Question

ABC is a triangle, and PQ is a straight line meeting AB in P and AC in Q. If AP= 1 cm, 1 BP = 3 cm, AQ = 1.5 cm, CQ = 4.5 cm. Prove that the area of Δ APQ = 1/16 (area of ΔABC).

Answer 1

Given: AP= 1 cm, 1 BP= 3 cm, AQ = 1.5 cm, CQ = 4.5 cm

Here, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3

⇒ PQ || BC [by converse of basic proportionality theorem]

In ΔABC an ΔAPQ

∠B = ∠P [∵ PQ || BC and AB is transversal, Corresponding angles are equal]

∠C = ∠Q [∵ PQ || BC and AC is transversal, Corresponding angles are equal]

∠BAC =∠PAQ [common angle]

∴ ΔABC ~ ΔAPQ [by AAA similarity]

Now, we know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

Hence Proved