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

March 23, 2019 by Fazal Leave a Comment

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.
RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1 img-1
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

RBSE Solutions

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 ?
RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1 img-2
Solution:
For first figure (cylinder)
r = \(\frac { 7 }{ 2 }\) meter
h = 7 meter
∴ Surface Area = 2πr (h + r)
RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1 img-3
= 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..

RBSE Solutions

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
RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1 img-4
r = 28 cm
h = 30 cm.
Area of curved surface = 2πrh ,
= 2. \(\frac { 22 }{ 7 }\). 28.30
= 5280 cm.²

RBSE Solutions

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

RBSE Solutions

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
RBSE Solutions for Class 8 Maths Chapter 15 Surface Area and Volume Ex 15.1 img-5
= \(\frac { 14500 }{ 100 }\)
= 145

RBSE Solutions

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

RBSE Solutions

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

RBSE Solutions

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

RBSE Solutions

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
= 6a2 = 6 x (4)2
= 6 x 16 = 96 cm.2

RBSE Solutions

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.

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