Question

ABCD a parallelogram. $E$ is a point on $AD$ and $CE$ is produced to meet $BA$ at $F$. If $AE=4$ $cm$, $AF=8$ $cm$ and $AB=12$ $cm$. Find the perimeter (in $cm$) of parallelogram $ABCD$.

Answer 1

In $△AFE$ and $△BFC$,
$\angle AFE=\angle BFC$ (Common angle)
$\angle AEF=\angle BCF$ (Corresponding angles of parallel lines)
$\angle FAE=\angle CBF$ (Corresponding angles of parallel lines)
Thus, $△AFE\sim △CFB$ (AAA rule)
Hence, $\frac{AF}{BF}=\frac{AE}{BC}$
$\frac{AF}{AF+AB}=\frac{AE}{BC}$
$\frac{8}{8+12}=\frac{4}{BC}$
$BC=\frac{4\times 20}{8}$
$BC=10$ cm
Perimeter of parallelogram = $AB+BC+BC+AD$
Perimeter of parallelogram = $2(AB+BC)$
Perimeter of parallelogram = $2(12+10)$ (Opposite sides are equal)
Perimeter of parallelogram = $44$ cm