RD SHARMA Solutions for Class 9 Maths Chapter 18 - Surface Areas and Volume of a Cuboid and Cube
Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.1
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(10 + 7.5) + 12] m2
= 47 m2
= [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
n = 100
Thus, 100 bricks can be painted out by the container.
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



= 19100 cm2
Area of front face = [85



= 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

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)


= (4200 + 2250) cm2
= 6450 cm2
Area to be painted in 3 rows = (3

Cost of painting 1 cm2 area = Rs 0.10
Cost of painting 19350 cm2 area = Rs (19350

Total expense required for polishing and painting the surface of the bookshelf
Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.2




1 m3 = 1000 litres
Thus, the tank can hold 135000 litres of water.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380 m3


10


h = 4.75
Thus, the height of the vessel should be 4.75 m.
Width (b) of the cuboidal pit = 6 m
Depth (h) of the cuboidal pit = 3 m
Volume of the cuboidal pit = l





Cost of digging 1 m3 = Rs 30
Cost of digging 144 m3 = Rs (144

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.

Breadth

Height

Volume of godown = l1






Length

Breadth

Height

Volume of a wooden crate =








Let n wooden crates be stored in the godown.
Volume of n wooden crates = volume of godown
0.9375


Thus, 10666 wooden crates can be stored in godown.

Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min

Thus, in 1 minute 4000

Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Exercise 18.35
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).
Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Exercise 18.36
Let edge = a
Volume, V = a3
If a' = 2a, then
V' = (a')3 = (2a)3 = 8a3
V' = 8 V
Hence, correct option is (d).
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).
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).
Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Exercise 18.37
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change