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# RD Sharma Solution for Class 9 Mathematics Chapter 18 - Surface Areas and Volume of a Cuboid and Cube

Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 18 - Surface Areas and Volume of a Cuboid and Cube.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## RD Sharma Solution for Class 9 Mathematics Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Page/Excercise 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

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 13

Solution 14

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 16

Solution 17

Solution 18

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

## RD Sharma Solution for Class 9 Mathematics Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Page/Excercise 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 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

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 17

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 19

Solution 20

Solution 21

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.

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

## RD Sharma Solution for Class 9 Mathematics Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Page/Excercise 18.35

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

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).

Solution 7

## RD Sharma Solution for Class 9 Mathematics Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Page/Excercise 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 9

Solution 10

Solution 11

Solution 12

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).

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

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