RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 15 Surface Area and Volume Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 15 |

Chapter Name |
Surface Area and Volume |

Exercise |
Additional Questions |

Number of Questions |
33 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 15 Surface Area and Volume Additional Questions

**I. Objective Type Questions**

Question 1.

The dimensions of a box are 20 cm. × 10 cm. × 5 cm. The total surface area of box is

(a) 20 × 10 × 5

(b) 2(20 × 10 + 10 × 5 + 5 × 20)

(c) 2(20 +10 + 5)

(d) 2(20 × 10 × 5)

Question 2.

Side of a cube is 10 m. Total surface area of cube is

(a) 600 m²

(b) 100 m²

(c) 500 m²

(d) 400 m²

Question 3.

Total surface (RBSESolutions.com)area of a cube is 36 cm.². Side of a cube is

(a) √9 cm.

(b) √8 cm.

(c) √6 cm.

(d) √10 cm.

Question 4.

The area of base of a cuboid is 140 m² and height is 3 m. Then volume of cuboid is

(a) 140 x 3 m^{3}

(b) \(\frac { 140 }{ 3 }\) m^{3}

(c) (140 + 3) m^{3}

(d) [(140)² – (3)²] m^{3}

Question 5.

The side of a cube is 4 cm. and volume of cube (in cm.^{3}) is

(a) (4)^{2}

(b) (4)^{3}

(c) (4)^{4}

(d) (4)^{5}

Question 6.

Curved surface area of a cylinder is

(a) πr²h

(b) 2πr(h + r)

(c) 2πrh

(d) 2πr

Question 7.

If the radius of a cylinder is 3 cm. and height is 10 cm., then volume is

(a) π(3)^{2} x (10) cm.^{3}

(b) 2π(3) (10) cm.^{2}

(c) 2π(3)^{2} (10) cm.^{3}

(d) π(3)^{2} cm ^{2}

Question 8.

Side of a cube is 1 cm. If we join 2 cubes of side 1 cm., then the(RBSESolutions.com)total surface area is

(a) 6 cm.²

(b) 10 cm.²

(c) 12 cm.²

(d) 20 cm.²

Answers

1. (b)

2. (a)

3. (c)

4. (a)

5. (b)

6. (c)

7. (a)

8. (b)

**II. Fill in the blanks**

Question 1.

Space occupied by any solid is called its___

Question 2.

The quantity of liquid in a three dimensional pot is called___

Question 3.

Area of four(RBSESolutions.com)walls of a room = 2 x (l + b) x ___

Question 4.

Surface area of a solid is equal to___of all areas of its faces.

Question 5.

1 m^{3} =___litre.

Answers

1. Volume

2. Capacity

3. h

4. Sum

5. 1000

**III. True/False Type Questions**

Question 1.

1 cm^{3} = 1 mL

Question 2.

1 L = 1000 cm^{3}

Question 3.

The \(\frac { 2 }{ 3 }\) part of volume of a tank of dimension 6m x 5m x 4m is 80 m^{3}.

Question 4.

The volume of a cylindrical poll, whose height is 7 m and diameter is 12 cm, is π (6)² x 7 cm^{3}.

Answers

1. True

2. True

3. True

4. False.

**IV. Matching Type Questions**

Part 1 | Part 2 |

1. Number of vertices of a cube | (a) 6 m |

2. Number of edges of a cube | (b) 2 |

3. Number of edges of cylinder | (c) 12 |

4. Side of a cube of volume 216 m³ | (d) 8 |

Answers

1. ↔ (d)

2. ↔ (c)

3. ↔ (b)

4. ↔ (a)

**V. Very Short Answer Type Questions**

Question 1.

Why is it wrong to say that given shape is a cylinder?

Solution

A cylinder has two edges and(RBSESolutions.com)congruent circles. Here circles are not congruent. Therefore this figure is not a cylinder.

Question 2.

The ratio of two cylindrical poles is 3 : 2 and ratio of their height is 2 : 3. Find the ratio of curved surface area of poles.

Solution

= \(\frac { 1 }{ 1 }\)

= 1 : 1

Question 3.

The area of three adjacent faces of a cuboid are p, q and r. The volume of cuboid is V. Prove that V² = pqr

Solution

Let l, b and h be the(RBSESolutions.com)dimensions of cuboid respectively. Then,

p = lb

q = bh

r = hl

∴ pqr = (lb) (bh) (hl)

= l²b²h²

= (lbh)²

= V²

⇒V² = pqr

**VI. Short Answer Type Questions**

Question 1.

If V is the volume of a cuboid of dimensions a, b and c and surface(RBSESolutions.com)area is S, then prove that

\(\frac { 1 }{ V } =\frac { 2 }{ S } \left( \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \right) \)

Solution

Question 2.

The population of a village is 2000. Every citizen needs 150 litre water per day. The dimensions of water tank is 20 m x 15 m x 6 m in village. For how many days this(RBSESolutions.com)tank is sufficient for village?

Solution

Volume of tank = 20 x 15 x 6 m^{3}

= 1800 m^{3} ;

= 1800 x 1000 litre

Water requirement of village per day = 150 x 2000

= 3,00,000 litre

∴ Required number of days

Question 3.

How many bricks are required to make a wall of dimensions 12 m. length x 6 decimeter breadth x 4.5 m height. The dimension of one brick is 18 cm x 12 cm x 10 cm(RBSESolutions.com)and \(\frac { 1 }{ 10 }\) Volume is covered by cement?

Solution—Volume of one brick

= 18 x 12 x 10

= 2160 cm.^{3}

Volume of wall = 12 x 0.6 x 4.5

= 32.4 m^{3}

| 6 decimeter = \(\frac { 6 }{ 10 }\) meter = 0.6 meter

∴Volume covered by cement

= \(\frac { 1 }{ 10 }\) x 32.4 = 3.24 m^{3}

∴Volume of bricks

= 32.4 – 3.24

= 29.16 m^{3}

= 29.16 x 1000000 cm.^{3}

∴Number of bricks

= 13500

Question 4.

A closed iron tank of dimensions 12 m length, 9 m breadth and 4 m depth is to be made. The iron sheet is used whose cost is Rs 50 per meter and it is 2 m wide.

Solution

Surface area of tank

= 2 (12 x 9 + 9 x 4 + 4 x 12)

= 2 (108 + 36 + 48)

= 2 (192)

= 384 m²

Width of iron sheet = 2 meter

Length of iron sheet = \(\frac { 384 }{ 2 }\)

= 192 meter

Cost of iron sheet = 192 x 50

= Rs 9600

Question 5.

A cylinder, whose height is 3 m, is open(RBSESolutions.com)from top. The circumference of its base is 22 m. Find its total surface area.

Solution

Let the radius of base be r m. Then,

2πr = 22

⇒ 2 x \(\frac { 22 }{ 7 }\) x r = 22

\(r=\frac { 7 }{ 2 }\)

h = 3 m

Total Surface Area

= 2 πrh + πr^{2}

= \(2.\frac { 22 }{ 7 } .\frac { 7 }{ 2 } .3+\frac { 22 }{ 7 } .\frac { 7 }{ 2 } .\frac { 7 }{ 2 } \)

= 66 + 38.5

= 104.5 m²

Question 6.

A cylindrical pipe, which is open from both sides, is made of iron sheet of thickness 2 cm. If external diameter is 16 cm and height is 100 cm, then find the total iron used in making the pipe?

Solution

External diameter = 16 cm.

External(RBSESolutions.com)radius (R) = \(\frac { 16 }{ 2 }\) cm.

= 8 cm.

Thickness of sheet = 2 cm.

∴ Inner radius (r) = 8 – 2 = 6 cm.

Height (h) = 100 cm.

∴ Used iron

= External Volume – Internal Volume

= πR²h – πr^{2}h

= π (R² – r²) h

= \(\frac { 22 }{ 7 }\) {(8)² – (6)²} 100

= \(\frac { 22 }{ 7 }\) x 28 x 100

= 8800 cm.^{3}

Question 7.

Find the length of longest rod, which can be put in a room of dimensions 12 m x 9 m x 8 m.

Solution

For room

l = 12 m

b = 9 m

h = 8 m

∴ Length of longest rod = Length of diagonal

Question 8.

The curved surface area of a cylindrical tank is 440 m². Its height is 4 m. Find volume of the tank.

Solution

Height of cylinder (h) = 4 m.

Curved surface area of cylinder = 440 m²

To find = Volume of cylinder

∵ Curved(RBSESolutions.com)Surface Area of cylinder = 440 m²

⇒ 2πrh = 440

= 3850 m²

Question 9.

The radius of a cylinder is 7 cm. and its height is 10 cm. Find the curved surface area and volume of the cylinder, (use π = \(\frac { 22 }{ 7 }\))

Solution

Radius of cylinder (r) = 7 cm.

Height (h) = 10 cm.

Curved(RBSESolutions.com)Surface Area of cylinder = 2πrh

Question 10.

Curved surface area of a cylinder is 880 m², whose height is 10 m. Find volume of cylinder. (use π = \(\frac { 22 }{ 7 }\))

Solution

The curved(RBSESolutions.com)surface area of cylinder = 880 sq. m

and height (h) = 10 meter

Hence 2πrh = 880 m

Question 11.

Determine side of a cube whose total surface area is 1014 cm². Find its volume also.

Solution

Let the side of cube be a cm.

Then the total surface area of cube

= 6a² .sq. m

According to question

6a² = 1014

a^{2} = \(\frac { 1014 }{ 6 }\) = 169

a = √169 = 13 cm

Hence the side of cube = 13 cm

Volume(RBSESolutions.com)of cube = (side)^{3}

Hence 13 x 13 x 13 = 2197 cm^{3}

Question 12.

Radius and total surface area of a cylinder are 7 cm and 968 cm^{2} respectively. Find its height.(use π = \(\frac { 22 }{ 7 }\))

Solution

Given,

Radius of cylinder (r) = 7 cm

Total Surface Area = 968 sq. cm.

Height (h) = ?

Total(RBSESolutions.com)surface area of cylinder = 2πr(h + r)

By putting values

968 = 2 x \(\frac { 22 }{ 7 }\) 7(h + 7)

968 = 44(h + 7)

\(\frac { 968 }{ 44 }\) = h + 7

22 = h + 7

h = 22 – 7

h = 15 cm.

∴ Height of cylinder = 15 cm.

We hope the given RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 15 Surface Area and Volume Additional Questions, drop a comment below and we will get back to you at the earliest.

## Leave a Reply