Question

Let \( A\left(3, 3\sqrt{3}\right) \), \( B(6, 0) \) and \( C(\alpha, \beta) \) be the vertices of an equilateral triangle. Let its incentre be \( D \), and \( P \) and \( Q \) be the feet of perpendiculars drawn from the point \( D \) on \( AC \) and \( BC \), respectively. Then the area of quadrilateral \( CPDQ \) is :

• A. \( 2\sqrt{3} \)
• B. \( 3\sqrt{3} \)
• C. \( 6\sqrt{3} \)
• D. \( 8\sqrt{3} \)

Hint:

The side is 6, the inradius is \( \sqrt{3} \); the quadrilateral is made of two congruent right triangles.

Answer 1

The Correct Option is B

Side \( AB = \sqrt{(6 - 3)^2 + (0 - 3\sqrt{3})^2} = \sqrt{9 + 27} = 6 \), so the triangle is equilateral with side 6.
The incentre D is also the centroid, and the inradius is \( r = \frac{\text{side}}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \sqrt{3} \). So \( DP = DQ = \sqrt{3} \).
D lies on the bisector of angle C (60 degrees), so \( \angle DCP = 30^\circ \) and \( CP = \frac{DP}{\tan 30^\circ} = \sqrt{3} \times \sqrt{3} = 3 \).
Triangles CPD and CQD are congruent right triangles, each with area \( \frac{1}{2} \times 3 \times \sqrt{3} = \frac{3\sqrt{3}}{2} \).
Area of CPDQ = \( 2 \times \frac{3\sqrt{3}}{2} = 3\sqrt{3} \).
Answer: (2) \( 3\sqrt{3} \)