find the area of pentagon $A B C D E A$ in which $B L ⊥ A C, D M ⊥ A C$ and $E N ⊥ A C$ such that AC=18cm, AM= 14 cm, AN = 6 cm, BL = 4 Cm , DM =12 cm and EN=9 cm

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Solution

The given details are,
$B L ⊥ A C, D M ⊥ A C$ and $E N ⊥ A C$
$A C = 18 c m, A M = 14 c m, A N = 6 c m, B L = 4 C m, D M = 12 c m, E N = 9 c m$
$M C = A C - A M = 18 - 14 = 4 c m$
$M N = A M - A N = 14 - 6 = 8 c m$
By using the formula,
$A r e a (p e n t . A B C D E)$ $= a r e a (△ A E N) + a r e a (△ D M C) + a r e a (△ A B C) + a r e a (T r a p . D M N E)$

Now, we know that

$A r e a o f t r i a n g l e$ $= 1 / 2 \times b a s e \times h e i g h t$
and

$A r e a o f t r a p e z i u m$ $= 1 / 2 \times (s u m o f p a r a l l e l s i d e s) \times h e i g h t$
$∴$ lets calculate each of the areas,

$A r e a (△ A E N) = \frac{1}{2} (A N) \times (E N) = \frac{1}{2} \times 6 \times 9 = 27 c m^{2}$

$A r e a (△ D M C) = \frac{1}{2} (M C) \times (D M) = \frac{1}{2} \times 4 \times 12 = 24 c m^{2}$

$A r e a (△ A B C) = \frac{1}{2} (A C) \times (B L) = \frac{1}{2} \times 18 \times 4 = 36 c m^{2}$

$A r e a (T r a p . D M N E) = \frac{1}{2} \times (D M + E N) \times M N = \frac{1}{2} \times (12 + 9) \times 8 = 84 c m^{2}$

$∴ A r e a (p e n t A B C D E) = a r e a (△ A E N) + a r e a (△ D M C) + a r e a (△ A B C) + a r e a (T r a p . D M N E)$

$= 27 + 24 + 36 + 84$
$= 171 c m^{2}$

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