Chapter 13 Surface Areas and Volumes

Chapter 13 Surface Areas and Volumes

 

Ex 13.1

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 = 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 = = 6
Thus, the required height of the hall = 6 m

Ex 13.1 Class 9 Maths 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:
Total area that can be painted = 9.375 m²
Here, Length of a brick (l) = 22.5 cm
Breadth of a brick (b) = 10 cm
Height of a brick (h) = 7.5 cm
Since a brick is like a cuboid, then
Total surface area of a brick = 2[lb + bh + hl]
= 2[(225 x 1(0) + (10 x 7.5) + (7.5 x 22.5)] cm²
= 2[(225) + (75) + (168.75)] cm²
= 2[468.75] cm² = 937.5 cm² = m²
Let the required number of bricks be n
∴ Total surface area of n bricks = n x m²

Thus, the required number of bricks = 100

Ex 13.1 Class 9 Maths 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:
For the cubical box with edge (a) = 10 cm
Lateral surface area = 4a² = 4 x 10² cm²
= 400 cm²
Total surface area = 6a² = 6 x 10² cm²
= 600 cm²
For the cuboidal box with dimensions,
Length (l) = 12.5 cm,
Breadth (b) = 10 cm,
Height (h) = 8 cm
∴ Lateral surface area = 2[l + b] x h = 2[12.5 + 10] x 8 cm² = 360 cm²
Total surface area = 2[lb + bh + hl]
= 2[(12.5 x 10) + (10 x 8) + (8 x 12.5)] cm²
= 2[125 + 80 + 100] cm²
= 2[305] cm²
= 610 cm²
(i) A cubical box has the greater lateral surface area by (400 – 360) cm² = 40 cm².
(ii) Total surface area of a cubical box is smaller than the cuboidal box by (610 – 600) cm² = 10 cm².

Ex 13.1 Class 9 Maths 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:
The herbarium is like a cuboid.
Here, length (l) = 30 cm,
breadth (b) = 25 cm,
height (h) = 25 cm
(i) Surface area of the herbarium (glass)
= 2[lb + bh + hl]
= 2[(30 x 25) + (25 x 25) + (25 x 30)] cm² – 2[750 + 625 + 750] cm²
= 2[2125] cm²
= 4250 cm²
Thus, the required area of the glass = 4250 cm²

(ii) Total length of 12 edges = 4l + 4b + 4h
= 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4 x 80 cm = 320 cm
Thus, the required length of tape = 320 cm

Ex 13.1 Class 9 Maths 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:
For bigger box:
Length (l) = 25 cm,
Breadth (b) = 20 cm,
Height (h) = 5 cm
Total surface area of a box = 2(lb + bh + hl)
= 2[(25 x 20) + (20 x 5) + (5 x t25)] cm²
= 2 [500 + 100 + 125] cm²
= 2[725] cm²
= 1450 cm²
Total surface area of 250 boxes = (250 x 1450) cm² = 362500 cm²

For smaller box:
l = 15 cm, b = 12 cm, h = 5 cm
Total surface area of a box = 2 [lb + bh + hl]
= 2[(15 x 12) + (12 x 5) + (5 x 15)] cm²
= 2[180 + 60 + 75] cm² = 2[315] cm² = 630 cm²
∴ Total surface area of 250 boxes = (250 x 630) cm² = 157500 cm²
Now, total surface area of both type of boxes = 362500 cm² +157500 cm² = 520000 cm² Area for overlaps = 5% of [total surface area]
= x 520000 cm² = 26000 cm²
∴ Total surface area of the cardboard required = [Total surface area of 250 boxes of each type] + [Area for overlaps]
= 520000 cm² + 26000 cm² = 546000 cm²
∵ Cost of 1000 cm² cardboard = Rs. 4
∴ Cost of 546000 cm² cardboard
= Rs. = Rs. 2184

Ex 13.1 Class 9 Maths 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:
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.