Question

ABCD is a trapezoid where BC is parallel to AD and perpendicular to AB . Kindly note that BC<AD . P is a point on AD such that CPD is an equilateral triangle. Q is a point on BC such that AQ is parallel to PC . If the area of the triangle CPD is 4√3. Find the area of the triangle ABQ.

• A. 2√3
• B. 4√3
• C. 4
• D. 8√3
• E. None of the above

Answer 1

The Correct Option is A

Step 1: Find the side length of \(\triangle CPD\). 

The formula for the area of an equilateral triangle of side \(s\) is:

\[ \text{Area} = \frac{\sqrt{3}}{4}s^2 \]

Given area = \(4\sqrt{3}\), we have:

\[ \frac{\sqrt{3}}{4}s^2 = 4\sqrt{3} \]

\[ s^2 = 16 \quad \Rightarrow \quad s = 4 \]

So, each side of the equilateral triangle \(CPD\) is \(4\).

Step 2: Find the altitude of \(\triangle CPD\).

The altitude is:

\[ h = \frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{2}\times 4 = 2\sqrt{3} \]

Step 3: Analyze trapezium conditions.

• \(BC \parallel AD\)
• \(BC \perp AB\)
• \(AQ \parallel PC\)

Since \(PC = 4\), line \(AQ\) is also parallel and equal to \(PC\). This makes \(\triangle ABQ\) a right triangle where the height corresponds to the perpendicular condition.

Step 4: Compute the area of \(\triangle ABQ\).

Using base–height reasoning, we get:

\[ \text{Area} = \frac{1}{2}\times \text{base}\times \text{height} = 2\sqrt{3} \]

Final Answer:

\[ \boxed{2\sqrt{3}} \]