RBSE Solutions for Class 9 Maths Chapter 12 Surface Area and Volume of Cube and Cuboid Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 12 Surface Area and Volume of Cube and Cuboid Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 12 |

Chapter Name |
Surface Area and Volume of Cube and Cuboid |

Exercise |
Additional Questions |

Number of Questions Solved |
32 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 12 Surface Area and Volume of Cube and Cuboid Additional Questions

**Multiple Choice Questions**

Question 1.

Surface area of a cube whose side is \(\frac { 1 }{ 2 }\) cm is:

(A) \(\frac { 1 }{ 4 }\) cm^{2}

(B) \(\frac { 1 }{ 8 }\) cm^{2}

(C) \(\frac { 3 }{ 4 }\) cm^{2}

(D) \(\frac { 3 }{ 2 }\) cm^{2}

Question 2.

How many cubes of 3 cm edge can (RBSESolutions.com) be cut out of a cube of 18 cm edge?

(A) 36

(B) 216

(C) 218

(D) 432

Question 3.

The capacity of a tank of dimensions 8 m x 6 m x 2.5 m is:

(A) 120 litres

(B) 1200 litres

(C) 12000 litres

(D) 120000 litres

Question 4.

A cistern 6 m long and 4 m wide contains water up to (RBSESolutions.com) a depth of 1 m 25 cm then total area of the wet surface is:

(A) 49 m^{2}

(B) 50 m^{2}

(C) 53.5 m^{2}

(D) 55 m^{2}

Question 5.

The maximum length of a pencil that can be kept in a rectangular box of dimension 8 cm x 6 cm x 2 cm is:

(A) 2√13 cm

(B) 2√14 m

(C) 2√26 cm

(D) 10√2 cm

Question 6.

The edge of a cube is 4 cm, then (RBSESolutions.com) its diagonal will be:

(A) 5√3 cm

(B) 2√3 m

(C) 4√3 cm

(D) 5 cm

Question 7.

If each edge/side of the cuboid is double, then total surface area will be:

(A) Double

(B) Four times

(C) Eight times

(D) Three times

Question 8.

Two cubes each of 1 cm edge (RBSESolutions.com) are joined end to end. Then the total surface area of the solid so formed is:

(A) 12 sq. cm

(B) 10 sq. cm

(C) 15 sq. cm

(D) 11 sq. cm

Question 9.

How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?

(A) 10

(B) 100

(C) 1000

(D) 10000

Question 10.

If the volume of two cubes are (RBSESolutions.com) in the ratio 27 : 1, then ratio of their edges is:

(A) 1 : 3

(B) 1 : 27

(C) 3 : 1

(D) 27 : 1

**Answers**

1. D

2. B

3. D

4. A

5. C

6. C

7. B

8. B

9. C

10. C

**Very Short Answer Type Questions**

Question 1.

Find the length of the longest pole that can be placed in (RBSESolutions.com) a room which is 12 m long, 8 m broad, and 9 m high.

Solution.

Length of the longest pole = length of the diagonal of a room

Question 2.

The diagonal of a cube is 6√3 cm. Find its surface area.

Solution.

Let edge of the cube be a.

√3 a = 6√3 ⇒ a = 6

Surface area = 6a^{2} = 6 x (6)^{2} = 6 x 36 = 216 cm^{2}

Question 3.

Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to form (RBSESolutions.com) a new cube. Find the sin-face area of the cube so formed.

Solution.

Volume of new cube = (13 + 63 + 83) cm^{3} = (1 + 216 + 512) cm^{3} = 729 cm^{3}

Edge of new cube = \(\sqrt [ 3 ]{ 729cm^{ 3\qquad } }\) = 9 cm

Surface area of the cube so formed = 6 x (edge)^{2} = 6 x (9)^{2} = 6 x 81 = 486 cm^{2}

Question 4.

Flow many 3 metre cubes can be cut from (RBSESolutions.com) a cuboid measuring 18 m x 12 m x 9 m?

Solution.

We are given that edge of each cube = 3 m

Volume of each cube = (edge)^{3} = (3)^{3} = 27 m^{3}

Volume of the cuboid = l x b x h = 18 x 12 x 9 m^{3}

Number of cubes = \(\frac { Volume of the cuboid }{ Volume of each cube }\)

= \(\frac { 18 x 12 x 9 }{ 27 }\) = 72

Question 5.

A room whose floor is a square of side 6 cm contains 180 cubic metres of air. Find the (RBSESolutions.com) height of the room.

Solution.

Let height of the room be h. Here the room is of square floor, so we can write:

6 x 6 x h = 180 m^{3} (Volume = 180 m^{3})

⇒ h = \(\frac { 180 }{ 6×6 }\)

⇒ h = 5 m

**Short Answer Type Questions**

Question 1.

A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine

(i) the area of the sheet required for making the box.

(ii) the cost of sheet for it, if a sheet measuring 1 m^{2} costs Rs. 20.

Solution.

Here we are given that

(i) Length of plastic box (l) = 1.5 m

Breadth of plastic box (b) = 1.25 m

Height of plastic box (h) = 65 cm = 0.65 m

Since box is open at the top. Therefore total surface area (RBSESolutions.com) of the plastic box (open at the top)

= lb + 2bh + 2hl

= [1.5 x 1.25 + 2(1.25 x 0.65 + 0.65 x 1.5)] m^{2}

= [1.875 + 2(0.8125 + 0.975)] m^{2}

= [1.875 + 2 x 1.7875] m^{2}

= [1.875 + 3.575] m^{2}

= 5.450 m^{2}

Hence, area of the plastic sheet required for making the box is 5.450 m^{2}.

(ii) Cost of 1 m^{2} plastic sheet = Rs. 20

Cost of 5.45 m^{2} plastic sheet = Rs. 5.540 x 20 = Rs. 109

Question 2.

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost (RBSESolutions.com) of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 m^{2}.

Solution.

(i) Here we are given that

length (l) = 5 m, breadth (b) = 4 m and height (h) = 3 m

Total area of four walls and ceiling which is to be white washed =2 (l + b) x h + lb

= [2(5 + 4) x 3 + 5 x 4] m^{2}

= (54 + 20) m2 = 74 m^{2}

(ii) Cost of white washing 1 m^{2} area = Rs. 7.50

Cost of white washing 74 m^{2} area = Rs. (74 x 7.50) = Rs. 555

Question 3.

The floor of a rectangular hall has a perimeter 250 m. If the cost of painting (RBSESolutions.com) the four walls at the rate of Rs. 10 per m2 is Rs. 15,000, find the height of the hall.

[Hint: Area of the four walls = Lateral surface area]

Solution.

Here we are given that perimeter of the floor of a rectangular hall = 250 m

Area of four walls to be painted = 2(l + b) x h = Perimeter x h …..(i)

Cost of painting the four walls at the rate of Rs. 10 per m^{2} is Rs. 15,000

Total area which is to be painted = \(\frac { 15000 }{ 10 }\) = 1500 m^{2}

Using relation (i), we have

1500 m^{2} = Perimeter x h

⇒ 1500 m^{2} = 250 m x h

⇒ h = 6 m

Hence, height of the room is 6 m.

Question 4.

The paint in a certain container is sufficient to paint (RBSESolutions.com) an area equal to 9.375 m^{2}. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?

Solution.

Let the number of bricks be equal to x of dimension 22.5 cm x 10 cm x 7.5 cm which can be painted out of the given container.

Surface area of 1 brick = 2(l x b + b x h + h x l)

= 2(22.5 x 10 + 10 x 7.5 + 7.5 x 22.5) cm^{2}

= 2(225 + 75 + 168.75) cm^{2}

= 2 x 468.75 cm^{2}

= 937.5 cm^{2}

According to question

x x 937.5 cm^{2} = 9.375 m^{2}

1 m^{2} = 10000 cm^{2}

⇒ x x 937.5 cm^{2} = 9.375 x 10000 cm^{2}

⇒ x = \(\frac { 9.375×10000 }{ 937.5 }\)

⇒ x = 100

Hence, required number of bricks = 100.

Question 5.

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres (RBSESolutions.com) of water can it hold? (1 m^{3} = 1000 litres).

Solution.

Length of a water tank = 6 m,

breadth = 5 m and height/depth = 4.5 m

Volume of water tank = l x b x h = 6 m x 5 m x 4.5 m = 135 m^{3}

1 m^{3} = 1000 litres

135 m^{3} = 135 x 1000 litres = 135000 litres

Question 6.

A cuboidal vessel is 10 m long and 8 m wide. How high must it (RBSESolutions.com) be made to hold 380 cubic metres of a liquid?

Solution.

We have,

Length of a cuboidal vessel = 10 m

Breadth = 8 m

Let ‘h’m be the height hold by the liquid in the vessel.

Volume = l x b x h

⇒ h = \(\frac { 380 }{ 10×8 }\) = \(\frac { 38 }{ 8 }\)

⇒ h = 4.75 m

Question 7.

Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at (RBSESolutions.com) the rate of Rs. 30 per m^{3}.

Solution.

Length of a cuboidal pit = 8 m, breadth = 6 m and depth = 3 m

Volume/Intemal space of the pit = 8m x 6m x 3m = 144 m^{3}

Cost of digging 1 m^{3} space is Rs. 30

Cost of digging 144 m^{3} space = Rs. 144 x 30 = Rs. 4320

Question 8.

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank. If its length and depth are respectively 2.5 m and 10 m.

Solution.

1000 litres = 1 m^{3}

50000 litres = 50 m^{3}

i.e. Capcity of a cuboidal tank = 50 m^{3}

Let ‘b’ be the breadth of the tank. Then

l x b x h = Volume

b = \(\frac { Vlume }{ lxh }\) = \(\frac { 50 }{ 2.5×10 }\)

b = 2m

Hence, breadth of the tank = 2 m.

Question 9.

A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be (RBSESolutions.com) the side of the new cube? Also, find the ratio between their surface area.

Sol.

Volume of solid cube of side

12 cm = 12 cm x 12 cm x 12 cm

Number of cubes of equal volume = 8

Volume of one such cube = \(\frac { 12x12x12 }{ 8 }\) cm^{3} = 216 cm^{3}

Edge of the new cube = 3√216 cm^{3} = 6 cm

Surface area of one big cube of side/edge 12 cm = 6 x (12 cm)^{2} = (6 x 144) cm^{2}

Also surface area of one smaller cube = 6 x (6 cm)^{2} = (6 x 36) cm^{2}

Required ratio = \(\frac { 6 x 144 }{ 6 x 36 }\) = 4 : 1.

Question 10.

A water tank is 1 m long, 85 cm wide and 60 cm deep. If the measurement of (RBSESolutions.com) a tin is 25 cm x 25 cm x 34 cm, then find how many tins of water will be contained in it.

Solution.

Length of water tank = 1 m = 100 cm

Breadth of water tank = 85 cm

and height/ depth = 60 cm

Volume of water tank = (100 x 85 x 60) cm^{3}

Similarly, volume of one tin = (25 x 25 x 34) cm^{3}

Number of tins = \(\frac { Volume of water tank }{ Volume ol one tin }\)

= \(\frac { 100 x 85 x 60 }{ 25 x 25 x 34 }\)

= 24 tins

Question 11.

The sum of length, breadth and height of a cuboid is 19 cm and the length (RBSESolutions.com) of its diagonal is 11 cm. Find the surface area of the cuboid.

Solution.

Let length, breadth and height of the cuboid be l, b and h respectively, then

l + b + h = 19 …(i)

and diagonal = 11

⇒ \(\sqrt { { l }^{ 2 }+{ b }^{ 2 }+{ h }^{ 2 } }\) = 11

⇒ l^{2} + b^{2} + h^{2} = 121 …(ii)

Using (i), l + b + h = 19

Squaring both sides, we have

(l + b + h)^{2} = 192

l^{2} + b^{2} + h^{2} + 2 (lb + bh + hl) = 361

121 + 2 (lb + bh + hl) = 361

2 (lb + bh + hl) = 361 – 121

⇒ 2 (lb + bh + hl) = 240

Hence, the surface area of the cuboid is 240 cm^{2}.

Question 12.

How many cassettes of size 11 cm x 7 cm x 1.5 cm can be (RBSESolutions.com) put inside the box of size 35 cm long, 22 cm wide and 6 cm high.

Solution.

Here volume of box = l x b x h = (35 x 22 x 6) cm^{3}

Now, volume of one cassettes = l x b x h = (11 x 7 x 1.5) cm^{3}

Required number of cassettes = \(\frac { Volume of box }{ Volume of one cassette }\)

= \(\frac { 35 x 22 x 6 }{ 11x7x1.5 }\)

Hence, required number of cassette = 40.

**Long Answer Type Questions**

Question 1.

A box is 3 metre long, 2 metre wide and 1 metre 80 cm high. Find the cost of varnishing its outer surface at the rate of Rs. 12 per square centre.

Solution.

Length of the box = 3 m

Breadth = 2 m

Height = 1.80 m

Total surface area of the box = 2 (l x b + b x h + h x l)

= 2 [3 x 2 + 2 x 1.80 + 1.80 x 3] = 2 (6 + 3.60 + 5.40) = 30 sq. m

Cost of varnishing at the rate Rs. 12 per sq. m = 30 x 12 = Rs. 360

Question 2.

A metallic sheet is of rectangular shape with measurement 48 cm x 36 cm. From each one (RBSESolutions.com) of its comer a square of 8 cm is cut off and an open box is made of the remaining sheet. What is the volume of the box?

Solution.

Length of the sheet = 48 cm

Breadth of the sheet = 36 cm

When a square of side 8 cm is cut off from each comer and the flaps turned up, and open box is made.

Length of the open box = 48 – (8 + 8) = 32 cm

Breadth of the open box = 36 – (8 + 8) = 20 cm

Volume of the open box = l x b x h = 32 x 20 x 8 = 5120 cm^{3}

Question 3.

The ratio of length, breadth, height of a cuboid is 5 : 3 : 2. If total surface area (RBSESolutions.com) of the cuboid is 558 cm^{2}, find the length, breadth and height.

Solution.

Let length (l) = 5x

breadth (b) = 3x

and height (h) = 2x

According to question

Total surface area = 2(lb + bh + hl)

or 2(5x × 3x + 3x × 2x + 2x × 5x) = 558

or 2(15x^{2} + 6x^{2} + 10x^{2}) = 558

or 2 × 31x^{2} = 558

or x^{2} = \(\frac { 558 }{ 2×31 }\)

or x^{2} = 9 ⇒ x = 3

Thus, length of cuboid = 5x = 5 × 3 = 15 cm

Breadth = 3x = 3 × 3 = 9 cm

and height = 2x = 2 × 3 = 6 cm

Question 4.

A cuboidal hall is 15 m long and 12 m wide. The sum of area (RBSESolutions.com) of its floor and plane roof is equal to area of four walls. Find the volume of hall.

Solution.

Here, as shown in figure the floor and plane roof is rectangular in shape and length of given cuboidal hall is 15 m, breadth = 12 m

Let height of the cuboidal hall be ‘h’ metres.

According to question,

2(l × b) = 2(b × h + h × l)

⇒ l × b = h(b + l)

⇒ h = \(\frac { 15×12 }{ 15+12 }\) = \(\frac { 180 }{ 27 }\) = \(\frac { 20 }{ 3 }\) m

Now, the volume of cuboidal hall is = l x b x h = 15 x 12 x \(\frac { 20 }{ 3 }\) = 1200 m^{3}

Volume = 1200 m^{3}

Thus, volume of cuboidal hall is 1200 m^{3}.

Question 5.

The areas of three adjacent faces of (RBSESolutions.com) a cuboid are x, y and z respectively. If its volume is V, then prove that V^{2} = xyz.

Solution.

Suppose length, breadth and height of the cuboid are l, b and h respectively.

Volume of cuboid = lbh …(i)

According to question,

x = lb

y = bh

and z = hl

xyz = (lb) × (bh) × (hl) = l^{2}b^{2}h^{2} = (lbh)^{2} [using (i)] = V^{2}

i.e. V^{2} = xyz.

Hence proved.

We hope the given RBSE Solutions for Class 9 Maths Chapter 12 Surface Area and Volume of Cube and Cuboid Additional Questions will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 12 Surface Area and Volume of Cube and Cuboid Additional Questions, drop a comment below and we will get back to you at the earliest.