Class 10 - Chapter 12: Surface Areas and Volumes

Exercise 12.1

Q1. Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Volume of each cube = 64 cm³

Let the side of each cube be a cm.

a³ = 64

a = 4 cm

When two cubes are joined end to end:

Length = 8 cm

Breadth = 4 cm

Height = 4 cm

Surface Area of cuboid

= 2(lb + bh + hl)

= 2[(8×4) + (4×4) + (8×4)]

= 2(32 + 16 + 32)

= 160 cm²

Surface Area = 160 cm²

Q2. 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.

Diameter of hemisphere = 14 cm

Radius (r) = 7 cm

Total height = 13 cm

Height of cylinder (h)

= 13 − 7

= 6 cm

Inner Surface Area

= Curved Surface Area of Hemisphere + Curved Surface Area of Cylinder

= 2πr² + 2πrh

= 2 × (22/7) × 7² + 2 × (22/7) × 7 × 6

= 308 + 264

= 572 cm²

Inner Surface Area = 572 cm²

Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Radius (r) = 3.5 cm

Total height = 15.5 cm

Height of cone (h)

= 15.5 − 3.5

= 12 cm

Slant height of cone (l)

= √(h² + r²)

= √(12² + 3.5²)

= √(144 + 12.25)

= √156.25

= 12.5 cm

Total Surface Area

= CSA of Cone + CSA of Hemisphere

= πrl + 2πr²

= (22/7 × 3.5 × 12.5) + 2 × (22/7 × 3.5 × 3.5)

= 137.5 + 77

= 214.5 cm²

Total Surface Area = 214.5 cm²

Q4. 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.

Side of cube = 7 cm

Greatest diameter of hemisphere = side of cube

= 7 cm

Radius (r) = 3.5 cm

Surface Area of Solid

= Surface Area of Cube + CSA of Hemisphere − Area of Circular Base

= 6a² + 2πr² − πr²

= 6(7²) + π(3.5²)

= 294 + 38.5

= 332.5 cm²

Greatest Diameter = 7 cm

Surface Area = 332.5 cm²

Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Let the edge of the cube be a cm.

Radius of hemisphere

r = a/2

Surface Area of Remaining Solid

= Surface Area of Cube − Area of Circular Portion + CSA of Hemisphere

= 6a² − πr² + 2πr²

= 6a² + πr²

= 6a² + π(a/2)²

= 6a² + πa²/4

Using π = 22/7,

Surface Area = a²(6 + 11/14)

= 95a²/14

Surface Area of remaining solid = 95a²/14 square units

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Diameter = 5 mm

Radius (r) = 2.5 mm

Total length = 14 mm

Length of cylindrical portion (h)

= 14 − 5

= 9 mm

Surface Area of Capsule

= CSA of Cylinder + Surface Area of Sphere

= 2πrh + 4πr²

= 2 × (22/7) × 2.5 × 9 + 4 × (22/7) × (2.5)²

= 141.43 + 78.57

= 220 mm²

Surface Area of Capsule = 220 mm²

Q7. 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 at the rate of ₹500 per m².

Given:

Height of cylinder (h) = 2.1 m

Diameter = 4 m

Radius (r) = 2 m

Slant height of cone (l) = 2.8 m

Since the base of the tent is not covered,

Canvas required

= Curved Surface Area of Cylinder + Curved Surface Area of Cone

= 2πrh + πrl

= 2 × (22/7) × 2 × 2.1 + (22/7) × 2 × 2.8

= 26.4 + 17.6

= 44 m²

Cost of canvas

= 44 × 500

= ₹22,000

Area of canvas used = 44 m²

Cost of canvas = ₹22,000

Q8. 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.

Height of cylinder = 2.4 cm

Diameter = 1.4 cm

Radius (r) = 0.7 cm

Height of cone = 2.4 cm

Slant height of cone:

l = √(r² + h²)

= √(0.7² + 2.4²)

= √(0.49 + 5.76)

= √6.25

= 2.5 cm

Total Surface Area of remaining solid

= CSA of Cylinder + Area of Bottom Circular Base + CSA of Conical Cavity

= 2πrh + πr² + πrl

= 2 × (22/7) × 0.7 × 2.4

+ (22/7) × (0.7)²

+ (22/7) × 0.7 × 2.5

= 10.56 + 1.54 + 5.50

= 17.60 cm²

To the nearest square centimetre,

Total Surface Area = 18 cm²

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base radius is 3.5 cm, find the total surface area of the article.

Given:

Height of cylinder (h) = 10 cm

Radius (r) = 3.5 cm

The surface consists of:

• Curved Surface Area of Cylinder
• Curved Surface Area of two hemispherical hollows

CSA of Cylinder

= 2πrh

= 2 × (22/7) × 3.5 × 10

= 220 cm²

CSA of one hemisphere

= 2πr²

= 2 × (22/7) × 3.5 × 3.5

= 77 cm²

CSA of two hemispheres

= 2 × 77

= 154 cm²

Total Surface Area

= 220 + 154

= 374 cm²

Total Surface Area of the article = 374 cm²