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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

with Solutions - page 4
Qstn# B-16 Prvs-QstnNext-Qstn
  • #16
    The length of a hall is 18 m and its width is 13.5 m. Find the least number of square tiles, each of side 25 cm, required to cover the floor of the hall, (i) without leaving any margin. (ii) leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of Rs. 6 per tile (i) without leaving any margin. (ii) leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of Rs. 6 per tile
    Ans : (i) Length of hall (l) = 18 m and breadth
    (b) = 13.5 m

    ∴ Area of the floor = l × b
    = 18 × 13.5 m2
    = 243.0 m2
    Side of each square tiles
    (a) = 25 cm
    = 25/100
    = ¼ m
    ∴ Area of one tile = a2 = ¼ × ¼
    = 1/16 m2
    No. of tiles required = 243 ÷ 1/16
    = (243 × 16)/1
    = 3888
    Rate of tiles = Rs. 6 per tile
    ∴ Total cost = Rs. 3888 × 6
    = Rs. 233.28 (ii) Width of margin left in side = 1.5 m
    ∴ Inner lemgth = 18 - 2 × 1.5 = 18 - 3 = 15 cm
    And breadth = 13.5 - 2 × 1.5
    = 13.5 - 3
    = 10.5 m
    ∴ Inner area = 15 × 10.5 m2
    = 157.5 m2
    ∴ No. of tiles = 157.5 ÷ 1/16
    = 157.5 × 16
    = 2520
    ∴ Cost of tiles = 2520 × 6
    = Rs. 15120 (i) Length of hall (l) = 18 m and breadth
    (b) = 13.5 m

    ∴ Area of the floor = l × b
    = 18 × 13.5 m2
    = 243.0 m2
    Side of each square tiles
    (a) = 25 cm
    = 25/100
    = ¼ m
    ∴ Area of one tile = a2 = ¼ × ¼
    = 1/16 m2
    No. of tiles required = 243 ÷ 1/16
    = (243 × 16)/1
    = 3888
    Rate of tiles = Rs. 6 per tile
    ∴ Total cost = Rs. 3888 × 6
    = Rs. 233.28 (ii) Width of margin left in side = 1.5 m
    ∴ Inner lemgth = 18 - 2 × 1.5 = 18 - 3 = 15 cm
    And breadth = 13.5 - 2 × 1.5
    = 13.5 - 3
    = 10.5 m
    ∴ Inner area = 15 × 10.5 m2
    = 157.5 m2
    ∴ No. of tiles = 157.5 ÷ 1/16
    = 157.5 × 16
    = 2520
    ∴ Cost of tiles = 2520 × 6
    = Rs. 15120
  • #16-i
    without leaving any margin.
    Ans : Length of hall (l) = 18 m and breadth
    (b) = 13.5 m

    ∴ Area of the floor = l × b
    = 18 × 13.5 m2
    = 243.0 m2
    Side of each square tiles
    (a) = 25 cm
    = 25/100
    = ¼ m
    ∴ Area of one tile = a2 = ¼ × ¼
    = 1/16 m2
    No. of tiles required = 243 ÷ 1/16
    = (243 × 16)/1
    = 3888
    Rate of tiles = Rs. 6 per tile
    ∴ Total cost = Rs. 3888 × 6
    = Rs. 233.28
  • #16-ii
    leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of Rs. 6 per tile
    Ans : Width of margin left in side = 1.5 m
    ∴ Inner lemgth = 18 - 2 × 1.5 = 18 - 3 = 15 cm
    And breadth = 13.5 - 2 × 1.5
    = 13.5 - 3
    = 10.5 m
    ∴ Inner area = 15 × 10.5 m2
    = 157.5 m2
    ∴ No. of tiles = 157.5 ÷ 1/16
    = 157.5 × 16
    = 2520
    ∴ Cost of tiles = 2520 × 6
    = Rs. 15120