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NCERT Class 9 Maths Textbook Chapter 13 With Answer Book PDF Free Download
Chapter 13: Surface Areas and Volumes
13.1 Introduction
Wherever we look, usually we see solids. So far, in all our studies, we have been dealing with figures that can be easily drawn on our notebooks or blackboards.
These are called plane figures. We have understood what rectangles, squares, and circles are, what we mean by their perimeters and areas, and how we can find them.
We have learned these in earlier classes. It would be interesting to see what happens if we cut
out many of these plane figures of the same shape and size from cardboard sheets and
stack them up in a vertical pile.
By this process, we shall obtain some solid figures (briefly called solids) such as a cuboid, a cylinder, etc.
In the earlier classes, you have also learned to find the surface areas and volumes of cuboids, cubes, and cylinders.
We shall now learn to find the surface areas and volumes of cuboids and cylinders in detail and extend this study to some other solids such as cones and spheres.
13.4 Surface Area of a Right Circular Cone
So far, we have been generating solids by stacking up congruent figures. Incidentally, such figures are called prisms.
Now let us look at another kind of solid that is not a prism (These kinds of solids are called pyramids.). Let us see how we can generate them.
| Author | NCERT |
| Language | English |
| No. of Pages | 30 |
| PDF Size | 3 MB |
| Category | Mathematics |
| Source/ Credits | ncert.nic.in |
NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes
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 m2 is 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.
5. A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
From the question statement, we have
Edge of a cube = 10cm
Length, l = 12.5 cm
Breadth, b = 10cm
Height, h = 8 cm
(i) Find the lateral surface area for both the figures
Lateral surface area of cubical box = 4 (edge)2
= 4(10)2
= 400 cm2 …(1)
Lateral surface area of cuboidal box = 2[lh+bh]
= [2(12.5×8+10×8)]
= (2×180) = 360
Therefore, the Lateral surface area of the cuboidal box is 360 cm2. …(2)
From (1) and (2), the lateral surface area of the cubical box is more than the lateral surface area of the cuboidal box. The difference between both the lateral surfaces is, 40 cm2.
(Lateral surface area of cubical box – Lateral surface area of cuboidal box=400cm2–360cm2 = 40 cm2)
(ii) Find the total surface area for both the figures
The total surface area of the cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2…(3)
Total surface area of cuboidal box
= 2[lh+bh+lb]
= [2(12.5×8+10×8+12.5×100)]
= 610
This implies Total surface area of a cuboidal box is 610 cm2..(4)
From (3) and (4), the total surface area of the cubical box is smaller than that of the cuboidal box. And their difference is 10cm2.
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2
6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25 cm wide and 25 cm high.
(i)What is the area of the glass?
(ii)How much tape is needed for all the 12 edges?
Solution:
Length of greenhouse, say l = 30cm
Breadth of greenhouse, say b = 25 cm
Height of greenhouse, say h = 25 cm
(i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh]
= [2(30×25+30×25+25×25)]
= [2(750+750+625)]
= (2×2125) = 4250
Total surface area of the glass is 4250 cm2
(ii)
From figure, tape is required along sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF.
Total length of tape = 4(l+b+h)
= [4(30+25+25)] (after substituting the values)
= 320
Therefore, 320 cm tape is required for all the 12 edges
NCERT Class 9 Maths Textbook Chapter 13 With Answer Book PDF Free Download