building.

61. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a

cone with the same radius as that of the cylinder. The diameter and height of the cylinder are

9 cm and 15 cm, respectively. If the slant height of the conical portion is 7.5 cm, find the total

surface area and volume of the rocket.

62. A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains

54 10 m3 of air. If the internal diameter of dome is equal to 4 of the total height above the floor,
21 5

find the height of the building.

63. A pen stand made of wood is in the shape of a cuboid with three conical depressions and a

cubical depression to hold the pens and pins respectively. The dimension of the cuboid are

8 cm × 6 cm × 4 cm. The radius of each of the conical depression is 0.5 cm and the depth is

1.8 cm. The edge of the cubical depression is 2.5 cm. Find the volume of the wood in the entire

stand.

64. Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/ sec in an

empty cylindrical tank, the radius of whose base is 40 cm . What is the rise of water level in tank

in half an hour?

65. From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same

height and same diameter is hollowed out. Find the total surface area of the remaining solid.

<Take r = 22 F [CBSE Delhi 2014]
7
66. A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its

lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can

completely fill the container at the rate of ™ 21 per litre. <Use r = 22 F [ CBSE (AI) 2014]
7

67. A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the

radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find:
(i) the volume of water which can completely fill the bucket.

(ii) the area of the metal sheet used to make the bucket. <Use r = 22 F [CBSE (F) 2014]
7

Surface Areas and Volumes 385

68. A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well

to form a 40 cm high embankment. Find the width of the embankment. [CBSE Delhi 2015]

69. A vessel full of water is in the form of an inverted cone of height 8 cm and the radius of its top,

which is open, is 5 cm. 100 spherical lead balls are dropped into the vessel. One-fourth of the

water flows out of the vessel. Find the radius of a spherical ball. [CBSE (F) 2015]

70. A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The

radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of

the bucket and the area of metal sheet used in making the bucket. (Use p = 3.14)

[CBSE Delhi 2016]

71. The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base

at the middle of its height. Find the ratio of the volumes of the two parts. [CBSE Delhi 2017]

Answers

1. (i) (c) (ii) (a) (iii) (b) (iv) (a) (v) (d)

2. (i) cone (ii) rrl + 2rrh + rr2 (iii) 3rr2 (iv) one-eighth

(v) 1 πh[R2 + r2 + rR]
3

3. 1 4. Hemisphere and cone 5. Frustum and Cone 6. 2r
9

7. 21 cm 8. 18 cubes 10. False 11. True 12. 7 cm 13. 3 : 10

14. 2 :1 15. 36p cm3 16. 153 cm2 17. 380 cm3 18. 374 cm2 19. 680.17 cm3

20. 12 cm 21. 6 cm; 144 πcm2 22. 100 minutes or 1 hr 40 minutes 23. 672

24. 2640 cm2 25. 490 cm2 26. 216 cm2 27. 1100 cm2 28. 2400 – 2200 (1 – 5) cm2
7

29. 56 30. 150 31. 16 32. 12960 33. 1 : 7 34. 21 cm

35. 14 cm, 7 cm, 132 cm3, 66 cm3, 396 cm3

36. 327.4 cm3 37. ™ 480.48 38. 126 39. 1 cm 40. 1980 second or 33 minutes

41. 854.85 cm2 42. (i) 1711.3 cm2 (ii) we should avoid use of plastic because it is non- degradable.

43. 24631.20 44. 892.262 kg 45. radius = 3 cm 46. 235.40 cm3

47. 229.88 cm2 , 2160.32 cm2, h =15 cm, 48. ™ 522.50, ™ 195.93

49. volume 48510 cm3, CSA=330 274 cm2 50. h =1 1 m 51. ™ 2764.80 52. 0.5 cm
8

53. 3960 cm3, 29.7 kg 54. 0.7875 m 55. ™ 2745.60 56. 10 cones 50r
57. 166.83 cm3, 690.55 cm3 3
61. 594 cm2, 1081.9 cm3 58. 770 cm2 59. 28 cm, 5082 cm2 60.( 14 + 4 2 )pm2, m3

62. 5 m 63. 175 cm3 (approx) 64. 90 cm

65. 73.92 cm2 66. ™329.49 67. (i) 8.624L (ii) 1804 cm2

68. 10 m 69. 1 cm 70. 15 cm, 2160.32 cm2 71. 4 : 7
2

386 Xam idea Mathematics–X

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) A cylindrical pencil sharpened at one edge is the combination of

(a) a cone and a cylinder (b) frustum of a cone and a cylinder

(c) a hemisphere and a cylinder (d) two cylinders

(ii) A plumbline (sahul) in the given Fig. 13.49 is the combination of

(a) a cone and a cylinder (b) a hemisphere and a cone

(c) frustum of a cone and a cylinder (d) sphere and cylinder Fig. 13.49

(iii) The shape of a glass (tumbler) in given Fig. 13.50 is usually in the form of

(a) a cone (b) frustum of cone Fig. 13.50
(c) a cylinder (d) a sphere

(iv) A solid consists of a circular cylinder with an exact fitting right circular cone placed at the

top. If the height of the cone is h and the total volume of the solid is 3 times the volume of

the cone, then the height of the circular cylinder is

(a) 2 h (b) 2h (c) 3h (d) 4 h
3 2

2. Fill in the blanks. (3 × 1 = 3)

(i) A solid sphere of radius r cm is melted and recast into the shape of a solid cone of height r.
Then the radius of the base of cone is _______________ .

(ii) The length of the longest rod that can be placed in a 12 m × 9 m × 8 m rod is _____________.

(iii) The total surface area of a solid hemisphere having radius r is _______________.

3. Solve the following questions. (3 × 1 = 3)

(i) The total surface area of a sphere and a cube are equal. What is the ratio of their volumes?

(ii) If the height of a right circular cone is y and radius of the base is x, then write the relation
which gives its volume.

(iii) What is the curved surface area of a right circular cone of height 24 cm and base diameter
14 cm?

Section B

QQ Solve the following questions. (3 × 2 = 6)

4. Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a
single cube whose diagonal is 12 3 cm. Find the edges of the three cubes.

5. Water in a canal, 5·4 m wide and 1·8 m deep, is flowing with a speed of 25 km/hour. How much
area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?
[CBSE (AI) 2017]

Surface Areas and Volumes 387

6. A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all

around it in the shape of a circular ring of width 3 m to form an embankment. Find the height

of the embankment. [CBSE Delhi 2016]

QQ Solve the following questions. (3 × 3 = 9)

7. A solid metallic right circular cone 20 cm high and whose vertical angle is 60°, is cut into two
parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be

drawn into a wire of diameter 1 cm, find the length of the wire. [CBSE Delhi 2014]
12
8. In Fig. 13.51, is a decorative block, made up of two solids – a cube and a hemisphere. The base

of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm.

Find the total surface area of the block. ;Use r = 22 E [CBSE Delhi 2016]
7

Fig. 13.51

9. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full

barrel of ink in the pen is used up on writing 3300 words on an average. How many words can

be written in a bottle of ink containing one-fifth of a litre? [NCERT Exemplar]

QQ Solve the following questions (3 × 5 = 15)

10. Water is flowing at the rate of 15km/h through a pipe of diameter 14 cm into a cuboidal pond
which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm?

[NCERT Exemplar]

11. In Fig. 13.52, from a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a

cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block.

[Use r = 22 ] [CBSE Delhi 2015]
7

Fig. 13.52

388 Xam idea Mathematics–X

12. A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its

lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can
22
completely fill the container at the rate of ™ 21 per litre. [Use r = 7 ] [CBSE (AI) 2014]

OR

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the

radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find:
(i) the volume of water which can completely fill the bucket.

(ii) the area of the metal sheet used to make the bucket. [Use r = 22 ] [CBSE (F) 2014]
7

Answers

1. (i) (a) (ii) (b) (iii) (b) (iv) (b)

2. (i) 2r (ii) 17 m (iii) 3πr2

3. (i) 6: r (ii) rx2 y (iii) 550 cm2
3

4. 6, 8, 10 cm 5. 162 hectares 6. h = 4 m 7. 4480 m

8. 225.625 cm2 9. 480000 words

10. 2 hours 11. 583 cm2 12. `329.49 OR (i) 8624 cm3 (ii) 1804 cm2

z z z

Surface Areas and Volumes 389

14 Statistics

BASIC CONCEPTS – A FLOW CHART

390 Xam idea Mathematics–X

Statistics 391

MORE POINTS TO REMEMBER

 There are three methods to find the mean of a grouped data. So the choice of method
to be used depends on the numerical values of xi and fi. If xi and fi are sufficiently small,
then the direct method is used. If xi and fi are numerically large, then we can go for the
assumed mean method or step-deviation method. If the class sizes are unequal and xi are
large numerically, we can still apply the step-deviation method by taking h to be a suitable
divisor of all the di’s.

 Modal class: A class interval in a group data is called modal class which have maximum
frequency.

 Mean is the arithmetic average of observations of the data but mode is the average of that
observations of data which belong to the group having maximum frequency, i.e., mode is
most frequent value or most popular observation.

 Median class: A class interval, in which cumulative frequency is greater than and nearest

to n is called median class.
2
 The median of a group data can be obtained graphically as the x coordinate of the point
of intersection of more and less than ogive.

Fig. 14.1

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. While computing mean of a grouped data, we assume that the frequencies are

[NCERT Exemplar]
(a) centered at the lower limits of the classes

(b) centered at the upper limits of the classes

(c) centered at the class marks of the classes

(d) evenly distributed over all the classes

2. The graphical representation of a cumulative frequency distribution is called

[NCERT Exemplar]

(a) bar graph (b) histogram (c) frequency polygon (d) ogive

3. Construction of a cumulative frequency table is useful in determining the [NCERT Exemplar]

(a) mean (b) median (c) mode (d) all of the above

392 Xam idea Mathematics–X