Chapter 11 MENSURATION
Exercise 11.3
Q.1 There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Sol. (a) Given, length (l) = 60 cm, breadth (b) = 40 cm and height (h) = 50 cm
Now, surface area of cuboidal box = 2 (lb + bh + hl)
= 2 (60 x 40 + 40 x 50 + 50 x 60) cm^{2}
= 2 (2400 + 2000 + 3000) cm^{2}
= 14800 cm^{2}
(b) Given, length (l) = 50 cm
Now, surface area of cube box = 6 (l)^{2}
= 6 (50)^{2} cm^{2}
= 15000 cm^{2}
Here, surface area of cuboidal box (a) is less than cube box (b).
Hence, cuboidal box (a) will require less amount of material to make.
Q.2 A suitcase with measures 80 cm Ã— 48 cm Ã— 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Sol. Given, suitcase is of length (l) = 80 cm, breadth (b) = 48 cm and height (h) = 24 cm
Now, surface area of suitcase = 2 (lb + bh + hl)
= 2 (80 x 48 + 48 x 24 + 24 x 80) cm^{2}
= 2 (3840 + 1152 + 1920) cm^{2}
= 13824 cm^{2}
Therefore, total surface area of 100 suitcases = (13824 x 100) cm^{2 }= 1382400 cm^{2}
Now, area of tarpaulin cloth = total surface area of suitcase
Length x Breadth = 1382400 cm^{2}
Length = (1382400/96)cm
= 14400 cm = 144 m
Hence, 144 metres of tarpaulin is required to cover 100 suitcases.
Q.3 Find the side of a cube whose surface area is 600 cm^{2}.
Sol. Given, surface area of cube = 600 cm^{2}
Let the length of each side of cube be l.
Now, surface area of cube box = 6 (side)^{2}
600 cm^{2} = 6 (l)^{2}
(l)^{2} = 100 cm^{2}
l = 10 cm.
Hence, the side of the cube is 10 cm.
Q.4 Rukhsar painted the outside of the cabinet of measure 1 m Ã— 2 m Ã— 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.
Sol. Given, cabinet is of length (l) = 2 m, breadth (b) = 1 m and height (h) = 1.5 m
Now, surface area of cabinet = 2h (l + b) + lb
= [2x1.5 (2 + 1) + 2 x 1] m^{2}
= [3 (3) + 2] m^{2}
= (9 + 2) m^{2}
= 11 m^{2 }
Q.5 Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m^{2} of area is painted. How many cans of paint will she need to paint the room?
Sol. Given, length (l) = 15 m, breadth (b) = 10 m and height (h) = 7 m
Here, area of the hall to be painted = area of the wall + area of the ceiling
= 2h (l + b) + lb
= [2x7 (15 + 10) + 15 x 10] m^{2}
= [14 (25) + 150] m^{2}
=500 m^{2 }
Given, 100 m^{2} area can be painted from each can.
Therefore, number of cans required to paint 500 m^{2} = 500/100 = 5
Hence, 5 cans will be needed to paint the room.
Q.6 Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
Sol. It can be seen from the figure that the height of both the figures is same.
Also, it can be seen that the shape of both the figures are different one is a cylinder and the other is a cube.
Now, lateral surface area of the cube = 4 l^{2} = 4 (7 cm)^{2} = 196 cm^{2}
Lateral surface area of the cylinder = 2Ï€rh = [2 x 22/7 x 7/2 x 7]cm^{2Â }= 154Â cm^{2Â }
Here, lateral surface area of the cube is greater than that of the cylinder.
Hence, cube box has larger lateral surface area.
Q.7 A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Sol. Given, tank is of radius (r) = 7 m and height (h) = 3 m
Now, total surface area of cylinder =Â 2Ï€r(r+h)
= [2 x 22/7 x 7(7+3)] m^{2}
= 440 m^{2}
Therefore, 440 m^{2 }sheet of metal is required.
Q.8 The lateral surface area of a hollow cylinder is 4224 cm^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Sol. Given, lateral surface area of a hollow cylinder is 4224 cm^{2} and height (h) = 33 cm
Now, area of cylinder = area of rectangular sheet
4224 cm^{2} = 33 cm x length
Length = 4224cm^{2} / 33cm = 128 cm
Now, perimeter of the rectangular sheet = 2 (length + width)
= [2 (128 + 33)] cm
= 322 cm
Hence, the perimeter of rectangular sheet is 322 cm.
Q.9 A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.
Sol. Given, diameter of road roller = 84 cm
Therefore, radius of road roller (r) = d/2 = 84/2 = 42 cm
So, in 1 revolution, area of the road covered =Â 2Ï€rh
= 2 x 22/7 x 42 cm x 1 m
=Â 2 x 22/7 x 42/100 m x 1 m
= 264/100 m^{2}
Now, in 750 revolutions, area of the road covered
= (750 x 264/100) m^{2}
= 1980Â m^{2}
Hence, the area of the road is 1980 m^{2}
Q.10 A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.
Sol. Given, container is of diameter (d) = 14 cm and height (h) = 20 cm
Here, the height of the label = 20 cm â€“ 2 cm â€“ 2 cm = 16 cm
Radius of the label (r) = d/2 = 14/2 = 7 cm
Now, area of the label =Â 2Ï€rh
= 2 x 22/7 x 7 x 16 cm^{2}
= 704 cm^{2}
Hence, the area of the label is 704 cm^{2}.
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