RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1 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 Exercise 15.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 15 |

Chapter Name |
Surface Area and Volume |

Exercise |
Exercise 15.1 |

Number of Questions |
10 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1

Question 1.

On the basis of given measure determine surface area of cuboidal wooden log, cuboidal brick and box.

Solution:

(i) Surface area of(RBSESolutions.com)cubical wooden log

= 6 a²

= 6 x (8)²

= 6 x 64

= 384 cm.²

(ii) Surface area of brick

= 2 (lb + bh + hl)

= 2(9 x 5 + 5 x 3 + 3 x 9)

= 2 (45 + 15 + 27)

= 2 (87)

= 174 cm.²

(iii) Surface area of box

= 2 (lb + bh + hl)

= 2 (150 x 40 + 40 x 70 + 70 x 150)

= 2 (6000 + 2800 + 10500)

= 2 (19300)

= 38600 cm.²

Question 2.

Determine side of a cube whose total surface area is 600 square cm.

Solution:

Let a be the length of side of cube.

Then, total surface area of cube = 6a² cm.²

According to question,

6 a² = 600

a² = \(\frac { 600 }{ 6 }\) = 100

a = √100 = 10 cm

Hence, required(RBSESolutions.com)length of side of cube is 10 cm.

Question 3.

In the given figure whose surface area is more ?

Solution:

For first figure (cylinder)

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

h = 7 meter

∴ Surface Area = 2πr (h + r)

= 231 m²

For second figure (cube) –

a = 7 meter

∴Surface Area = 6 a²

= 6 (7)²

= 6 x 49

= 294 m²

Hence, Surface area of cube is greater than surface area of cylinder..

Question 4.

Find the area of curved surface if area of base of cylindrical tank is 176 cm.² and height is 30 cm.

Solution

Let r cm be the radius of base.

Then, circumference(RBSESolutions.com)of base = 2πr cm.

According to question,

2πr = 176

r = 28 cm

h = 30 cm.

Area of curved surface = 2πrh ,

= 2. \(\frac { 22 }{ 7 }\). 28.30

= 5280 cm.²

Question 5.

Form a sheet of 8 square meter, a closed cylindrical tank is formed which has one meter height and 140 cm diameter.How much sheet will be left after making tank?

Solution

For cylindrical tank diameter = 140 cm.

∴Radius (R) = \(\frac { 140 }{ 2 }\) cm. = 70 cm.

Height (H) = 1 meter = 100 cm.

∴ Surface area of cylindrical tank = 2 πR (H + R)

= 2 . \(\frac { 22 }{ 7 }\) . 70 (100 + 70)

= 74800 cm.²

= \(\frac { 74800 }{ 1000 }\) m²

= 7.48 m²

Surface area(RBSESolutions.com)of metal sheet = 8 m²

∴ Surface area of remaining sheet = 8 – 7.48

= 0.52 m²

Question 6.

How many paint tins having spread capacity of 100 cm.² will be required to paint external surface of box having dimensions 80 cm x 50 cm x 25 cm.

Solution

Area of outer surface of box

= 2 (lb + bh + hl)

= 2 (80 x 50 + 50 x 25 + 25 x 80)

= 2 (4000 + 1250 + 2000)

= 2 (7250)

= 14500 cm.²

Expansion capacity of 1 paint box = 100 cm.²

∴Required number of boxes

= \(\frac { 14500 }{ 100 }\)

= 145

Question 7.

There are 25 cylindrical pillars in a building. Each pillar has radius of 28 cm and height of 4 m. find expenditure of painting curved surface area of all(RBSESolutions.com)pillars at the rate of Rs. 8 per meter square.

Solution

For one cylindrical pillar,

Radius (r) = 28 cm.

Height (h) = 4 m = 4 x 100 cm.

= 400 cm.

∴ Curved surface area of 1 pillar = 2πrh

= 2 x \(\frac { 22 }{ 7 }\) x 28 x 400

= 70,400 cm.²

∴ Curves surface area of 25 cylindrical pillars = 70400 x 25

= 1760000 cm.²

= \(\frac { 1760000 }{ 1000 }\) m²

= 176 m²

∴ Cost of 25 cylindrical pillars painting = 176 x 8

= Rs 1408

Question 8.

Curved surface area of a hollow cylinder is 4224 cm². A rectangular sheet having width 33 cm is formed cutting it along its height Find perimeter of sheet

Solution

Let r meter be the radius of base of cylinder and height be the h m. Then,

Surface Area = 2πrh cm²

According(RBSESolutions.com)to question,

⇒ 2πrh = 4224

⇒ 2πr (33) = 4224 | ∵ h = 33cm

⇒ 2πr = \(\frac { 4224 }{ 33 }\)

⇒ 2πr = 128

Perimeter of base = 128cm

Length of sheet (l) = 128 cm

Breadth of sheet (b) = 33 cm

Perimeter of rectangular sheet = 2 (l + b)

= 2 (128 + 33)

= 2 (161)

= 322 cm

Question 9.

To make a road plain a roller has to complete 750 rounds. If the diameter of roller is 84 cm and length 1 meter then find the area of road.

Solution

For Rollar

Diameter = 84 cm.

Radius (r) = \(\frac { 84 }{ 2 }\) cm. = 42 cm.

Length (h) = 1 meter = 100 cm.

∴ Curved(RBSESolutions.com)surface area = 2πrh

= 2 . \(\frac { 22 }{ 7 }\) . 42 . 100

= 26400 cm.²

∴ Area of road made plane in one round = 26400 cm.²

∴ Area of road = 26400 x 750

= 19800000 cm.²

= \(\frac { 19800000 }{ 10000 }\) m²

= 1980 m²

Question 10.

A cube is made by arranging 64 cubes having side of 1 cm, find total surface area of cube so formed.

Solution

Side of one cube = 1 cm.

∴Volume of one cube = a³ = (1)³

= 1 cm³

∴ Volume of 64 cubes = 64 x 1 = 64 cm.³

Let the side of new cube = x cm.

∴ Volume(RBSESolutions.com)of new cube = Volume of 64 cubes

∴ x³ = 64

=> x = (64)^{1/3} = 4 cm.

On comparing

x = 4

∴ Total surface area of new cube

= 6a^{2} = 6 x (4)^{2}

= 6 x 16 = 96 cm.^{2}

We hope the given RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 15 Surface Area and Volume Exercise 15.1, drop a comment below and we will get back to you at the earliest.