Question

In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.

Answer 1

GIVEN: In ΔABC, P divides the side AB such that AP : PB = 1 : 2, Q is a point on AC such that PQ || BC.

TO FIND: The ratio of the areas of ΔAPQ and the trapezium BPQC.

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.

Let Area of ΔAPQ= 1 sq. units and Area of ΔABC = 9x sq. units

Now,

$\frac{\mathrm{ar}\left(∆\mathrm{APQ}\right)}{\mathrm{ar}\left(\mathrm{trapBCED}\right)}=\frac{x\mathrm{sq}\mathrm{units}}{8x\mathrm{sq}\mathrm{units}}=\frac{1}{8}$