Question : $ABC$ is a triangle and $D$ is a point on the side $BC$. If $BC = 16\mathrm{~cm}$, $BD = 11 \mathrm{~cm}$ and $\angle ADC = \angle BAC$, then the length of $AC$ is equal to:
Option 1: $4 \sqrt{5} \mathrm{~cm}$
Option 2: $4 \mathrm{~cm}$
Option 3: $3 \sqrt{5} \mathrm{~cm}$
Option 4: $5 \mathrm{~cm}$
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Correct Answer: $4 \sqrt{5} \mathrm{~cm}$
Solution :
We have, The $ABC$ is a triangle and $D$ is a point on the side $BC$. If $BC = 16 \mathrm{~cm}$, $BD = 11 \mathrm{~cm}$, $\angle ADC = \angle BAC$ (given), $\angle C$ is common angle and $AC$ is common side. So, $\triangle BAC \sim \triangle ADC$ Also, $DC = BC – BD ==16-11= 5 \mathrm{~cm}$ Now, $\frac{BC}{AC}=\frac{AC}{DC}$ ⇒ $\frac{16}{AC}=\frac{AC}{5}$ ⇒ $AC = 4\sqrt{5}$ Hence, the correct answer is $4 \sqrt{5}\mathrm{~cm}$.
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