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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

with Solutions - page 2
Qstn# A-9 Prvs-QstnNext-Qstn
  • #9
    A field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40m, DC = 50 m and angle A = 90°. Find the area of the field.
    Ans : Since ∠A = 90°
    By Pythagoras Theorem,
    In ∆ABD,


    Now, area of ∆ABD = ½(18)(24)
    = (18)(12)
    = 216 m2
    Again in ∆BCD;
    ⇒ By Pythagoras Theorem √CBD = 90˚
    [∴ DC2 = BD2 + BC2, Since (50)2 = (30)2 + (40)2]
    ∴ Area of ∆BCD = 1/2(40)(30)
    = 600 m2
    Hence, area of quadrilateral ABCD = Area of ∆ABD + area of ∆BCD
    = 216 + 600
    = 816 m2