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

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions 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 ₹ 20.
Answer:
We have : Length l = 1.5 m, Breadth b = 1.25 m and Depth = Height h = 65 cm = 0.65 m.
(i) Since, the plastic box is open at the top, therefore,
plastic sheet required for making such a box
= [2(l + b) × h + lb] m²
= [2(1.5 + 1.25) × 65 + 1.5 × 1.25]m²
= [2 × 2.75 × 65 + 1.875] m²
= (3.575 + 1.875) m² = 5.45 m²

(ii) Cost of 1 m² of sheet = ₹ 20
∴ Total cost of 5.45 m² of 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 whitewashing the walls of the room and the ceiling at the rate of × 7.50 per m².
Answer:
Here, l = 5m, 6 = 4 m and h = 3 m.
Area of the four walls including ceiling = [2(l + b) × h + lb] m²
= [2(5 + 4) × 3 + 5 × 4] m²
= (2 × 9 × 3 + 20) m²
= (54 + 20) m²
= 74 m²
Cost of whitewashing is ₹ 7.50 per square metre.
Cost of whitewashing = ₹ (74 × 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.]
Answer:
Cost of painting the four walls = ₹ 15000.
Rate of painting is ₹ 10 per m²
∴ Area of four walls = m² = 1500 m²
So, 2(l + b)h = 1500
or Perimeter × Height = 1500
or 250 × Height = 1500
or Height = \(\left(\frac{15000}{10}\right)\) = 6
Hence, the height of the hall is 6 metres.

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:
Surface area of one brick= 2(lb + bh + hl)
= 2\(\left(\frac{22.5}{100} \times \frac{10}{100}+\frac{10}{100} \times \frac{7.5}{100}+\frac{7.5}{100} \times \frac{22.5}{100}\right)\)m²
= 2 × \(\frac{1}{100} \times \frac{1}{100}\) (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) m²
= \(\frac{1}{5000}\) × (225 + 75 + 168.75) m²
= \(\frac{1}{5000}\) × 468.75 m² = 0.09375 m²
Area for which the paint is just sufficient is 9.375 m².
Total surface area of cuboidal box = 2(lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm2²
= 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:
Here, l = 30 cm, b = 25 cm and h = 25 cm.
(i) Area of the glass = Total surface area
= 2 (lb + bh + hl)
= 2(30 × 25 + 25 × 25 + 25 × 30) cm²
= 2(750 + 625 + 750) cm²
= (2 × 2125) cm² = 4250 cm²

(ii) Tap needed for all the 12 edges = The sum of all the edges
= 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:
In case of bigger box :
l = 25 cm, b = 20 cm and h = 5 cm.
Total surface area = 2(lb + bh + hl)
= 2(25 × 20 + 20 × 5 + 5 × 25) cm²
= 2(500 + 100 + 125) cm²
= (2 × 725) cm²
= 1450 cm²

In case of smaller box : l = 15 cm, b = 12 cm and h = 5 cm.
Total surface area = 2(lb + bh + hl)
= 2(15 × 12 + 12 × 5 + 5 × 15) cm²
= 2(180 + 60 + 75) cm²
= (2 × 315) cm²
= 630 cm²

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

Cardboard required (i.e. including 5% extra for overlaps, etc.)
= (520000 × \frac{105}{100}) cm² = 546000 cm²
Cost of 1000 cm² of cardboard = ₹ 4
Total cost of cardboard = ₹(\frac{546000}{1000} × 4) = ₹ 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 he required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
Answer:
Dimensions of the box-like structure are Z = 4m, 6 = 3m and h = 2.5 m. Since, there is no tarpaulin for the floor, therefore
Tarpaulin required = [2(l + b) × h + lb] m²
= [2(4 + 3) × 2.5 + 4 × 3] m²
= (2 × 7 × 2.5 + 12) m²
= (35 + 12) m² = 47 m².