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

Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Questions and Answers.

BSEB Bihar Board Class 9th 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 to be 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 Rs 20.
Solution:
We have, Length, l = 1.5 m, Breadth, b = 1.25 m and Depth = Height.
h = 65 cm = .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 = Rs 20
∴ Total cost of 5.45 m² of sheet = Rs (5.45 × 20)
= Rs 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 Rs 7.50 per m².
Solution:
Here, Z = 5 m, 6 = 4 m and h = 3 m
Area of 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 white washing is Rs 7.50 per square metre.
∴ Cost of white washing = Rs (74 × 7.50) = Rs 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 Rs 10 per m² is Rs 15000, find the height of the hall.
Solution:
Cost of painting the four walls = Rs 15000
Rate of painting is Rs 10 per m²
∴ Area of four walls = (\(\frac {15000}{10}\)) m² = 1500 m²
⇒ 2(l + b)h = 1500
⇒ Perimeter × Height = 1500
⇒ 250 × Height = 1500
⇒ Height = \(\frac {1500}{250}\) = 6
Hence, the height of the hall = 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?
Solution:
Surface area of one brick
= 2(lb + bh + hl)
= 2(\(\frac {22.5}{100}\) × \(\frac {10}{100}\) + \(\frac {10}{100}\) × \(\frac {7.5}{100}\) + \(\frac {7.5}{100}\) × \(\frac {22.5}{100}\)) m²
= 2 × \(\frac {1}{100}\) × \(\frac {1}{100}\) (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
∴ Number of bricks that can be painted with the available Paint = \(\frac {9.375}{0.09375}\) = 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?
Solution:
(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 and is more by (400 – 360) cm² i.e., 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 + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm²
= 2(125 + 80 + 100) cm²
= (2 × 305) cm²
= 610 cm²
Thus, total surface area of cuboidal box is greater 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 ?
Solution:
Here, Z = 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 Rs 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each \kind.
Solution:
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
= Rs 4
∴ Total cost of cardboard
= Rs (\(\frac {546000}{1000}\) × 4) = Rs 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?
Solution:
Dimensions of the box-like structure are l = 4 m, b = 3 m and h = 2.5 m.
Since there is no tarpaulin for the floor.
∴ 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