Question

ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that the area of ∆APQ is one-sixteenth of the area of ∆ABC.

Answer 1

Given: In ΔABC, PQ is a line segment intersecting AB at P, and AC at Q. AP = 1cm, PB = 3cm, AQ = 1.5cm and QC = 4.5cm.

To find:

In ΔABC,

According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence, PQ || BC

In ΔAPQ and ΔABC,

$\angle \mathrm{APQ}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle \mathrm{PAQ}=\angle \mathrm{BAC}\left(\mathrm{Common}\right)$

So, $∆\mathrm{APQ}~∆\mathrm{ABC}$ (AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence