Class 9 - Mathematics
Surface Areas and Volumes - Exercise 13.1
Top Block 1
Exercise 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.
Answer :
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 + 2 (1.25 * 0.65 + 0.65 * 1.5)
= 1.875 + 2 (1.7875)
= (1.875 + 3.575)
= 5.45 m2
Hence, 5.45 m2 of sheet is required.
(ii) Cost of 1 m2 of the sheet = Rs 20
So, cost of 5.45 m2 of the sheet = Rs 20 * 5.45 = 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.
Answer :
Here, l = 5 m, b = 4 m, h = 3 m
Surface area of the walls of the room and the ceiling = 2h (l + b) + lb
= [2 * 3 (5 + 4) + 5 * 4]
= (6 * 9 + 20)
= 74 m2
Cost of white washing = Rs 7.50 per m2
So, total cost of white washing the walls and the ceiling of the room = 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.
Answer :
Let 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) ………….1
Also, area of the four walls of the hall = 15000/10 = 1500 m2 ………….2
From equation 1 and 2, we have,
2h (l + b) = 1500
⇒ h * 250 = 1500 [Since 2(l + b) = 250]
⇒ h = 1500/250
⇒ h = 6
Hence, 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 * 10 cm * 7.5 cm can be painted out of this container?
Answer :
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)
= 2(225 + 75 + 168.75) cm2
= 937.5 cm2
= 9375/(100 * 100) m2
= 9375/10000
= 0.09375 m2
So, required number of bricks = 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 :
Here, a = 10 cm, l = 12.5 cm, b = 10 cm, h = 8 cm
(i) Lateral surface area of the cubical box = 4a2
= 4 * (10)2
= 4 * 100
= 400 cm2
Lateral surface area of the cuboidal box = 2h(l + b)
= 2 * 8 (12.5 + 10)
= 16 × 22.5
= 360 cm2
Difference in the lateral surface areas of the two boxes = (400 – 360) = 40 cm2.
Hence, the cubical box has greater lateral surface area by 40 cm2.
(ii) Total surface area of the cubical box = 6a2
= 6 * (10)2
= 6 * 100
= 600 cm2
Total surface area of the cuboidal box = 2(lb + bh + hl)
= 2(12.5 * 10 + 10 * 8 + 8 * 12.5)
= 2(125 + 80 + 100)
= 2 × 305
= 610 cm2
Difference in the total surface areas of the two boxes = (610 – 600) = 10 cm2
Hence, the cubical box has smaller total surface area 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?
Sol. Here, l = 30 cm, b = 25 cm, h = 25 cm.
(i) Total surface area of the herbarium = 2(lb + bh + hl)
= 2(30 × 25 + 25 × 25 + 25 × 30)
= 2(750 + 625 + 750)
= 2 × 2125
= 4250 cm2
Hence, area of the glass = 4250 cm2
(ii) A cuboid has 12 edges. These consist of 4 lengths, 4 breadths and 4 heights.
Hence, length of the tape required = 4l + 4b + 4h
= (4 × 30 + 4 × 25 + 4 × 25)
= (120 + 100 + 100) cm
= 320 cm
Mddle block 1
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.
Answer :
For bigger boxes: l = 25 cm, b = 20 cm, h = 5 cm
Total surface area of 1 bigger box = 2(lb + bh + hl)
= 2(25 * 20 + 20 * 5 + 5 * 25)
= 2 (500 + 100 + 125)
= 1450 cm2
Area of cardboard required for overlaps = 5% of 1450
= (5/100) * 1450
= 7250/100
= 72.5 cm2
Total area of cardboard needed for 1 bigger box = 1450 + 72.5 = 1522.5 cm2
Total area of cardboard needed for 250 bigger boxes = 1522.5 * 250 = 380625 cm2.
For smaller boxes: l = 15 cm, b = 12 cm, h = 5 cm
Total surface area of 1 smaller box = 2 (lb + bh + hl)
= 2(15 * 12 + 12 * 5 + 5 * 15)
= 2 (180 + 60 + 75)
= 630 cm2
Area of cardboard required for overlaps = 5% of 630
= (5/100) * 630
= 3150/100
= 31.5 cm2
Total area of cardboard needed for 1 smaller box = 630 + 31.5
= 661.5 cm2
Total area of cardboard needed for 250 smaller boxes = 661.5 * 250 = 165375 cm2
Now, total area of cardboard needed for 500 boxes (250 bigger and 250 smaller boxes)
= 380625 + 165375 = 546000 cm2
Cost of 1000 cm2 of cardboard = Rs 4
So, Cost of 546000 cm2 of cardboard = Rs (4/1000) * 546000
= 4 * 546
= 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?
Answer :
Here, l = 4 m, b = 3 m, h = 2.5 m
The tarpaulin is needed to cover 5 faces only (excluding the floor) Surface area of the shelter
= lb + 2(bh + hl)
= 4 * 3 + 2(3 * 2.5 + 2.5 * 4)
= 12 + 2(7.5 + 10)
= 12 + 35
= 47 m2
Hence, 47 m2 of tarpaulin is required to make the shelter.