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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

with Solutions - page 4

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  • #13-i
    (ii)
    Ans : Area of the shaded portion
    = Area of the rectangle PQRS - Area of square ABCD
    = 3.2 × 1.8 - (1.4)2 (∵ PQ = 3.2 and PS = 1.8)Side of square AB = 1.4 m
    = 5.76 - 1.96= 3.8 m2
    (ii) Area of the shaded portion = Area of square ABCD - Area of rectangle PQRS
    = 6 × 6 - (3.6) (4.8)= 36 - 17.28= 18.72 m2
  • Qstn #14
    A path of uniform width, 3 m, runs around the outside of a square field of side 21 m. Find the area of the path.
    Ans : According to the given information the figure will be as shown alongside.
    Clearly, length of the square field excluding path = 21 m.
    Area of the square side excluding the path = 21× 21 = 441 m2
    Again, length of the square field including the path = 21 + 3 + 3 = 27 m
    Area of the square field including the path = 27× 27 = 729 m2
    Area of the path = 729 - 441 = 288 m2
  • Qstn #15
    A path of uniform width, 2.5 m, runs around the inside of a rectangular field 30 m by 27 m. Find the area of the path.
    Ans : According to the given statement the figure will be as shown alongside.
    Clearly, the length of the rectangular field including the path = 30 m.
    Breadth = 27 m.
    Its Area = 30× 27 = 810 m2
    Width of the path = 2.5 m
    Length of the rectangular field including the path = 30 - 2.5 - 2.5 = 25 m.
    Breadth = 27 - 2.5 - 2.5 = 22m
    Area of the rectangular field including the path = 25× 22 = 550 m2
    Hence, area of the path = 810 - 550 = 260 m2
  • 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,
  • #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
  • Qstn #17
    A rectangular field is 30 m in length and 22m in width. Two mutually perpendicular roads, each 2.5 m wide, are drawn inside the field so that one road is parallel to the length of the field and the other road is parallel to its width. Calculate the area of the crossroads.
    Ans : Length of rectangular field (l) = 30 m and breadth
    (b) = 22m
    width of parallel roads perpendicular to each other inside the field = 2.5m

    Area of cross roads = width of roads (Length + breadth) - area of middle square
    = 2.5 (30 + 22) - (2.5)2
    = 2.5× 52 - 6.25 m2= (130 - 6.25) m = 123.75 m2
  • Qstn #18
    The length and the breadth of a rectangular field are in the ratio 5:4 and its area is 3380 m2. Find the cost of fencing it at the rate of ₹75 per m.
    Ans : Ratio in length and breadth = 5 : 4
    Area of rectangular field = 3380 m2
    Let length = 5x and breadth = 4x
    5x × 4x = 3380
    ⇒ 20x2 = 3380
    ⇒ x2 = 3380/20 = 169 = (13)2
    ⇒ x = 13
    Length = 13× 5 = 65 m
    Breadth =13× 4 = 52 m
    dlt23851564568179465016">Perimeter = (l + b) = 2× (65 + 52) m = 2× 117 = 234 m
    Rate of fencing = ₹ 75 per m
    Total cost = 234× 75 = ₹ 17550
  • Qstn #19
    The length and the breadth of a conference hall are in the ratio 7 : 4 and its perimeter is 110 m. Find:
    Ans : Ratio in length and breadth = 7 : 4
    Perimeter = 110 m
    ∴ Length + Breadth = 110/2 = 55m
    Sum of ratios = 7 + 4 = 11
    ∴ Length = (55 × 7)/11
    = 35 m
    And Breadth = (55 × 4)/11
    = 20m
  • #19-i
    area of the floor of the hall.
    Ans : Area of floor = l × b
    = 35 × 20
    = 700 m2
  • #19-ii
    number of tiles, each a rectangle of size 25 cm x 20 cm, required for flooring of the hall.
    Ans : Size of tile = 25 cm × 20 cm
    = (25 × 20)/(100 × 100)
    = 1/20 m2
    ∴ Number of tiles = Area of floor/Area of one tile
    = (700 × 20)/1
    = 14000
  • #19-iii
    the cost of the tiles at the rate of ₹ 1,400 per hundred tiles.
    Ans : Cost of tiles = ₹1400 per 100 tiles
    ∴ Total cost = (14000 × 1400)/100
    = ₹196000
  • #
    Section : C
  • Qstn #1
    The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape.
    If the width DC, of the swimming pool is 6.4cm, depth (AD) at the shallow end is 80 cm and depth (BC) at deepest end is 2.4m, find Its area of the cross-section.
    Ans : Area of the cross-section = Area of trapezium ABCD
    = ½ × (Sum of parallel sides) × height
    = ½ (80 + 20) × 6.4
    = (320)(3.2)
    = (32)(32)
    = 1024 cm2 or 10.24 sq. m.
  • Qstn #2
    The parallel sides of a trapezium are in the ratio 3: 4. If the distance between the parallel sides is 9 dm and its area is 126 dm2 ; find the lengths of its parallel sides.
    Ans :
    Let parallel sides of trapezium bea = 3x
    b = 4x
    Distance between parallel sides, h = 9 dm
    area of trapezium = 126 dm2
    ½ (a + b) × h = 126
    ⇒ ½ (3x + 4x) × 9 = 126
    ⇒ 7x × 9 = 126 × 2
    ⇒ x = (126 × 2)/(7 × 9)
    ⇒ x = 4
    a = 3x = 3 × 4 = 12 dm
    b = 4x = 4 × 4 = 12 dm
    12 dm, 16 dm
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