Surface Area And Volume Class 9 PDF

Surface Area And Volume Class 9 PDF Summary

Greetings to all, today we are going to upload the Surface Area And volume Class 9 PDF to assist you all. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes include the accurately prepared wide range of solved exercise questions for an outstanding understanding. These solutions in Maths for Class 9 are prepared considering the latest CBSE syllabus 2021-22 update of the Term II examination. NCERT for Class 9 Maths Solutions is specially created to be a guiding solution and good reference to assist students to clear doubts spontaneously and effectively. NCERT Solutions for Class 9 Maths is conditioned by the teaching faculties having vast teaching experience along with subject matter experts to serve the purpose.
Class 9 NCERT Solutions are formulated keeping in mind the concept-based approach along with the precise answering method for Term-II examinations. Guide to NCERT Solutions for Class 9 for a better score. It is an exact and well-structured solution for a good understanding of concept-based knowledge.

Surface Area And volume Class 9 PDF

1. A plastic box 1.5 m long, 1.25 m wide, and 65 cm deep, is to be made. It is to be open 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 the sheet for it, if a sheet measuring 1m² costs Rs. 20.
Solution:
Given: length (l) of box = 1.5m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65m
(i) Box is to be open at top
Area of sheet required.
= 2lh+2bh+lb
= [2×1.5×0.65+2×1.25×0.65+1.5×1.25]m²
= (1.95+1.625+1.875) m² = 5.45 m²
(ii) Cost of sheet per m² area = Rs.20.
Cost of sheet of 5.45 m² area = Rs (5.45×20)
= Rs.109.

2. The length, breadth, and height of a room are 5 m, 4 m, and 3 m respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs 7.50 per m².

Solution:
Length (l) of room = 5m
Breadth (b) of room = 4m
Height (h) of room = 3m
It can be observed that four walls and the ceiling of the room are to be whitewashed.
Total area to be whitewashed = Area of walls + Area of the ceiling of the room
= 2lh+2bh+lb
= [2×5×3+2×4×3+5×4]
= (30+24+20)
= 74
Area = 74 m²
Also,
Cost of whitewash per m² area = Rs.7.50 (Given)
Cost of whitewashing 74 m² area = Rs. (74×7.50)
= Rs. 555

3. The floor of a rectangular hall has a perimeter of 250 m. If the cost of painting the four walls at the rate of Rs.10 per m²is Rs.15000, find the height of the hall.

[Hint: Area of the four walls = Lateral surface area.]
Solution:
Let length, breadth, and height of the rectangular hall be l, b, and h respectively.
Area of four walls = 2lh+2bh
= 2(l+b)h
Perimeter of the floor of hall = 2(l+b)
= 250 m
Area of four walls = 2(l+b) h = 250h m²
Cost of painting per square meter area = Rs.10
Cost of painting 250h square meter area = Rs (250h×10) = Rs.2500h
However, it is given that the cost of painting the walls is Rs. 15000.
15000 = 2500h
Or h = 6
Therefore, the height of the hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Solution:
Total surface area of one brick = 2(lb +bh+lb)
= [2(22.5×10+10×7.5+22.5×7.5)] cm²
= 2(225+75+168.75) cm²
= (2×468.75) cm²
= 937.5 cm²
Let n bricks can be painted out by the paint of the container
Area of n bricks = (n×937.5) cm² = 937.5n cm²
As per the given instructions, an area that can be painted by the paint of the container = 9.375 m² = 93750 cm²
So, we have, 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
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