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

Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

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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 thick-ness 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 ₹ 20.
Answer:
Length of plastic box (l) = 1.5 m
Width of plastic box (b) = 1.25 m
Depth of plastic box (h) = 65 cm = 0.65 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box
= Lateral surface area + Area of the base
= 2(l + b) × h + (l × b)
= 2[( 1.5 + 1.25) × 0.65] + (1.5 × 1.25)
= (3.575 + 1.875) m²= 5.45 m²
The sheet required to make the box is 5.45 m².

(ii) ∵ Cost of 1 m² of sheet = ₹ 20
∴ Cost of 5.45 m² of sheet = ₹ (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 ₹ 7.50 per m².
Answer:
Length of the room, l = 5 m
Breadth of the room, b = 4 m
Height of the room, h = 3 m
Area of four walls including the ceiling = 2(l + b) × h + (l × b)
= 2(5 + 4) × 3 + (5 × 4)
= (54 + 20) m² = 74 m²
Cost of white washing = ₹ 7.50 per m²
Total cost = ₹ (74 x 7.50) =₹ 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.]
Ans. Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ₹ 15000
Rate per m² = ₹ 10
Area of four walls = 2(l + b) h
= (250 × h) m²
According to question,
(250 × h) × 10 = ₹ 15000
⇒ 2500 x h = ₹ 15000
⇒ h = \(\frac{15000}{2500}\) m
⇒ h = 6 m
Thus, 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?
Answer:
Area of paint
= 9.375 m² = 93750 cm² Dimensions of brick
= 22.5 cm × 10 cm × 7.5 cm Total surface area of a brick = 2(lb + bh + lh) cm²
=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

Number of bricks which can be painted = 93750

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?
Answer:
(i) Lateral surface area of cubical box of edge 10 cm = 4 × 102 cm² = 400 cm²
Lateral surface area of cuboidal box = 2(l + b) × h
= 2 × (12.5 + 10) × 8 cm²
= 2 × 22.5 × 8 cm² = 360 cm²

Thus, lateral surface area of the cubical box is greater by (400 – 360) cm² = 40 cm²

(ii) Total surface area of cubical box of edge 10 cm = 6 × 102 cm² = 600 cm²
Total surface area of cuboidal box = 2 (lb + bh + lh)
= 2(12.5 x 10+10 x 8 + 8 x 12.5) cm²
= 2(125 + 80 + 100) cm² = (2 × 305) cm² = 610 cm²
Thus, total surface area of cubical box is smaller by 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?
Answer:
(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30 × 25 + 25 × 25 + 25 × 30) cm²
= 2(750 + 625 + 750) cm² = 4250 cm²

(ii) Length of the tape needed
= 4(l + b + h) = 4(30 + 25 + 25) cm = 4 × 80 cm = 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 × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 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.
Answer:
Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25 × 20 + 20 × 5 + 25 × 5) cm² = 2(500 + 100 + 125) cm² = 1450 cm²

Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15 × 12 + 12 × 5 + 15 × 5) cm²
= 2(180 + 60 + 75) cm² = 630 cm²

Total surface area of 250 boxes of each type = 250(1450 + 630) cm²
= 250 × 2080 cm²
= 520000 cm²

Extra area required
= \(\frac{5}{100}\) × 520000 cm² = 26000 cm²

Total cardboard required =(520000 + 26000) cm² = 546000 cm²

Total cost of cardboard sheet = \(\frac{(546000 × 4)}{1000}\)

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 stitch-ing margins are very small, and there fore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions m × 3 m?
Answer:
Dimensions of the box- like structure
= 4m × 3m × 2.5m
Tarpaulin only required for all the four sides and top.
Thus, tarpaulin required = 2(l + b) × h + lb
= [2(4 + 3) × 2.5 + 4 × 3] m²
= (35 + 12) m²
= 47 m²