Find the area of pentagon ABCDE in which BL ⊥ AC,

Question:

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

Solution:

Area of pentagon $\mathrm{ABCDE}=($ Area of $\Delta \mathrm{ABC})+($ Area of $\Delta \mathrm{ACD})+($ Area of $\Delta \mathrm{ADE})$

$=\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)+\left(\frac{1}{2} \times \mathrm{AD} \times \mathrm{CM}\right)+\left(\frac{1}{2} \times \mathrm{AD} \times \mathrm{EM}\right)$

$=\left[\left(\frac{1}{2} \times 10 \times 3\right)+\left(\frac{1}{2} \times 12 \times 7\right)+\left(\frac{1}{2} \times 12 \times 5\right)\right] \mathrm{cm}^{2}$

$=(15+42+30) \mathrm{cm}^{2}$

$=87 \mathrm{~cm}^{2}$

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

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