Q8
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.
Answer: A
Given that, CPD is an equilateral triangle
∠CPD = ∠PDC = ∠DCP = 60°
AQ || PC ⇒ ∠CPD = ∠QAP = 60°
BC || AD ⇒ ∠QAP = ∠AQB = 60°
⇒ AQCP is a parallelogram, hence AQ = PC = a
∆AQB is a 30°-60°-90° triangle.
⇒ BQ = 1/2 × AQ = a/2
Comapring ∆AQB and ∆CPD
Base AQ = 1/2 × PD while the heights of both triangles is same.
⇒ Area of ∆AQB = 1/2 × Area of ∆CPD = 2√3
Hence, option (a).