Question : In an equilateral triangle ABC, P is the centroid of this triangle. Side of $\triangle A B C$ is $16 \sqrt{3} \ \text{cm}$. What is the distance of point P from side BC?
Option 1: 8 cm
Option 2: 12 cm
Option 3: 9 cm
Option 4: 10 cm
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Correct Answer: 8 cm
Solution :
Height $=$ AD $=\frac{\sqrt3}{2}\times \text{side} = \frac{\sqrt3}{2}\times 16\sqrt{3} = 24$ cm
Now, AP : PD = 2 : 1
Let AP = $2x$ and PD = $x$, Then, PD = $3x$
Here, $3x=24$
⇒ $x=8$ cm
Hence, the correct answer is 8 cm.
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