# Frank Solutions for Chapter 20 Mensuration II Class 10 ICSE Mathematics

### Exercise 20.1

1. Find the curved surface area, the total surface area and the volume of a cone if its:

(i) Height = 12 cm, radius = 5 cm

(ii) Height = 15 cm, radius = 8 cm

(iii) Height = 16 cm, diameter = 24 cm

(iv) Height = 8 cm, diameter = 12 cm

(i) From the question it is given that,

Height = 12 cm, radius = 5 cm

We know that, curved surface area = (Ï€r√(h2 + r2))

= (22/7) × 5 × √(122 + 52)

= (22/7) × 5 × √(169)

= (22/7) × 5 × 13

= 204.29 cm2

Then, total surface area = area of circular base + curved surface area

= Ï€r2 + (Ï€r√(h2 + r2))

= (22/7) × 52 + 204.29

= 78.57 + 204.29

Therefore, total surface area = 282.86 cm2

Volume = 1/3 × (Ï€r2) × h

= 1/3 × ((22/7) × 52) × 12

= 314.29 cm3

(ii) From the question it is given that,

Height = 15 cm, radius = 8 cm

We know that, curved surface area = (Ï€r√(h2 + r2))

= (22/7) × 8 × √(152 + 82)

= (22/7) × 8 × √(289)

= (22/7) × 8 × 17

= 427.43 cm2

Then, total surface area = area of circular base + curved surface area

= Ï€r2 + (Ï€r√(h2 + r2))

= (22/7) × 82 + 427.43

= 201.14 + 427.43

Therefore, total surface area = 628.57 cm2

Volume = 1/3 × (Ï€r2) × h

= 1/3 × ((22/7) × 82) × 15

= 1005.71 cm3

(iii) From the question it is given that,

Height = 16 cm, diameter = 24 cm

= 24/2

= 12 cm

We know that, curved surface area = (Ï€r√(h2 + r2))

= (22/7) × 12 × √(162 +122)

= (22/7) × 12 × √(400)

= (22/7) × 12 × 20

= 754.29 cm2

Then, total surface area = area of circular base + curved surface area

= Ï€r2 + (Ï€r√(h2 + r2))

= (22/7) × 122 + 754.29

= 452.57 + 754.29

Therefore, total surface area = 1206.86 cm2

Volume = 1/3 × (Ï€r2) × h

= 1/3 × ((22/7) × 122) × 16

= 2413.71 cm3

(iv) From the question it is given that,

Height = 8 cm, diameter = 12 cm

= 12/2

= 6 cm

We know that, curved surface area = (Ï€r√(h2 + r2))

= (22/7) × 6 × √(82 +62)

= (22/7) × 6 × √(100)

= (22/7) × 6 × 10

= 188.57 cm2

Then, total surface area = area of circular base + curved surface area

= Ï€r2 + (Ï€r√(h2 + r2))

= (22/7) × 62 + 188.57

= 113.14 + 188.57

Therefore, total surface area = 301.71 cm2

Volume = 1/3 × (Ï€r2) × h

= 1/3 × ((22/7) × 62) × 8

= 301.71 cm3

2. Find the radius of the circular base of the cone, if its volume is 154 cm3and the perpendicular height is 12 cm.

From the question it is given that,

Volume of circular based cone = 154 cm3

Perpendicular height of the circular based cone = 12 cm

We know that, volume of the cone = 1/3 × (Ï€r2) × h

So,

1/3 × (Ï€r2) × h = 154

⇒ 1/3 × ((22/7) × r2) × 12 = 154

⇒ r2 = (154 × 3 × 7)/(12 × 22)

⇒ r2 = 12.25

⇒ r = √12.25

⇒ r = 3.5 cm

Therefore, radius of the circular base of the cone is 3.5 cm.

3. Find the volume of the right circular cone whose height is 12 cm and slant height is 15 cm. (Ï€ = 3.14)

From the question it is given that,

Height of the right circular cone, h = 12 cm

Slant height of the right circular cone, l = 15 cm

Let us assume radius of the base = r

So, we know that l2 = h2 + r2

r2 = l2 – h2

⇒ r2 = 152 – 122

⇒ r2 = 225 – 144

⇒ r2 = 81

⇒ r = √81

⇒ r = 9 cm

Then, volume = 1/3 × (Ï€r2) × h

= 1/3 × (3.14 × 92) × 12

= 1017.36 cm3

Therefore, volume of the cone is 1017.36 cm3.

4. Find the curved surface area of a cone whose height is 8 cm and base diameter is 12 cm.

From the question it is given that,

Height of the cone = 8 cm

Base diameter of the cone = 12 cm

We know that, curved surface area = (Ï€r√(h2 + r2))

= (22/7) × 6 × √(82 +62)

= (22/7) × 6 × √(100)

= (22/7) × 6 × 10

= 188.57 cm2

Therefore, curved surface area cone is 188.57 cm2.

5. The diameter of a right circular is 12 m and the slant height is 10 m. Find its curved surface area and the total surface area.

From the question it is given that,

Diameter of a right circular = 12 m

Slant height of a right circular = 6 m

We know that, curved surface area = circumference of the base × height

= 2Ï€r × h

= 2 × (22/7) × 6 × 10

= 377.14 m2

Therefore, area of curved surface = 377.14 m2

Then, total surface area = curved surface area + (2 × base area)

= 2Ï€rh + 2Ï€r2

= 2Ï€r (h + r)

= 2 × (22/7) × 6 × (10 + 6)

= 2 × (22/7) × 6 × 16

= 603.42

Therefore, total surface area = 603.42 m2

6. The heights of two cones are in the ratio 1: 3 and their base radii are in the ratio 3: 1. Find the ratio of their volumes.

Let us assume radius of the first cone be 3r and height be h, then radius of second cone will be r and height will be 3h.

We know that, volume of cone = 1/3 × (Ï€r2) × h

Ratio of volumes of cone = (volume of first cone)/(Volume of second cone)

= [1/3 × (Ï€(3r)2) × h]/[1/3 × (Ï€r)2 × 3h]

= ((1/3) Ï€9r2h)/((1/3) Ï€r23h)

= 3

Therefore, ratio of volume of cone = 3 : 1.

7. The base circumference of two cones are the same. If their slant heights are in the ratio 5: 4, find the ratio of their curved surface areas.

From the question it is given that,

Ratio of slant heights of two cones = 5: 4

Let us assume radius be r,

Let us assume slant height of first cone = 5x

Slant height of second cone = 4x

Then, curved surface area of cone = Ï€rl

So, ratio of curved surface areas = (Ï€r × 5x)/(Ï€r × 4x)

= 5/4

Therefore ratio of curved surface areas = 5 : 4.

8. Find the height of the cone whose base radius is 5 cm and volume is 75Ï€ cm3.

From the question it is given that,

Base radius of the cone = 5 cm

Volume of cone = 75Ï€ cm3

We know that,

Volume of cone = 1/3 × (Ï€r2) × h

⇒ 75Ï€ = 1/3 × (Ï€ × 52) × h

⇒ h = 225/25

⇒ h = 9 cm

Therefore, height of the cone is 9 cm.

9. The curved surface area of a right circular cone of radius 11.3 cm is 710 cm2. What is the slant height of the cone?

From the question it is given that,

Radius of right circular cone, r = 11.3 cm

Curved surface area = 710 cm2

Let us assume slant height be l,

Then, Ï€rl = 710

⇒ 22/7 × 11.3 × l = 710

By cross multiplication we get,

l = (710 × 7)/(11.3 × 22)

⇒ l = 19.99 cm

Therefore, slant height of cone is approximately equal to 20 cm.

10. A conical tent requires 264 mof canvas. If the slant height is 12 m, find the vertical height of the cone.

From the question it is given that,

Curved surface area of the tent = 264 m2

Slant height of tent, l = 12 m

Then, Ï€rl = 264

⇒ 22/7 × r × 12 = 264

By cross multiplication we get,

r = (264 × 7)/(22 × 12)

⇒ r = 7 m

Let us assume h be the vertical height,

Then,

l2 = r2 + h2

⇒ h = √(l2 – r2)

⇒ h = √(122 – 72)

⇒ h = √(144 – 49)

⇒ h = √(95)

⇒ h = 9.75 m

Therefore, vertical height of cone = 9.75 m.

11. A conical tent with a capacity of 600 m3stands on a circular base of area 160 m2. Find in m2the area of the canvas.

From the question it is given that,

Volume of the conical tent = 600 m3

Area of circular base cone = 160 m2

Ï€r2 = 160

By cross multiplication we get,

r2 = 160/Ï€

⇒ r = √(160 × 7)/22

⇒ r = √50.909

⇒ r = 7.134 m

We know that,

Volume of cone = 1/3 × (Ï€r2) × h = 600

⇒ 1/3 × (Ï€ × 7.1342) × h = 600

⇒ h = (600 × 7 × 3)/(7.1342 × 22)

⇒ h = 11.265 m

Then, we know that slant height, l = √(r2 + h2)

l = √(7.1342 + 11.2652)

⇒ l = √(177.624)

⇒ l = 13.327 m

Therefore, curved surface area = Ï€rl

= 22/7 × 7.134 × 13.327

= 298.9 m2

Therefore, the area of the canvas is 298.9 m2

12. A hallow metallic cylindrical tube has an internal radius of 3.5 cm and height 21 cm. The thickness of the metal tube is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place.

From the question it is given that,

Internal radius of hallow metallic cylindrical tube, r = 3.5 cm

Height of hallow metallic cylindrical tube, h = 21 cm

Thickness of the metal = 0.5 cm

Then, outer radius of the tube, R = (3.5 + 0.5)cm = 4 cm

We know that volume of hallow metallic cylindrical tube = Ï€h(R2 – r2)

= (22/7) × 21 × (42 – 3.52)

= (22/7) × 21 × (16 – 12.25)

= (22/7) × 21 × 3.75

= 247.5 cm3

Therefore, volume of metal used = 247.5 cm3

Tube is melted and cast into a right circular cone of height 7 cm.

Now, let us assume r1 be the radius of right circular cone,

So, Volume = 1/3Ï€r12h = 247.5

1/3 × 22/7 × r12 × 7 = 247.5

By cross multiplication we get,

r12 = (247.5 × 3 × 7)/(22 × 7)

⇒ r12 = 33.75

⇒ r1 = √(33.75)

⇒ r1 = 5.8 cm

Therefore, radius of the cone is 5.8 cm.

13. A canvas tent is in the shape of a cylinder surrounded by a conical roof. The common diameter of the cone and the cylinder is 14 m. The height of the cylindrical part is 8 m and the height of the conical roof is 4 m. Find the area of the canvas used to make the tent.

From the question it is give that,

The common diameter of the cone and the cylinder = 14 m

The common radius of the cone and the cylinder = 14/2 = 7 m

Height of the cylindrical part, H = 8 m

Height of the conical part, h = 4 m

We know that, slant height of the cone l = √(r2 + h2)

l = √(72 + 42)

⇒ l = √65

⇒ l = 8.06 m

Then,

Area of the canvas used = curved surface area of cylinder + curved surface area of cone

= 2Ï€rH + Ï€rl

= (2 × 22/7 × 7 × 8) + (22/7 × 7 × 8.06)

= 352 + 177.32

= 529.32 m2

14. A circus tent is cylindrical to a height of 5 m and conical above it. If its diameter is 42 m and slant height of the cone is 53 m, calculate the total area of the canvas required.

From the question it is given that,

Height of circle tent, h = 5m

Diameter of circular tent = 42 m

Radius of circular tent = 42/2 = 21 m

Then,

Area of the canvas used = curved surface area of cylinder + curved surface area of cone

= 2Ï€rh + Ï€rl

= (2 × 22/7 × 21 × 5) + (22/7 × 21 × 53)

= 660 + 3498

= 4158 m2

15. Sand from a cylindrical bucket 32 cm in height and 18 cm in radius is poured onto the ground making a conical heap 24 cm high. Find the radius of the conical heap.

From the question it is given that,

Height of the cylinder, h1 = 32 cm

Radius of bucket, r1 = 18 cm

Height of the conical heap, h2 = 24 cm

Let us assume radius of conical heap be r2,

Then,

Volume of sand in the bucket = volume of sand in conical heap

Ï€ × r12 × h1 = 1/3 × Ï€ × r22 × h2

⇒ 18 × 18 × 32 = 1/3 × r2 × 24

⇒ r22 = (10368 × 3)/24

⇒ r22 = 1296

⇒ r2 = √(1296)

⇒ r2 = 36 cm

Therefore, radius of the conical heap is 36 cm.

### Exercise 20.2

1. Find the volume and the surface area of the spheres in the following :

(iii) Diameter = 6.3 cm

2. Find the diameter of the sphere in each of the following :

(i) Volume = 523.17/21 cm3

(ii) Volume = 72Ï€ cm3

(iii) Surface Area = 221. 76 cm2

(iv) Surface area = 576 Ï€ cm2

4. Find the radius of the sphere whose surface area is equal to its volume.

5. Find the radius of a sphere whose surface area is equal to the area of the circle of diameter 2.8 cm.

6. Find the length of the wire of diameter 4 m that can be drawn from a solid sphere of radius 9 m.

Radius of solid sphere = 9 m

7. The radius of a sphere is 9 cm. It is melted and drawn into aa wire of diameter 2mm. Find the length of the wire in metre.

8. A sphere and a cone have the same radii. If their volumes are also eqyual, prove that the height of the cone is twice its radius.

Let r be the radii of sphere and cone.

9. The radius and height of a cylinder, a cone and a sphere are same. Calculate the ratio of their volumes.

Let r, h be the radius and height of cylinder, Cone and sphere.

10. A cylindrical beaker of 7 cm diameter is a partly filled with water. Determine the number of spherical marbles of diameter 1.4 cm that are to be submerged in it to raise the water level by 5.6 cm.

11. A cylindrical bucket, whose base is 20 cm, is filled with water to a height of 25 cm. A heavy iron spherical ball of radius 10 cm is dropped to submerge completely in water in the bucket. Find the increase in the level in water.

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

13. Find the total surface area and volume of a hemisphere whose radius is 10 cm.

14. Find the cost of painting a hemispherical dome of diameter 10 m at the rat of Rs 1.40 per square metre.

15. A solid sphere metal is cut through its centre into 2 equal parts. If the diameter of the sphere is 3.1/3 cm, find the total surface of each part, correct to two decimal places.

16. A circular hall, surmounted by a hemispherical roof, contains 5236 m3 of air. If the internal diameter of the room is equal to the height of the highest point of the roof from the floor, find the height of the hall.

17. Find the volume of the hollow sphere whose inner diameter is 8 cm and the thickness of the material of which it is made is 1 cm.

18. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm, Find the height of the cone.

19. A hollow metallic sphere is 2 cm thick all around and has an external diameter of 12 cm. Find the radius of the solid sphere made by recasting this hollow sphere.

20. A buoy is made in the form of hemisphere surmounted by a right circular cone whose circular base coincides with the plate surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.