Find the area of pentagon $A B C D E$ in which $B L ⊥ A C, C M ⊥ A D$ and $E N ⊥ A D$ such that $A C = 10 c m, A D = 12 c m, B L = 3 c m, C M = 7 c m$ and $E N = 5 c m$ .

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Solution

Area of pentagon ABCDE= Area of $Δ A B C +$ Area of $Δ A C D +$ Area of $Δ A D E$

Here , $B L ⊥ A C, C M ⊥ A D, E N ⊥ A D$

So, BL,AM and EN perpendiculars to $Δ A B C, Δ A C D Δ A D E$ respectively.

Area of triangle with perpendicular $= \frac{1}{2} \times b a s e \times h e i g h t$
$∴ A r e a o f Δ A B C = 1 / 2 \times A C \times B L$
$A r e a o f Δ A C D = 1 / 2 \times A D \times C M$

$A r e a o f Δ A D E = 1 / 2 \times A D \times E N$

Now area of pentagon $A B C D E = 1 / 2 A C \times B L + 1 / 2 \times A D \times C M + 1 / 2 \times A D \times E N$
$= (1 / 2 \times 10 \times 3) + (1 / 2 \times 12 \times 7) + (1 / 2 \times 12 \times 5)$

$= 15 + 42 + 30$

$= 87$

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