# Class 9 RD SHARMA Solutions Maths Chapter 18 - Surface Areas and Volume of a Cuboid and Cube

## Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.1

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

Breadth of shelter = 3 m

Height of shelter = 2.5 m

The tarpaulin will be required for top and four sides of the shelter.

Area of Tarpaulin required = 2(lh + bh) + lb

= [2(4 2.5 + 3 2.5) + 4 3] m

^{2}

= [2(10 + 7.5) + 12] m

^{2}

= 47 m

^{2}

### Solution 13

### Solution 14

### Solution 15

= [2(22.5 × 10 + 10 × 7.5 + 22.5 × 7.5)]cm

^{2}

= 2(225 + 75 + 168.75)

= (2 × 468.75) cm

^{2}

= 937.5 cm

^{2}

Let n number of bricks be painted by the container.

Area of n bricks = 937.5n cm

^{2}

Area that can be painted by the container = 9.375 m

^{2}= 93750 cm

^{2}

93750 = 937.5n

n = 100

Thus, 100 bricks can be painted out by the container.

### Solution 16

### Solution 17

### Solution 18

### Solution 19

External breadth (b) of bookshelf = 25 cm

External height (h) of bookshelf = 110 cm

External surface area of shelf while leaving front face of shelf

= lh + 2 (lb + bh)

= [85 110 + 2 (85 25 + 25 110)] cm

^{2}

= 19100 cm

^{2}

Area of front face = [85 110 - 75 100 + 2 (75 5)] cm

^{2}

= 1850 + 750 cm

^{2}

= 2600 cm

^{2}

Area to be polished = (19100 + 2600) cm

^{2}= 21700 cm

^{2}

Cost of polishing 1 cm

^{2}area = Rs 0.20

Cost of polishing 21700 cm

^{2}area = Rs (21700 0.20) = Rs 4340

Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and

Area to be painted in 1 row = 2 (l + h) b + lh

= [2 (75 + 30) 20 + 75 30] cm

^{2}

= (4200 + 2250) cm

^{2}

= 6450 cm

^{2}

Area to be painted in 3 rows = (3 6450) cm

^{2}= 19350 cm

^{2}

Cost of painting 1 cm

^{2}area = Rs 0.10

Cost of painting 19350 cm

^{2}area = Rs (19350 0.10) = Rs 1935

Total expense required for polishing and painting the surface of the bookshelf

## Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.2

### Solution 1

^{3}= 135 m

^{3}

1 m

^{3}= 1000 litres

Thus, the tank can hold 135000 litres of water.

### Solution 2

Length (l) of vessel = 10 m

Width (b) of vessel = 8 m

Volume of vessel = 380 m

^{3}

l b h = 380

10 8 h = 380

h = 4.75

Thus, the height of the vessel should be 4.75 m.

### Solution 3

Width (b) of the cuboidal pit = 6 m

Depth (h) of the cuboidal pit = 3 m

Volume of the cuboidal pit = l b h = (8 6 3) = 144 m

^{3}

Cost of digging 1 m

^{3}= Rs 30

Cost of digging 144 m

^{3}= Rs (144 30) = Rs 4320

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

Breadth (b) of the cuboidal tank = 15 m

Height (h) of the cuboidal tank = 6 m

Capacity of tank = l × b × h = (20 × 15 × 6) m

^{3}= 1800 m

^{3}= 1800000 litres

Water consumed by people of village in 1 day = 4000 × 150 litres = 600000 litres

Let water of this tank lasts for n days.

Water consumed by all people of village in n days = capacity of tank

n × 600000 = 1800000

n = 3

Thus, the water of tank will last for 3 days.

### Solution 17

### Solution 18

Breadth of the godown = 25 m

Height of the godown = 10 m

Volume of godown = l

_{1}b

_{1}h

_{1}= (40 25 10) = 10000

Length of a wooden crate = 1.5 m

Breadth of a wooden crate = 1.25 m

Height of a wooden crate = 0.5 m

Volume of a wooden crate = = (1.5 1.25 0.5) m3 = 0.9375

Let n wooden crates be stored in the godown.

Volume of n wooden crates = volume of godown

0.9375 n = 10000

Thus, 10666 wooden crates can be stored in godown.

### Solution 19

### Solution 20

### Solution 21

### Solution 22

Depth (h) of river = 3 m

Width (b) of river = 40 m

Volume of water flowed in 1 min

Thus, in 1 minute 4000 = 4000000 litres of water will fall into the sea.

### Solution 23

### Solution 24

### Solution 25

### Solution 26

### Solution 27

### Solution 28

## Surface Areas and Volume of a Cuboid and Cube Exercise 18.35

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

Let,

Length = 3x,

Width = 2x

Height = x

Volume = 48 cm^{3}

L×W×H = 48 cm^{3}

3x × 2x × x = 48 cm^{3}

6x^{3} = 48 cm^{3}

x^{3} = 8 cm^{3}

x = 2 cm

Total Surface area

= 2(3x × 2x + 2x × x + 3x × x)

= 2(6x^{2} + 2x^{2} + 3x^{2})

= 2(11x^{2})

= 22x^{2}

= 22(4)

= 88 cm^{2}

Hence, correct option is (d).

### Solution 7

## Surface Areas and Volume of a Cuboid and Cube Exercise 18.36

### Solution 8

Let edge = a

Volume, V = a^{3}

If a' = 2a, then

V' = (a')^{3} = (2a)^{3} = 8a^{3}

V' = 8 V

Hence, correct option is (d).

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

Let the dimensions of Cuboid be a, b, c respectively.

Volume, V = abc = 12 cm^{3}

If a' = 2a, b' = 2b, c' = 2c, then

V' = a'b'c' = 8abc = 8 × 12 = 96 cm^{3}

Hence, correct option is (d).

### Solution 14

Let the edge of cube = a

Total no. of edge = 12

Sum of all edges = 12a

12a = 36cm

i.e. a = 3 cm

Volume = a^{3} = 3^{3} = 27 cm^{3}

Hence, correct option is (b).

### Solution 15

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### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

## Surface Areas and Volume of a Cuboid and Cube Exercise 18.37

### Solution 22