the sides of a quadrilateral abcd are 6cm,8cm,11cm and 12cm respectively,and the angle between the first two sides is a right angle.find its area.

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  1. draw the diagonal to show the right-angled triangle,
    the hypotenuse of that using Pythagoras is 10
    so the area of the right_angled triangle is
    (1/2)(6)(8) = 24

    for the other triangle with sides, 10, 11, and 12
    use Heron's formula
    Area = √(s(s-a)(s-b)(s-c))
    where s = (1/2)(perimeter) = 16.5
    s-a = 16.5-10 = 6.5
    s-b = 16.5 - 11 = 5.5
    s-c = 16.5 - 12 = 4.5
    Area = √(16.5*6.5*5.5*4.5) = 51.92

    total area = 24+51.92 = 75.52

    check my arithmetic

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  2. ABCD is a square and arc BEC is a semicircle. if AB=10.0 cm,find the area of the area of the figure.

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