Refer to the below diagram
Observe that triangle BPC and BQC are inscribed inside a semicircle. Hence,
∠BPC = ∠BQC = 90°
Therefore, we can say that BQ ⊥ AC and CP ⊥ AB.
Also, In triangle ABC,
Area of triangle = (1/2)×Base×Height = (1/2)×AB×CP = (1/2)×AC×BQ
⇒ BQ = AB×CP / AC = 30×20/25 = 24 cm.
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.