Answer :

Let consider a quadrilateral ABCD

In ∆ADC;

AC = = 25 cm

In ∆ABC

AB = a = 26 cm, BC = b = 27 cm, AC = c = 25 cm

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

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

A = where

= = 39

A_{1} =

A_{1} = = cm^{2}

In ∆ADC;

DA = a = 24 cm, CD = b = 7cm, AC = c = 25 cm

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]

= = 28

A_{2} =

A_{2} = = cm^{2}

Therefore area of quadrilateral ABCD = A_{1} + A_{2} = 291.85+84 = 375.85 cm^{2}

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