Page 290 - Start Up Mathematics_8 (Non CCE)

Example 22: The dimensions of an encyclopedia are 24 cm × 12 cm × 6 cm. It is to be covered with a
2
plastic sheet. If each encyclopedia requires 170 cm of extra sheet for folding, how much
plastic sheet is required to wrap 50 such encyclopedias? b = 12 cm
Solution: The encyclopedia is to be covered by a plastic sheet only
on three faces, i.e., top, bottom and spine (the bound side).
Area of top = l × b = 24 cm × 12 m = 288 cm 2 l = 24 cm
Area of bottom = l × b = 24 cm × 12 cm = 288 cm 2
Area of spine = l × h = 24 cm × 6 cm = 144 cm 2

Area of extra sheet required for folding = 170 cm 2 6 cm
2
Area of plastic sheet required for 1 encyclopedia = (288 + 288 + 144 + 170) cm = 890 cm 2
2
\ Area of plastic sheet required for 50 such encyclopedias = (50 × 890) cm = 44,500 cm 2
Example 23: The length of a hall is 3 times its height and its breadth is 2½ times its height. The cost of
2
whitewashing the walls at the rate of ` 2.50 per m is ` 440. Find the cost of tiling the floor
2
at the rate of ` 7.50 per m .
Solution: Let the height of the hall (h) be (x) m.

Ê 1 ˆ Ê 5 ˆ
Then, length (l) = (3x) m and breadth (b) = 2 x m = Á Ë 2 ¯ ˜
x m
Á
˜
Ë
2 ¯
Ê 5 ˆ
Area of four walls = 2h(l + b) = 2 × x × 3x + 2 ¯ ˜ 2
x m
Á
Ë
Ê 6x + 5xˆ 2 11x 2 2 2
= 2x × Á Ë 2 ˜ ¯ m = 2x × 2 m = 11x m
2
Cost of whitewashing 1 m = ` 2.50
2
2
2
\ Cost of whitewashing 11x m = ` (2.50 × 11x ) = ` 27.50x 2
But the cost of whitewashing is ` 440.
2
2
fi 27.50x = 440 fi x = 440 = 16
27 50.
2
fi x = 16 fi x = 16 = 4 m
\ Height of hall (h) = 4 m, Length of hall (l) = (3 × 4) = 12 m

Ê 5 ˆ
Breadth of hall (b) = Á 2 ¥ 4 m = 10 m
˜
Ë
¯
2
\ Area of floor of the room = (l × b) = (12 × 10) m = 120 m 2
2
Cost of tiling 1 m of floor = ` 7.50
2
\ Cost of tiling 120 m of floor = ` (7.50 × 120) = ` 900
2
Example 24: The area of four walls of a room is 105 m . If the room is 13 m long and 4.5 m broad, find
its volume.
Solution: Length of room (l) = 13 m, Breadth of room (b) = 4.5 m
Let the height of the room be h m.
\ Area of four walls = 2 × h × (l + b)
fi 105 = 2 × h × (13 m + 4.5 m)

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