NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 13
Chapter Name	Surface Areas and Volumes
Exercise	Ex 13.1
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

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 1 m² costs ?
Solution.
We have, a plastic box whose length (l) = 1.5 m, width (b) = 1.25 m
and depth (b) = 65 cm = $\cfrac { 65 }{ 100 }$ = 0.65 m
[ ∵ 1 cm = $\cfrac { 65 }{ 100 }$
Surface area of the box = 2(lb + bh + hi)
= 2 (1.5 x 1.25 +1.25 x 0.65 + 0.65 x 1.5)
= 2(1.875 + 0.8125 + 0.975)
= 2 (3.6625) = 7.325 m²

(i) Area of sheet required for making the box
= Surface area of the box – Area of opened the top
= 7.325 -l x b [since, box is opened at the top]
= 7.325 – (1.5 x 1.25)
= 7.325 – 1.875= 5.45m²

(ii) Given, cost of 1 m² sheet = Rs. 20
∴ Cost of 5.45 m² sheet = 20 x 5.45 = Rs. 109

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 Rs.7.50 per m².
Solution.
Given, dimensions of room are l = 5 m, b = 4m and h = 3 m
Required area for white washing = Area of the four walls + Area of ceiling = 2 (l + b) x h + (l x b)
= 2 (5 + 4) x 3 + (5 x 4) = 2 x 9 x 3 + 20 = 54 + 20 = 74 m²Given, cost of white washing 1 m² wall = RS. 7.50
Cost of white washing of 74 m² wall = 7.50 x 74 = RS. 555

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 Rs. 15000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area]
Solution.
Let l, b and h be the length, breadth and height of rectangular hall respectively.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

Question 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 x 10 cm x 7.5 cm can be painted out of this container?
Solution.
Given, dimensions of a brick are l = 22.5 cm, b = 10 cm and h = 7.5 cm
and total surface area of container = 9.375 m²Total surface area of brick = 2 (l x b + b x h + h x l)
= 2(22.5 x 10 +10 x 7.5 + 7.5 x 22.5)
= 2(225 + 75 + 168.75)
= 2 x 468.75 = 937.5 cm² NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes 1

Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm 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.
Given, length of the cubical box (l) = 10 cm
For cuboidal box, length (Z) = 12.5 cm, breadth (b) = 10 cm and height (h) = 8 cm
(i) Lateral surface area of cubical box
= 4l² =4 (10)² = 4 x 100 = 400 cm²Lateral surface area of cuboidal box
= 2(l + b)x h = 2 (12.5 +10) x 8
= 2 (22.5) x 8 = 45 x 8 = 360 cm²(Lateral surface area of cubical box) >(Lateral surface area of cuboidal box) [∵ 400 > 360]
∴ Required difference = 400 -360 = 40 cm²Hence, cubical box has 40 cm² more lateral surface area.

(ii) Total surface area of cubical box
= 6l² = 6(10)² = 6 x 100 = 600 cm²Total surface area of cuboidal box
= 2(l x b + b x h + h x l)
= 2 (12.5 x 10+10 x 8+8 x 12.5)
= 2 (125 + 80 +100) = 2 x 305 = 610 cm²(Total surface area of cuboidal box) > (Total surface area of cubical box) [∵ 610 >600]
∴ Required difference = 610 – 600 = 10 cm²Hence, cubical box has 10 cm² less total surface area.

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Solution.
Dimensions of herbarium are Z = 30 cm, b = 25 cm and h = 25 cm
(i) Area of the glass = Total surface area of herbarium
= 2 (Z x b + b x h + h x l)
= 2 (30 x 25 + 25 x 25 + 25 x 30)
= 2(750 + 625 + 750) = 2 (2125) = 4250 cm²

(ii) Herbarium is a shape of cuboid.
So, length of required tape will be equal to total length of edges of cuboid.
∴ Length of the tape = 4 (l + b + h) = 4(30 + 25 + 25)
= 4 x 80 = 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm², then find the cost of cardboard required for supplying 250 boxes of each kind.
Solution.
Dimensions of bigger box are l = 25 cm, b = 20 cm and h = 5 cm
∴ Total surface area of the bigger box = 2 (l x b + b x h + h x l)
= 2 (25 x 20 + 20 x 5 + 5 x 25)
= 2 (500 +100 +125) = 2(725) = 1450 cm²Dimensions of smaller box are 1 = 15 cm, b = 12 cm and h = 5 cm
∴ Total surface area of the smaller box
= 2(15 x 12+12 x 5 + 5 x 15)
= 2(180 + 60 + 75) = 2 (315) = 630 cm²Total surface area of both boxes = 1450 + 630 = 2080 cm²Area for all the overlaps = 5 % of total surface area
= 5% x 2080 = $\cfrac { 5 }{ 100 }$ x 2080 = 104 cm²Total surface area of both boxes with area of overlaps = 2080 +104 = 2184 cm²∴ Total surface area for such 250 boxes = 2184 x 250 cm²Now, cost of the cardboard for 1000 cm² = Rs. 4
Then, cost of the cardboard for 1 cm ² = Rs. $\cfrac { 4 }{ 1000 }$
∴ Cost of the cardboard for (2184 x 250) cm²= $\cfrac { 4\times 2184\times 250 }{ 1000 }$ = Rs.2184
Hence, required cost of cardboard is Rs. 2184.

Question 8.
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 he 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.
Dimensions of shelter are l = 4 m, b = 3 m and h = 2.5 m Required area of tarpaulin to make the shelter
= Area of 4 sides + Area of the top of the car = 2(l + b) x h + (l x b)
= 2 (4 + 3) x 2.5 + (4 x 3)
= (2 x 7 x 2.5)+12 = 35+ 12 = 47m²

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

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