Class nine NCERT solutions Mathematics Chap 13
Chapter 13: Surface Areas and Volumes
TEXTBOOK’S EXERCISE – 13.1
1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep are 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 the sheet for it, if a sheet measuring 1 m2 costs Rs 20.
Sol. Here, l = 1.5 m, b = 1.25 m,
h = 65 cm = 0.65 m.
Since the box is open at the top, it has only five faces.
(i) So, surface area of the box
= lb + 2(bh + hl)
= 1.5 × 1.25 m2 + 2 (1.25 × 0.65 + 0.65 × 1.5) m2
= 1.875 m2 + 2 (1.7875) m2
= (1.875 + 3.575) m2 = 5.45 m2
Hence, 5.45 m2 of sheet is required.
(ii) Cost of 1 m2 of the sheet = Rs20
∴ Cost of 5.45 m2 of the sheet
= Rs 20 × 5.45 m2 = Rs 109
2. 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 Rs 7.50 per m2.
Sol. Here, l = 5 m, b = 4 m, h = 3 m
The surface area of the walls of the room and the ceiling
= 2h (l + b) + lb
= [2 × 3 (5 + 4) + 5 × 4] m2
= (6 × 9 + 20) m2 = 74 m2
Cost of whitewashing = Rs 7.50 per m2
∴ The total cost of whitewashing the walls and the ceiling of the room
= Rs 74 × 7.50 = Rs 555.
TEXTBOOK’S EXERCISE – 13.1
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.
Sol. Let the length, breadth and height of the hall be l, b and h respectively.
Perimeter of the floor of the hall
= 2 (l + b) = 250 m.
Area of the four walls of the hall
= 2h (l + b) … (i)
Also, area of the four walls of the hall = 15000 10 m2 = 1500 m2 … (ii)
From (i) and (ii), we have
2h (l + b) = 1500
⇒ h × 250 = 1500 [since 2(l + b) = 250]
⇒ h = 1500/ 250 = 6
Hence, height of the hall is 6 m.
Class nine NCERT solutions Mathematics Chap 13
4. 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?
Sol. Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm.
Total surface area of 1 brick = 2 (lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm2
= 2(225 + 75 + 168.75) cm2 = 937.5 cm2
∴ Required number of bricks
= 9 375/0.09375 = 100