SEBA Class 10 Maths Chapter 12 Surface Area and Volumes

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SEBA Class 10 Maths Chapter 12 Surface Area and Volumes

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Surface Area and Volumes

Chapter: 12

Exercise 12.1

Unless stated otherwise, take π = 22/7

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area

of the resulting cuboids. 

Ans: 

Let side of cube = x cm 

Volume of cube = 64cm³

[Volume of cube = (side)³]

x³ = 64

x = 4 cm 

∴ side of cube = 4cm 

When cube are joined end to end cuboid is formed whose 

Length = 2x cm = 2(4) = 8 cm 

Width = x cm = 4 cm Height = x cm = 4 cm 

∴ Surface area of cuboid = 2[lb + bh + lh]

= 2(8 × 4 × 4 × 4 × 4 × 8)

= 2[32 + 16 + 32] = 2[80] = 160

∴ Surface area of cuboid = 160 cm²

2. 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.

Ans: 

It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm).

Height of hemispherical part = Radius = 7 cm

Total height of vessel = 13 cm

Height of cylindrical part (h) = 13 −7 = 6 cm

Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part 

= 2πrh + 2πr²

3. 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.

Ans: 

It can be observed that the radius of the conical part and the hemispherical part is the same (i.e., 3.5 cm).

Height of hemispherical part = Radius (r) = 3.5 = 7/2 cm

Height of conical part (h) = 15.5 −3.5 = 12 cm 

4. A cubical block of side 7cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. 

Ans: From the figure, it can be observed that the greatest diameter possible for such a hemisphere is equal to the cube’s edge, i.e., 7cm.

Radius (r) of hemispherical part = 7/2 = 3.5cm

Total surface area of solid = Surface area of cubical part + CSA of

hemispherical part − Area of base of hemispherical part

= 6 (edge)² +  2πr² – πr² = (edge)² + πr²

Total surface area of solid = 6(7)2 + 22/7 × 7/2 × 7/2

= 294 + 38.5 = 332.5 cm²

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.

Ans: Diameter of hemisphere = Edge of cube = l

Radius of hemisphere = l/2

Total surface area of solid = Surface area of cubical part + CSA of

hemispherical part − Area of base of hemispherical part

= 6 (edge)² +  2πr² – πr² = (edge)² + πr²

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see fig.12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. 

Ans: 

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part 

= Diameter of the capsule/2 = 5/2

Length of cylindrical part (h) = length of the entire capsule – 2 × r = 14 – 5 = 9 cm 

Surface area of capsule = 2 × CSA of hemispherical part + CSA of cylindrical part 

7. 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 Rs. 500 per m². (Note that the base of the tent will not be covered with canvas.) [Use 𝜋 = 22 7 ]

Ans:  

Given that,

Height (h) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m

Slant height (l) of conical part = 2.8 m

Area of canvas used = CSA of conical part + CSA of cylindrical part 

= πrl + 2πrh

= π × 2 × 2.8 + 2π × 2 × 2.1 

= 44 m²

Cost of 1 m²

canvas = Rs 500

Cost of 44 m²

canvas = 44 × 500 = 22000

Therefore, it will cost Rs. 22000 for making such a tent. 

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical

a cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Ans: 

Given that,

Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Therefore, radius (r) of the cylindrical part = 0.7 cm 

Total surface area of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

The total surface area of the remaining solid to the nearest cm2 is 18 cm²

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the given fig12.11. 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. [Use 𝜋 = 22/7]

Ans: 

Given that,

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm

Height of cylindrical part (h) = 10 cm

Surface area of article = CSA of cylindrical part + 2 × CSA of

hemispherical part 

2πrh + 2 + 2πr²

= 2π × 3.5 × 10 + 2 × 2π × 3.5 × 3.5 

= 70π + 49π

= 119π

= 17 × 22 = 374cm²

10. A solid is hemispherical at the bottom and the top is conical. If the surface area of the two parts are equal, then ratio of its radius to the height of its conical part is.

(a) 1 : √2

(b) √2 : 1

(c) 1 : √3

(d) √3 : 1

Ans: (a) 1 : √2