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

To find the area of triangle \( \triangle ABQ \) in the given trapezoid \( ABCD \), we need to use the relationship between the similar triangles \( \triangle CPD \) and \( \triangle AQB \), given that the area of \( \triangle CPD \) is \( 4\sqrt{3} \).

Here is a step-by-step outline for the calculation:

1. Firstly, considering \( \triangle CPD \) is an equilateral triangle, all of its sides are equal. Let's denote the side length of \( \triangle CPD \) as \( a \).
2. The area of an equilateral triangle is given by the formula:

\(Area = \frac{\sqrt{3}}{4}a^2\)

1. Given that the area of \( \triangle CPD \) is \( 4\sqrt{3} \), we can set up the equation:

\(4\sqrt{3} = \frac{\sqrt{3}}{4}a^2\)

1. Solving for \( a^2 \), multiply both sides by 4:

\(a^2 = 16\)

1. Thus, \( a = 4 \). So, each side of \( \triangle CPD \) is 4.
2. Since \( AQ \parallel PC \) and \( AP = PC = 4 \), by the properties of similar triangles and the parallel line theorem:

\(\triangle AQB \sim \triangle CPD\)

1. Since \( AQ \parallel PC \), the ratios of corresponding sides in \( \triangle AQB \) and \( \triangle CPD \) are equal, and the length of \( AQ \) equals the height of \( \triangle AQB \).

The base \( AB \) is also parallel and equal to \( PD = 4 \). Since all characteristics of \( \triangle AQB \) are \( 1/2 \) of that of \( \triangle CPD \), the height from \( A \) perpendicular to \( QC \) is \( 4/2 = 2 \).

1. Thus, the base of \( \triangle ABQ \) is half of side \( AD \), and its height is half of side \( CP \), leading to the area:

\(\text{Area of } \triangle ABQ = \frac{1}{2} \times 2 \times 2 = 2\sqrt{3}\)

Thus, the area of \( \triangle ABQ \) is \( 2\sqrt{3} \). Therefore, the correct answer is 2√3.

Answer 2

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:

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