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# Class 8 RD Sharma – Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) – Exercise 21.3 | Set 1

• Last Updated : 28 Dec, 2020

### Question 1. Find the surface area of a cuboid whose

(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2 m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.

Solution :

i) length = 10 cm, breadth = 12 cm, height = 14 cm

Given, the Length of cuboid = 10 cm

The Breadth of cuboid = 12 cm

The Height of cuboid = 14 cm

And, Surface area of cuboid = 2(l × b + b × h + h × l)

= 2(10 × 12 + 12 × 14 + 14 × 10) = 856 cm2

Hence, the surface area of given cuboid is 856 cm2

ii) length = 6 dm, breadth = 8 dm, height = 10 dm

Given, the Length of cuboid = 6 dm

The Breadth of cuboid = 8 dm

The Height of cuboid = 10 dm

And, Surface area of cuboid = 2(l × b + b × h + h × l)

= 2(6 × 8 + 8 × 10 + 10 × 6) = 376 dm2

Hence, the surface area of given cuboid is 376 dm2

iii) length = 2 m, breadth = 4 m, height = 5 m

Given, the Length of cuboid = 2 m

The Breadth of cuboid = 4 m

The Height of cuboid = 5 m

And, Surface area of cuboid = 2(l × b + b × h + h × l)

= 2(2 × 4 + 4 × 5 + 5 × 2) = 76 m2

Hence, the surface area of given cuboid is 76 m2

iv) length = 3.2 m, breadth = 30 dm, height = 250 cm

Given, the Length of cuboid = 3.2 m = 32 dm

The Breadth of cuboid = 30 dm

The Height of cuboid = 250 dm = 25 dm

And, Surface area of cuboid = 2(l × b + b × h + h × l)

= 2(32 × 30 + 30 × 25 + 25 × 32) = 5020 dm2

Hence, the surface area of given cuboid is 5020 dm2

### Question 2. Find the surface area of a cube whose edge is

(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m

Solution:

i) 1.2 m

Given, the Edge of cube = 1.2 m

And the Surface area of cube = 6 × edge2

= 6 × (1.2)2 = 8.64 m2

Hence, the surface area of a given cube is 8.64 m2

ii) 27 cm

Given, the Edge of cube = 27 cm

And the Surface area of cube = 6 × edge2

= 6 × (27)2 = 4374 cm2

Hence, the surface area of a given cube is 4374 cm2

iii) 3 cm

Given, the Edge of cube = 3 cm

And the Surface area of cube = 6 × edge2

= 6 × (3)2 = 54 cm2

Hence, the surface area of a given cube is 54 cm2

iv) 6 m

Given, the Edge of cube = 6 m

And the Surface area of cube = 6 × edge2

= 6 × (6)2 = 216 m2

Hence, the surface area of a given cube is 216 m2

v) 2.1 m

Given, the Edge of cube = 2.1 m

And the Surface area of cube = 6 × edge2

= 6 × (2.1)2 = 26.46 m2

Hence, the surface area of a given cube is 26.46 m2

### Question 3. A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area?

Solution :

Given, the Length of cuboidal box = 5 cm

The Breadth of cuboidal box = 5 cm

The Height of the cuboidal box = 4 cm

And, Surface area of cuboidal box = 2(l × b + b × h + h × l)

= 2(5 × 5 + 5 × 4 + 4 × 5) = 130 cm2

Hence, the surface area of the given cuboidal box is 130 cm2

### Question 4. Find the surface area of a cube whose volume is

(i) 343 m3
(ii) 216 dm3

Solution:

i) Given, the volume of cube = 343 m3

It means (side)3 = 343

So, side = 7 m

And, the Surface area of cube = 6 × side2

= 6 × (7)2 = 294 m2

Hence, the surface area of a given cube is 294 m2

ii) Given, the volume of cube = 216 dm3

It means (side)3 = 216

So, side = 6 dm

And, the Surface area of cube = 6 × side2

= 6 × (6)2 = 216 dm2

Hence, the surface area of a given cube is 216 dm2

### Question 5. Find the volume of a cube whose surface area is

(i) 96 cm2
(ii) 150 m2

Solution:

i) Given, the surface area of cube = 96 cm2

It means 6 × side2 = 96

So, side = 4 cm

And, the Volume of cube = (side)3

= (4)3 = 64 cm3

Hence, the volume of the given cube is 64 cm3

ii) Given, the surface area of cube = 150 m2

It means 6 × side2 = 150

So, side = 5 m

And, the Volume of cube = (side)3

= (5)3 = 125 m3

Hence, the volume of the given cube is 125 m3

### Question 6. The dimensions of a cuboid are in the ratio 5 : 3 : 1 and its total surface area is 414 m2. Find the dimensions?

Solution:

Given, the dimensions of a cuboid are in the ratio 5 : 3 : 1

The total surface area is 414 m2

So, the Length of cuboid = 5a

The Breadth of cuboid = 3a

The Height of cuboid = 1a

And, Surface area of cuboid = 2(l × b + b × h + h × l)

414 = 2(5a × 3a + 3a × a + a × 5a)

414 = 46a2

So, a = 3 m

We can conclude, Length = 5a = 15 m

The Breadth = 3a = 9 m

The Height = a = 3 m

Hence, the dimensions of a cuboid are 15 m, 9 m, and 3 m.

### Question 7. Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m, and height 15 cm?

Solution:

Given, the Length of Cuboidal cardboard = 25 cm

The Breadth of Cuboidal cardboard = 0.5 m = 50 cm

The Height of Cuboidal cardboard = 15 cm

Area of cardboard needed = Area of Cuboid

And, Surface area of cuboid = 2(l × b + b × h + h × l)

= 2(25 × 50 + 50 × 15 + 15 × 25) = 4750 cm2

Hence, the area of cardboard required is 4750 cm2

### Question 8. Find the surface area of a wooden box whose shape is of a cube, and if the edge of the box is 12 cm?

Solution :

Given, the Edge of cube = 12 cm

The surface area of a wooden box = Surface area of Cube

And, the Surface area of cube = 6 × edge2

= 6 × (12)2 = 864 cm2

Hence, the surface area of a wooden box is 864 cm2

### Question 9. The dimensions of an oil tin are 26 cm× 26 cm× 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square meter of the tin sheet costs Rs. 10, find the cost of the tin sheet used for these 20 tins?

Solution :

Given, the Length of oil tin = 26 cm

The Breadth of oil tin = 26 cm

The Height of oil tin = 45 cm

Area of tin sheet required to make 1 oil tin = Surface area of Cuboid

And, Surface area of cuboid = 2(l × b + b × h + h × l)

= 2(26 × 26 + 26 × 45 + 45 × 26) = 6032 cm2

And, Area of tin sheet required to make 20 oil tins = 20 × 6032 = 120640 cm2

= 12.064 m2

Since, 1 m2 Tin cost = Rs 10

So, 12.064 m2 Tin cost = Rs 10 × 12.064

= Rs 120.64

Hence, the area of tin required to make 20 oil tin is 12.064 m2 and the cost of making 20 oil tin is Rs 120.64

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