In the given figure $A B C D E, E Y = C Y = 4 c m, A X = B X = 6 c m, D E = D C = 5 c m, D X = 9 c m$ $D X$ is perpendicular to $E C$ and $A B$ . Find the area of $A B C D E$ .

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Solution

In $△ D E Y$ , $(D E)^{2} = (E Y)^{2} + (D Y)^{2}$
So, $5^{2} = 4^{2} + (D Y)^{2}$ $⟹ D Y = 3$
$D X = D Y + X Y = 3 + X Y$
So, $X Y = 6$ and $D X = 9$
$A r (A B C D E) = A r (A B C E) + A r (△ D C E)$
$A r (A B C D E) = \frac{1}{2} (A B + C E) \times X Y + \frac{1}{2} (E C) (D Y) = \frac{12 + 8}{2} (6) + \frac{1}{2} (8) (3) = 72 c m^{2}$

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