(Q8) ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm and BP = 3 cm, AQ =1.5 cm, CQ = 4.5 cm. Prove that area of ∆APQ =

1

16

(area of ∆ABC).

Solution :

Given : ∆ABC and PQ - a line segment meeting AB in P and AC in Q.

AP = 1 cm; AQ =1.5 cm;

BP = 3 cm; CQ = 4.5 cm.

From (1) and (2),

[i.e., PQ divides AB and AC in the same ratio - By converse of Basic proportionality theorem]

Hence, PQ ∥ BC

Now in ∆APQ and ∆ABC

∠A = ∠ (Common)

∠ = ∠B [∵ Corresponding angles for the parallel lines PQ and BC]

∠Q =∠

∴ ∆APQ ~ ∆ABC [∵ A.A.A similarity condition]

[∵ Ratio of two similar triangles is equal to the ratio of the squares of their corresponding sides].