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

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

∠APQ=∠B              Corresponding angles

∠PAQ=∠BAC       Common

So, ∆APQ~∆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.

$\frac{A r(\triangle \mathrm{APQ})}{A r(\triangle \mathrm{ABC})}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}$

$\frac{\operatorname{Ar}(\Delta \mathrm{APQ})}{A r(\Delta \mathrm{ABC})}=\frac{1 x^{2}}{(1 x+2 x)^{2}}$

$\frac{A r(\triangle \mathrm{APQ})}{A r(\triangle \mathrm{ABC})}=\frac{1}{9}$

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

$A r[\operatorname{trapBCED}]=A r(\triangle \mathrm{ABC})-A r(\triangle \mathrm{APQ})$

$=9 x-1 x$

$=8 x$ sq units

Now,

ar∆APQartrapBCED=x sq units8x sq units=18