Answer :

Let consider a quadrilateral ABCD

In ∆ABC;

AC = = 13 m

AB = a = 5 m, BC = b = 12 m, AC = c = 13 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 15

A_{1} =

A_{1} = = m^{2}

In ∆ADC;

DA = a = 15 m, CD = b = 14 m, AC = c = 13 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 21

A_{2} =

A_{2} = = m^{2}

Therefore area of quadrilateral ABCD = A_{1} + A_{2} = 30+84 = 114 m^{2}

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