Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.

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Solution

$Area of pentagon ABCDE = (Area of ∆ ABC) + (Area of ∆ ACD) + (Area of ∆ ADE)$
$= (\frac{1}{2} \times AC \times BL) + (\frac{1}{2} \times AD \times CM) + (\frac{1}{2} \times AD \times EM) = [(\frac{1}{2} \times 10 \times 3) + (\frac{1}{2} \times 12 \times 7) + (\frac{1}{2} \times 12 \times 5)] {cm}^{2} = (15 + 42 + 30) {cm}^{2} = 87 {cm}^{2}$

$Hence, the area of the pentagon is 87 {cm}^{2} .$

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