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 area of $△$ $APQ$ is one- sixteenth of the area of $△$ $ABC$.

Answer 1

$AP=1cm,PB=3cm,AQ=1.5cm,QC=4.5cm$ [ Given ]

Then, $AB=AP+PB=1+3=4cm$

In $△APQ$ and $△ABC$,

$\angle A=\angle A$ [ Common angle ]

$\frac{AP}{AB}=\frac{AQ}{AC}$ [ Each equal to $\frac{1}{4}$ ]

$\therefore$ $△APQ\sim △ABC$ [ By SAS similarity ]

$\therefore$ $\frac{area\left(△APQ\right)}{area\left(△ABC\right)}=\frac{A{P}^2}{A{B}^2}$ [ By area of similar triangle theorem ]

$\therefore$ $\frac{area\left(△APQ\right)}{area\left(△ABC\right)}=\frac{1^2}{4^2}$

$\therefore$ $\frac{area\left(△APQ\right)}{area\left(△ABC\right)}=\frac{1}{16}$

$\therefore$ $area\left(△APQ\right)=\frac{1}{16}\times area\left(△ABC\right)$