Tag: 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.

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 1m² costs ₹20.
Solution:
We have a plastic box of
l = length = 15 m
b = width = 1.25 m
h = depth = 65 cm
= \(\frac { 65 }{ 100 }\) m= 0.65 m (∵ 1 m = 100cm)
Surface area of the box = 2 (lb + bh + hl)
= 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 the sheet required for making the box
= 7.325 – l x b (∵ BOX is opened at the top)
= 7.325 – 1.5 x 1.25
= 7.325 – 1.875 = 5.45 m²
(ii) A sheet measuring 1 m² costs = ₹20
∴ Sheet measuring 5.45 m² costs = ₹20 x 5.45 = ₹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 ₹17.50 per m².
Solution:
We have a room of l = 5m
b = 4m
h = 3m
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)
= 2x 9x 3 + 20
= 54+20
= 74 m2
White washing 1 m² costs = ₹7.50
White washing 74 m² costs = ₹7.50x 74 = ₹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 ₹15000, find the height of the hall.
[Hint Area of the four walls = Lateral surface area]
Solution:
Let the rectangular hall of length = l, breadth = b, height = h

Hence, the height of the hall is 6 m.

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
l = 22.5 cm, b = 10 cm
and b = 7.5cm
Total surface area of bricks = 2 ( l x b + b x h + h x l)
= 2(22.5 x 10 + 10 x 75 + 75 x 225)
= 2(225 + 75 + 168.75)
= 2 x 468.75 cm²
= 9375 cm²

Number of bricks that painted out of this container

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:
We have l₁ for cubical box = 10 cm
For cuboidal box l= 12.5 cm
b = 10 cm
h = 8 cm
(i) Lateral surface area of cubical box = 4l² = 4(10)²
= 4 x 100
= 400 cm²
Lateral surface area of cuboidal box = 4 (l + b) x h
= 2 (125 + 10) x 8
= 2 (225) x 8
= 45 x 8 = 360 cm²
(∵ Lateral surface area of cuboidal box) > (Lateral surface area of cuboidal box) (∵ 400 >360)
∴ Required area = (400 – 360) cm2 = 40 cm²
(ii) Total surface area of cubical box = 6l2 = 6(10)2 = 6x 100= 600 cm²
Total surface area of cuboidal box = 2 ( l x b + b x h + h x l)
= 2(125 x 10 + 10 x 8 + 8 x 125)
= 2(125 + 80+ 100)
= 2 x 305
= 610 cm²
∴ (Area of cuboidal box) > (Area of cubical box) (∵ 610 > 600)
Required area = (610 – 600)cm² = 10 cm²

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:
Dimension for herbarium are
l = 30 cm, b = 25 cm and h = 25 cm
Area of the glass = 2 (l x b + b x h + h x l)
= 2 ( 30 x 25 + 25 x 25 + 25 x 30)
= 2(750 + 625 + 750) = 2 (2125) = 4250cm²
∴ Length of the tape = 4 (l + b + h) = 4(30 + 25 + 25)
[∵ Herbarium is a shape of cuboid length = 4 (1+ b + h)] = 4×80= 320cm

Question 7.
Shanti Sweets Stalll 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 ₹4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Dimension for bigger box, l = 25 cm, b = 20 cm and b = 5 cm
Total surface area of the bigger size 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 cm²
Dimension for smaller box, l = 15 cm b = 12 cm and h = 5 cm
Total surface area of the smaller size box = 2(15 x 12 + 12 x 5 + 5 x 15)
= 2 (180 + 60 + 75)= 2 (315)= 630 cm²
Area for all the overlaps = 5% x 2080 cm² = \(\frac { 5 }{ 100 }\) – x 2080 cm² = 104 cm²
Total surface area of both boxes and area of overlaps = (2080 + 104) cm² = 2184 cm²
Total surface area for 250 boxes = 2184 x 250 cm²
Cost of the cardboard for 1000 cm2 = ₹4
Costs of the cardboard for 1 cm = ₹ \(\frac { 4 }{ 1000 }\)
Cost of the cardboard for 2184 x 250 cm² = ₹ \(\frac { 4x 2184 x 250 }{ 1000 }\) = ₹ 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 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:
Dimension for shetter, l = 4 m, b = 3 m and h = 25 cm
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 25 + (4 x 3)
= (2 x 7 x 25) + 12 = 35 + 12
= 47 m²

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