RBSE Maths Chapter 12 – Surface Area and Volume of Cube and Cuboid Class 9 Important questions and solutions are given here. The important questions and solutions of Chapter 12 available at BYJU’S are given by the expert faculty, which contain detailed explanations and required formulas for each and every question. Also, RBSE Class 9 solutions are provided to help the students, so that they can improve their problem solving skills.

Chapter 12 of the RBSE Class 9 Maths has formulas and different types of problems on cube and cuboid. After practicing all these important questions, students will be able to solve the problems of finding the surface area of the cube and cuboid as well as problems on the volume of the cube and cuboid. Also, several word problems are provided here, which will help the students to improve their analytical skills.

## RBSE Maths Chapter 12: Exercise 12.1 Textbook Important Questions and Solutions

**Question 1: The length, breadth and height of a wooden box is 1 m, 60 cm and 40 cm, respectively. Find the outer surface area of the wooden box.**

**Solution:**

Given,

Length of wooden box = l = 1 m = 100 cm

Breadth of the box = 60 cm

Height of the box = 40 cm

Outer surface area of the wooden box = 2 (lb + bh + hl)

= 2(100 × 60 + 60 × 40 + 40 × 100)

= 2(6000 + 2400 + 4000)

= 2 × 12400

= 24800 cm^{2}

= 2.48 m^{2}

**Question 2: The measures of a box are 40 cm, 30 cm and 20 cm respectively. How much square cm cloth is required to make the cover of the box?**

**Solution:**

Given,

Length of the box = l = 40 cm

Breadth of the box = b = 30 cm

Height of the box = h = 20 cm

Area of cloth which will covers the whole box = 2 (lb + bh + hl)

= 2(40 × 30 + 30 × 20 + 20 × 40)

= 2(1200 + 600 + 800)

= 2 × 2600

= 5200 cm^{2}

Therefore, 5200 cm^{2} of cloth is required to cover the whole box.

**Question 3: The length of a room is 5 m, breadth is 3.5 m and height is 4 m. Find the total expenditure of white washing on the four walls and roof at the rate of Rs. 15 per square metre.**

**Solution:**

Given dimensions of the room are:

Length = l = 5 m

Breadth = b = 3.5 m

Height = h = 4 m

Area of four walls = 2(l + b) × h = 2[5 + 3.5] × 4 = 8.5 × 8 = 68.0 sq.m

Area of roof = l × b = 5 × 3.5 = 17.5 sq.m

Total area = 17.5 + 68 = 85.5 sq.m

Cost of expenditure for white washing 1 square metre = Rs. 15

Cost of expenditure for white washing of 85.5 sq. m = 85.5 × Rs. 15 = Rs. 1282.5

**Question 4: The side of a cubical chalk box is 4 cm. Find the total surface area of the chalk box and length of its diagonal.**

**Solution:**

Given,

Side of the cubical chalk box = a = 4 cm

Total surface area of the cubical chalk box = 6a^{2}

= 6 × (4)^{2}

= 6 × 16

= 96 cm^{2}

Diagonal of a cube = √ 3 a

= √ 3 × 4 = 4√ 3 cm

**Question 5: Total surface area of a cube is 1014 m ^{2}. Find the length of the side of the cube.**

**Solution:**

Let a be the side of the cube.

Given,

Total surface area of a cube = 1014 cm^{2}

6a^{2} = 1014

a^{2} = 1014/6 = 169

a = √169 = 13

Therefore, the length of the side of the cube = 13 cm

**Question 6: A wooden box with a lid is made of 2.5 cm thick wood. Inner length, breadth and height of box are 1 m, 65 cm and 55 cm. Find the total expenditure of colouring its outer surface area at the rate of Rs. 15 per square meter.**

**Solution:**

Given,

The inner length of the wooden box = l = 1 m = 100 cm

Inner breadth = b = 65 cm

Inner height = h = 55 cm

Thickness of the wood = 2.5 cm

Outer length = L = 100 + 2 × 2.5 = 100 + 5 = 105 cm = 1.05 m

Outer breadth = B = 65 + 2 × 2.5 = 65 + 5 = 70 cm = 0.70 m

Outer height = H = 55 + 2 × 2.5 = 55 + 5 = 60 cm = 0.60 m

Total outer surface area = 2(LB + BH + HL)

= 2(1.05 × 0.70 + 0.70 × 0.60 + 0.60 × 1.05)

= 2(0.735 + 0.42 + 0.630)

= 2(1.785)

= 3.570 sq. m

Cost of colouring 1 sq. m = Rs. 15.

Cost of colouring 3.570 sq. m = Rs. 15 × 3.570 = Rs. 53.55

**Question 7: Each face of a cube is 100 cm ^{2}. If a cube is cut by a plane, parallel to its base, in two equal parts, then find the total surface area of each equal part.**

**Solution:**

Given,

Area of each face of a cube = 100 cm^{2}

If the surface of a cube is cut by plane which is parallel to the base into two equal parts, then the area of the upper and lower surface of a cube remains the same, whereas the area of the remaining four faces becomes half.

Surface area of each equal part of the separated portion = 100 + 100 + (50 + 50 + 50 + 50)

= 200 + 200

= 400 cm^{2}

Hence, the surface area of each equal part is 400 cm^{2}.

**Question 8: A box without a lid, is made of 3 cm thick wood whose external length, breadth and height are 146 cm, 116 cm and 83 cm, respectively. Find the total surface area of the inner side.**

**Solution:**

Outer length of the box = L = 146 cm

Outer breadth = B = 116 cm

Outer height = H = 83 cm

Thickness of the wood = 3 cm

Inner length =l = 146 – 3 × 2 = 140 cm

Inner breadth = b = 116 – 2 × 3 = 110 cm

Inner height =h = 83 – 3 × 1 (box is open)

Inner area of four walls = 2 (l + b) × h

= 2 (140 + 110) × 80

= 250 × 160

= 40,000 cm^{2}

Area of bottom surface of the wooden box = l × b = 140 × 110 = 15400 cm^{2}

Total surface area of inner side of the box = 40000 + 15400 = 55400 cm^{2}

## RBSE Maths Chapter 12: Exercise 12.2 Textbook Important Questions and Solutions

**Question 9: The measure of a match box is 3 cm × 2 cm × 1 cm. Find the volume of a packet of such 12 match boxes.**

**Solution:**

Given,

Dimensions of a match box (cuboid) are l x b x h = 3 cm x 2 cm x 1 cm, respectively.

Volume of match box = l x b x h = 3 x 2 x 1 = 6 cm^{3}

Volume of 12 such match boxes = 6 x 12 = 72 cm3

**Question 10: Perimeter of a face of a cube is 24 cm. Find the volume of the cube.**

**Solution:**

Let a be the edge of a cube.

Given,

Perimeter of face of a cube = 24 cm

4a = 24

a = 24/4 = 6 cm

Volume of a cube = a^{3}

= 63 = 216 cm^{3}

**Question 11: Core of three cubes of metal are 3 cm, 5 cm and 4 cm, respectively. By melting all there, a new cube is made. Find the volume of the new cube and the length of the core of this cube.**

**Solution:**

Given, edges of the three metal cubes are 3 cm, 5 cm and 4 cm, respectively.

Let x be the edge of the new cube.

According to the given,

Volume of the new cube = Sum of the volume of three cubes

x^{3} = (3)^{3} + (5)^{3} + (4)^{3}

x^{3} = 27 + 64 + 125

x^{3} = 216

x = 6 cm

Therefore, the length of the core of the cubes is 6 cm and volume is 216 cm3.

**Question 12: The length of a water tank is 2.5 m and breadth is 2 m. It contains 1500 litres of water in it. Find the depth of the tank.**

**Solution:**

Given,

Length of the water tank = l = 2.5 m

Breadth of the water tank = b = 2 m

Let h m be the depth of the water tank.

Volume (capacity) of the tank = 1500 litres (1 m^{3} = 1000 litres)

= 1500/1000 = 1.5 m^{3}

According to the given,

lbh = 1.5

2.5 × 2 × h = 1.5

5h = 1.5

h = 1.5/5 = 0.3 m

h = 30 cm

Therefore, depth of the water tank = 30 cm

**Question 13: The length of a wall is 4 m, breadth 15 cm and height us 3 m. How many bricks of size 20 cm × 10 cm × 8 cm are required to make a wall? If the cost of bricks is Rs. 120 per thousand, then find the total cost of bricks.**

**Solution:**

Given,

Length of wall = L = 4 m = 400 cm

Breadth of wall = B = 15 cm

Height of wall = H = 3 m = 300 cm

Volume of wall = L × B × H = 400 × 15 × 300

Length of bricks = l = 20 cm

Breadth of bricks = b = 10 cm

Height of bricks = h = 8 cm

Volume of one brick = l × b × h = 20 × 10 × 8

Number of bricks used = Volume of wall/Volume of one brick

= (400 × 15 × 300)/(20 × 10 × 8)

= 1125

Cost of thousand bricks = Rs. 120

Cost of 1 brick = Rs. 120/1000 = Rs. 0.12

Total cost of 1125 bricks = Rs. 0.12 × 1125 = Rs. 135

Question 14: There is a water tank of size 20 m × 15 m × 6 m in a village. How many litres of water can be filled in it? If 1000 litres of water is used daily from it, then for how many days will it be sufficient?

Solution:

Given,

Length, breadth and height of a water tank are 20 m, 15 m and 6 m, respectively.

Quantity of water required per day = 1000 litres

Capacity of the water tank = 20 × 15 × 6 = 1800 m^{3} (1 m^{3} = 1000 litres)

= 1800 × 1000

= 1800000 litres

Number of days for which the water of the tank will last

= 1800000/1000 = 1800

Hence, a water tank can serve for 1800 days.

Question 15: The length of a wall is 8 m and height is 4 m. Wall is 35 cm thick. There is one door of size 2 m × 1 m and two windows each of size 1.20 m × 1 m. Find the expenditure of making the wall at the rate of Rs. 1500 per cubic metre.

Solution:

Given,

Length of the wall = 8 metres

Thickness of the wall = 35 cm = 35/100 m

Height of the wall = 4 metres

Volume of wall = 8 × (35/100) × 4 = 11.2 cubic metres

Space occupied by 1 door

= 2 × 1 × (35/100)

= 70/100 = 0.70 cubic metres

Space occupied by two windows = 2 × 1.2 × 1 × (35/100)

= 21/25 = 0.84 cubic metres

The remaining volume of the wall = 11.2 – (0.70 + 0.84)

= 11.2 – 1.54

= 9.66 cubic metres

Cost of making of 1 cubic metre of wall = Rs. 1500

Cost of making 9.66 cubic metres of wall = 9.66 × Rs. 1500 = Rs. 14,490.

**Question 16: How many bricks of size 25 cm × 15 cm × 6 cm are required to make a wall 5 m long, 2m high and 50 cm thick, if 10 % of the place is occupied by cement?**

**Solution:**

Given,

Length of the wall = l = 5 m

Breadth of the wall = b = 2 m

Thickness of the wall = h = 50 cm = 0.50 m

Volume of wall = l x b x h

= 5 x 2 x 0.50

= 5 m^{3}

Space occupied by cement = 10% of 5 = 0.5 m^{3}

Actual space required for bricks = 5 – 0.5 = 4.5 m^{3}

Volume of 1 brick = 0.25 x 0.15 x 0.06

= 0.002250 m^{3}

Total number of bricks

= 4.5/0.002250

= (4.5 × 1000000)/2250

= 2000

**Question 17: The mud taken out from a pond, is spread equally in a field. If the pit dug out in a pond is 200 m long, 50 m wide and 0.75 m deep, then how high will be the end of the field? The area of the field is 1500 square meter.**

**Solution:**

Given,

Dimensions of the field are 200 m, 50 m and 0.75 m, respectively.

i.e. length of the pond = L = 200 m

and breadth of the pond = B = 50 m

Depth of the pond = H = 0.75 m

Now volume of the earth taken out from the pit = L x B x H

= 200 x 50 x 0.75

= 7500 m3

Let h metres be the level of field raised due to 7500 m3 of earth dug.

Volume of the field = Volume of earth taken out from the pit

1500 x h = 7500

⇒ h = 7500/1500 = 5 m

Hence, the level of the field be raised by 5 m.

**Question 18: The outer length, breadth and height of the wooden box with lid is 1.25 m, 0.80 m and 0.55 m. Thickness of the wood is 2.5 cm. If the weight of 1 cubic metre wood is 250 kg, then find the total weight of the box.**

**Solution:**

Given,

Outer length of the box (L) = 1.25 m = 125 cm

Outer breadth of box (B) = 0.80 m = 80 cm

Outer height of box (H) = 0.55 m = 55 cm

Thickness of the wood = 2.5 cm

Inner length of the box = l = 125 – 2 x 2.5 = 125 – 5 = 120 cm

Inner breadth of the box = b = 80 – 2 x 2.5 = 80 – 5 = 75 cm

Inner height of the box =h = 55 – 2 x 2.5 = 55 – 5 = 50 cm

Outer volume = L x B x H = 125 x 80 x 55 = 550000 cm^{3}

Inner volume = l x b x h = 120 x 75 x 50 = 450000 cm^{3}

Volume of wood = outer volume – inner volume

= 550000 – 450000

= 100000 cubic cm

= 1/10 cubic metre

Weight of 1 cubic metre of wood = 250 kg

Weight of 1/10 cubic metre of wood = 1/10 x 250 = 25 kg

### RBSE Maths Chapter 12: Additional Important Questions and Solutions

**Question 1: If the volume of a cube is 125 cm ^{3}, then side of a cube will be:**

**(A) 7 m**

**(B) 6 m**

**(C) 5 m**

**(D) 2 m**

**Solution: **

Correct answer: (C)

Let a be the side of the cube.

Given,

Volume of a cube = 125 cm^{3}

a^{3} = 125

a^{3} = 5^{3}

⇒ a = 5

Therefore, the side of the cube is 5 cm.

**Question 2: If the volume of a cube is 1331 cubic centimeter, then surface area of cube is:**

**(A) 762 cm ^{2}**

**(B) 726 cm ^{2}**

**(C) 426 cm ^{2}**

**(D) 468 cm ^{2}**

**Solution:**

Correct answer: (B)

Let a be the side of the cube.

Given,

Volume of a cube = 1331 cubic centimeters

a^{3} = 1331

a^{3} = 11^{3}

⇒ a = 11

Therefore, the side of the cube is 11 cm.

Surface area of the cube = 6a^{2}

= 6 × (11)^{2}

= 6 × 121

= 726 cm^{2}

**Question 3: Length, breadth and height of a cuboid is 4 m, 3 m and 2 m, surface area of cuboid will be**

**(A) 25 m ^{2}**

**(B) 26 m ^{2}**

**(C) 52 m ^{2}**

**(D) 62 m ^{2}**

**Solution:**

Correct answer: (C)

Given dimensions of the cuboid are:

Length = l = 4 m

Breadth = b = 3 m

Height = h = 2 m

Surface area of cuboid = 2(lb + bh + hl)

= 2(4 × 3 + 3 × 2 + 2 × 4)

= 2(12 + 6 + 8)

= 2 × 26

= 52 cm^{2}

**Question 4: Diagonal of a cuboid of measure 8 m × 7 m × 6 m is:**

**(A) 12.2 m**

**(B) 12.02 m**

**(C) 14.2 m**

**(D) 14.02 m**

**Solution:**

Correct answer: (A)

Given dimensions of the cuboid are:

Length = l = 8 m

Breadth = b = 7 m

Height = h = 6 m

Diagonal of the cuboid = √(l^{2} + b^{2} + h^{2})

= √(8^{2} + 7^{2} + 6^{2})

= √(64 + 49 + 36)

= √149

= 12.2 m

**Question 5: Side of a cube is 5 cm. Diagonal of cube is **

**(A) 4√3 cm**

**(B) 2√3 cm**

**(C) 5√3 cm**

**(D) 5 cm**

**Solution:**

Correct answer: (C)

Given,

Side of the cube = a = 5 cm

Diagonal of the cube = √3 a

= √3 × 5

= 5√3 cm

**Question 6: If the volume of a cuboid is 400 cm ^{3} and area of its base is 80 cm^{2}, then height of cuboid is**

**(A) 7 cm**

**(B) 6 cm**

**(C) 4 cm**

**(D) 5 cm**

**Solution:**

Correct answer: (D)

Given,

Volume of a cuboid = lbh = 400 cm^{3}

Area of the base = lb = 80 cm^{2}

Height of the cuboid = Volume/Area of the base

= 400/80

= 5 cm

**Question 7: Measures of a cuboid are 15 cm × 12 cm × 6 cm. How many cubes of side 3 cm can be made by melting this cuboid?**

**Solution:**

Given dimensions of cuboid are:

Length = l = 15 cm

Breadth = b = 12 cm

Height = h = 6 cm

Side of the cube = a = 3 cm

Required number of cubes = Volume of cuboid/Volume of cube

= lbh/a3

= (15 × 12 × 6)/ (3 × 3 × 3)

= 5 × 4 × 2

= 40

**Question 8: Edge of two cubical dice is 2 cm. A solid is made by pasting a face of each dice. Find the total surface area of solid.**

**Solution:**

Given,

Edge of two cubical dice = 2 cm

Length of the solid = l = 2 + 2 = 4 cm

Breadth = b = 2 cm

Height = h = 2 cm

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

= 2(4 × 2 + 2 × 2 + 2 × 4)

= 2(8 + 4 + 8)

= 2 × 20

= 40 cm^{2}

**Question 9: An empty cistern is 4 m long and 3 m wide. How many cubic meters of water is filled in it so that the height of water becomes 2 m?**

**Solution:**

Given,

Length = l = 4 m

Breadth = b = 3 m

Depth = h = 2 m

Volume of empty cistern = l × b × h

= 4 × 3 × 2

= 24 m3

Therefore, 24 m3 of water must be filled in the empty cistern so that depth of water in the cistern becomes 2 m.

**Question 10: There is 8 litres of water in a cubical pot. Find the total surface area of the pot.**

**Solution:**

Given,

Volume of pot = 8 litres (1 litre = 1000 cm^{3})

8 litres = 8000 cm^{3}

Let a be the side of the cubical pot.

Volume = a^{3}

⇒ a^{3} = 8000

⇒ a^{3} = (20)^{3}

⇒ a = 20 cm

Total surface area of the pot = 6a^{2}

= 6 × (20)^{2}

= 6 × 400

= 2400 cm^{2}

**Question 11: If the length, breadth and height of a right angular parallel hexagon are in the ratio 6 : 5 : 4 and its total surface area is 33300 square centimeters. Find the volume of the right angular parallel hexagon.**

**Solution:**

Given,

Ratio of dimensions = 6 : 5 : 4

Let 6x, 5x and 4x be the length, breadth and height.

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

= 2[(6x)(5x) + (5x)(4x) + (4x)(6x)]

= 2(30x^{2} + 20x^{2} + 24x^{2})

= 2(74x^{2})

= 148x^{2}

According to the given,

148x^{2} = 33300 cm^{2}

x^{2} = 33300/148

x^{2} = 225

x = √225

x = 15 cm

Dimensions of regular parallel hexagon:

6x = 6(15) = 90 cm

5x = 5(15) = 75 cm

4x = 4(15) = 60 cm

Volume of regular parallel hexagon (cuboid) = 90 × 75 × 60 = 405000 cm^{3}

**Question 12: A plot is 20 m long and 15 m wide. Digging the land 10 m, breadth 6 m and depth 5 m, from outside of the plot is spread over the plot. Find the height of the mud spread over in the plot.**

**Solution:**

Dimension of the plot are 20 m and 15 m respectively

Length of the plot = L = 20 m

and breadth of the plot = B = 15 m

Volume of the mud spreaded on the plot = l × b × h

= 10 × 6 × 5

= 300 m^{3}

Let H metres of the height of the mud spread over the plot.

Volume of the plot = Volume of mud

⇒ 20 x 15 x H = 300

⇒ h = 300/300 = 1 m

Hence, the level of the plot be raised by 1 m.

**Question 13: Write the formula for length of diagonal of cuboid.**

**Solution:**

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

Diagonal of the cuboid = √(l^{2} + b^{2} + h^{2})