Question

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

Answer 1

If a line is drawn parallel to one side of a triangle to intersect the

other two sides in distinct points, the other two sides are divided in the same ratio.

As $PQ\parallel BC$

So $\frac{AP}{PB}=\frac{AQ}{QC}$

$\angle AQP=\angle ACB$

$\angle APQ=\angle ABC$

So by $AAA$ $△AQP\sim △ACB$

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Hence $\frac{Area\left(APQ\right)}{Area\left(ABC\right)}=\frac{(AP{)}^2}{(AB{)}^2}$

$\frac{Area\left(APQ\right)}{Area\left(ABC\right)}=\frac{(AP{)}^2}{(AP+PB{)}^2}$

$\frac{Area\left(APQ\right)}{Area\left(ABC\right)}=\frac{(x{)}^2}{(3x{)}^2}$

$\frac{Area\left(APQ\right)}{Area\left(ABC\right)}=\frac{1}{9}$

Let $Area\left(APQ\right)=k$

$Area\left(ABC\right)=9k$

$Area\left(BPQC\right)=Area\left(ABC\right)-Area\left(APQ\right)=9k-k=8k$

$\frac{Area\left(APQ\right)}{Area\left(BPQC\right)}=\frac{1}{8}$

$\therefore$ the ratio of the $△APQ$ and trapezium $BPQC$ $=\frac{1}{8}$