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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB 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 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 sheet for it, if a sheet measuring 1 m² costs ₹ 20.
Answer:
The plastic box to be made is open at the top. Hence, the plastic sheet is required for the lateral surfaces and the base.
Here, for the box to be made,
length l = 1.5 m;
breadth b = 1.25 m and
height h = 65 cm = 0.65 m.
Area of the plastic sheet required for open box = Lateral surface area + Area of base
= 2 h(l + b) + l × b
= 2 × 0.65 (1.5 + 1.25) + 1.5 × 1.25 m²
= 1.3 × 2.75 + 1.875 m²
= 3.575 + 1.875 m²
= 5.45 m²
Cost of 1 m² sheet = ₹ 20
∴ Cost of 5.45 m² sheet = ₹ (5.45 × ₹ 20)
= ₹ 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:
For the given room, length 1 = 5 m; breadth b = 4 m and height h = 3 m.
Area of the region to be white washed
= Area of four walls + Area of ceiling
= 2 h(l + b) + l × b
= 2 × 3 (5 + 4) + 5 × 4 m²
= 54 + 20 m²
= 74 m²
Cost of white washing 1 m² region = ₹ 7.5
∴ Cost of white washing 74 m² region
= ₹ (74 × 7.5)
= ₹ 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 ₹ 15,000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
Answer:
Area painted at the cost of ₹ 10 = 1 m²
∴ Area painted at the cost of ₹ 15,000
= \(\frac{15000}{10}\)
= 1500 m²
∴ Area of the four walls = 1500m²
∴ Lateral surface area = 1500 m²
∴ Perimeter Of the floor × Height = 1500 m²
∴ 250 m × Height = 1500 m²
∴ Height = \(\frac{15000}{250}\)
∴ Height = 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 × 10 cm× 7.5 cm can be painted out of this container?
Answer:
For each brick, length l = 22.5 cm; breadth b = 10 cm and height h = 7.5 cm.
Total surface area of one brick
= 2 (lb + bh + hl)
= 2 (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2 (225 + 75 + 168.75) cm²
= 2 (468.75) cm²
= 937.5 cm²
= \(\frac{937.5}{10000}\) m² = 0.09375 m²
No. of bricks that can be painted with paint sufficient to paint 0.09375 m² area = 1
∴ No. of bricks that can be painted with paint sufficient to paint 9.375 m² area 9.375
= \(\frac{9.375}{0.09375}\) = 100

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:
For the cubical box, edge a = 10 cm and for the cuboidal box, length l = 12.5 cm; breadth b = 10 cm and height h = 8 cm
(i) Lateral surface area of cubical box
= 4a²
= 4 (10)² cm²
= 400 cm
Lateral surface area of cuboidal box
= 2h(l + b)
= 2 × 8(12.5 + 10) cm²
= 16 × 22.5 cm²
= 360 cm²
Thus, the lateral surface area of cubical box is greater by 40 cm² (400 – 360).

(ii) Total surface area of cubical box = 6a2
= 6 (10)² cm²
= 600 cm²
Total surface area of cuboidal box
= 2 (lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm²
= 2 (125 + 80 + 100) cm²
= 2 (305) cm²
= 610 cm²
Thus, the total surface area of cubical box is smaller by 10 cm² (610 – 600).

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) For the cuboidal greenhouse, length l = 30 cm; breadth fa = 25 cm and height h = 25 cm.
Area of glass used
= Total surface area of cuboid
= 2 (lb + bh + hl)
= 2 (30 × 25 + 25 × 25 + 25 × 30) cm²
= 2 (750 + 625 + 750) cm²
= 2 (2125) cm²
= 4250 cm²

(ii) 12 edges of the cuboidal greenhouse is made-up of 4 lengths, 4 breadths and 4 heights.
∴ Length of tape needed for 12 edges
= 4l + 4b + 4h
= 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:
For bigger cuboidal boxes, length l = 25 cm;
breadth b = 20 cm and height h = 5 cm.
Total surface area of a bigger box
= 2 (lb + bh + hl)
= 2 (25 × 20 + 20 × 5 + 5 × 25) cm²
= 2 (500 + 100 + 125) cm²
= 1450 cm²
Area of cardboard required for overlap
= 5 % of 1450 cm²
= 72.5 cm²
Thus, the total area of cardboard required for 1 bigger box = 1450 + 72.5 cm²
= 1522.5 cm²
∴ The total area of cardboard required for 250 bigger boxes = (1522.5 × 250) cm²
For smaller cuboidal boxes, length l = 15 cm; breadth b = 12 cm and height h = 5 cm.
Total surface area of a smaller box
= 2 (lb + bh + hl)
= 2 (15 × 12 + 12 × 5 + 5 × 15) cm²
= 2(180 +60 + 75) cm²
= 2 (315) cm²
= 630 cm²
Area of cardboard required for overlap
= 5% of 630 cm²
= 31.5 cm²
Thus, the total area of cardboard required for 1 smaller box = 630 + 31.5 cm² = 661.5 cm²
∴ The total area of cardboard required for 250 smaller boxes = (661.5 × 250) cm²
Now, the total area of cardboard required for all the boxes
= (1522.5 × 250) + (661.5 × 250) cm²
= 250(1522.5 + 661.5) cm²
= 250 × 2184 cm²
Cost of 1000 cm² cardboard = ₹ 4
∴ Cost of 250 × 2184 cm² cardboard
= ₹ \(\left(\frac{4 \times 250 \times 2184}{1000}\right)\)
= ₹ 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 × 3 m ?
Answer:
For the box-like structure without base, length
l = 4m; breadth b = 3m and height h = 2.5m.
Area of tarpaulin required
= Area of lateral surfaces + Area of top
= 2 h(l + b) + l × b
= 2 × 2.5 (4 + 3) + 4 × 3 m²
= 35 + 12 m²
= 47 m²