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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes are prepared by CBSE student eCARE expert to score good marks in class 9. This chapter contain many topics which are very important to score good in class 9. There are some important topic/Activity mentioned below:
Parveen wanted to make a temporary shelter, for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?
Solution:
Here, length (l) = 4 m,
breadth (b) = 3m
and height (h) = 2.5 m
The structure is like a cuboid.
∴ The surface area of the cuboid, excluding the base
=[Lateral surface area] + [Area of ceiling]
= [2(l + b)h] + [lb]
= [2(4 + 3) x 2.5] m2 + [4 x 3] m2
= 35 m2 + 12 m2 = 47 m2
Thus, 47 m2 tarpaulin would be required.
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Some questions are asked by class 9 students related to Maths
Ex 13.1 Class 9 Maths 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 1m2 costs ?20.
Solution:
(i) Here, length (l) = 1.5 m, bread th(b) = 1 .25 m
and height (h) = 65 cm = 65100 m = 0.65 m
? It is open from the top.
∴ Its surface area
= [Lateral surface area] + [Base area]
= [2(1 + b)h] + [lb]
= [2(1.50 + 1.25)0.65] m2 + [1.50 x 1.25] m2
= [2 x 2.75 x 0.65] m2 + [1.875] m2
= 3.575 m2+ 1.875 m2 = 5.45 m2
∴ Area of the sheet required for making the box = 5.45 m2
(ii) Cost of 1 m2 sheet = Rs. 20
Cost of 5.45 m2 sheet = Rs. (20 x 5.45)
= Rs. 109
Hence, cost of the required sheet = Rs. 109
Ex 13.1 Class 9 Maths Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ?17.50 per m2.
Solution:
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
The room is like a cuboid whose four walls (lateral surface) and ceiling are to be white washed.
∴ Area for white washing
= [Lateral surface area] + [Area of the ceiling]
= [2(l + b)h] + [l x b]
= [2(5 + 4) x 3] m2 + [5 x 4] m2 = 54 m2 + 20 m2 = 74 m2
Cost of white washing for 1 m2 area = Rs. 7.50
∴ Cost of white washing for 74 m2 area = Rs. (7.50 x 74) = Rs. 555
Thus, the required cost of white washing = Rs. 555
Ex 13.1 Class 9 Maths Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ?10 per m2 is ?15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area]
Solution:
A rectangular hall means a cuboid.
Let the length and breadth of the hall be l and b respectively.
∴ Perimeter of the floor = 2(l + b)
⇒ 2(l + b) = 250 m
? Area of four walls = Lateral surface area = 2(1 + b) x h, where h is the height of the hall = 250 h m2
Cost of painting the four walls
= Rs. (10 x 250 h) = Rs. 2500h
⇒ 2500 h = 15000 ⇒ h = 150002500 = 6
Thus, the required height of the hall = 6 m