Question : In $\triangle \mathrm{ABC}, \overline{\mathrm{BD}} \perp \overline{\mathrm{AC}}$, intersecting $\overline{\mathrm{AC}}$ at $\mathrm{D}$. Also, $\mathrm{BD}=12 \mathrm{~cm}$. If $\mathrm{m}(\overline{\mathrm{AD}})=6 \mathrm{~cm}$ and $\mathrm{m}(\overline{\mathrm{CD}})=4 \mathrm{~cm}$, find the area $\left(\right.$in $\left.\mathrm{cm}^2\right)$ of $\triangle \mathrm{ABC}$.
Option 1: 45
Option 2: 50
Option 3: 60
Option 4: 75
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Correct Answer: 60
Solution :
Given, $\overline{\mathrm{BD}} \perp \overline{\mathrm{AC}}$
$\mathrm{BD}=12 \mathrm{~cm}$, $\mathrm{m}(\overline{\mathrm{AD}})=6 \mathrm{~cm}$ and $\mathrm{m}(\overline{\mathrm{CD}})=4 \mathrm{~cm}$
$AC = AD + CD = 10$ cm
Area of $\Delta ABC$ = $\frac{1}{2}\times \text{base}\times \text{height}$
= $\frac{1}{2}\times AC\times BD$
= $\frac{1}{2}\times (6+4)\times 12$
= $60$ cm
2
Hence, the correct answer is 60.
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