Request a call back

# 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 12

Length of shelter = 4 m
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] m2
= [2(10 + 7.5) + 12] m2
= 47 m2

### Solution 15

Total surface area of one brick = 2(lb + bh + lh)
= [2(22.5 × 10 + 10 × 7.5 + 22.5 × 7.5)]cm2
= 2(225 + 75 + 168.75)
= (2 × 468.75) cm2
= 937.5 cm2
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n cm2
Area that can be painted by the container = 9.375 m2 = 93750 cm2
93750 = 937.5n
n = 100
Thus, 100 bricks can be painted out by the container.

### Solution 19

External length (l) of bookshelf = 85 cm
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)] cm2
= 19100 cm2
Area of front face = [85 110 - 75 100 + 2 (75 5)] cm2
= 1850 + 750 cm2
= 2600 cm2
Area to be polished = (19100 + 2600) cm2 = 21700 cm2
Cost of polishing 1 cm2 area = Rs 0.20
Cost of polishing 21700 cm2 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
30cm  respectively.
Area to be painted in 1 row = 2 (l + h) b + lh
= [2 (75 + 30) 20 + 75 30] cm2
= (4200 + 2250) cm2
= 6450 cm2
Area to be painted in 3 rows = (3 6450) cm2 = 19350 cm2
Cost of painting 1 cm2 area = Rs 0.10
Cost of painting 19350 cm2 area = Rs (19350 0.10) = Rs 1935
Total expense required for polishing and painting the surface of the bookshelf
= Rs(4340 + 1935) = Rs 6275

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

### Solution 1

Volume of tank = l  b h = (6  5 4.5) m3 = 135 m3
It is given that:
1 m3 = 1000 litres

Thus, the tank can hold 135000 litres of water.

### Solution 2

Let height of cuboidal vessel be h.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380 m3
b h = 380

10 8 h = 380
h = 4.75

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

### Solution 3

Length (l) of the cuboidal pit = 8 m
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 m3
Cost of digging 1 m3 = Rs 30
Cost of digging 144 m3 = Rs (144 30) = Rs 4320

### Solution 16

Length (l) of the cuboidal tank = 20 m
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) m3 = 1800 m3 = 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 18

Length  of the godown = 40 m
Breadth  of the godown = 25 m
Height  of the godown = 10 m

Volume of godown = l1 b1 h1 = (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 22

Rate of water flow = 2 km per hour
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.

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

### Solution 6

Let,

Length = 3x,

Width = 2x

Height = x

Volume = 48 cm3

L×W×H = 48 cm3

3x × 2x × x = 48 cm3

6x3 = 48 cm3

x3 = 8 cm3

x = 2 cm

Total Surface area

= 2(3x × 2x + 2x × x + 3x × x)

= 2(6x2 + 2x2 + 3x2

= 2(11x2)

= 22x2

= 22(4)

= 88 cm2

Hence, correct option is (d).

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

### Solution 8

Let edge = a

Volume, V = a3

If a' = 2a, then

V' = (a')3 = (2a)3 = 8a3

V' = 8 V

Hence, correct option is (d).

### Solution 13

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

Volume, V = abc = 12 cm3

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

V' = a'b'c' = 8abc = 8 × 12 = 96 cm3

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 = a3 = 33 = 27 cm3

Hence, correct option is (b).