Question

Rhombus $PQRB$ is inscribed in $△$ABC such that $\angle$B is one of its angle. $P,Q$ and $R$ lie on $AB,AC$ and $BC$ respectively. If $AB=12$ cm and $BC=6$ cm find the sides of rhombus $PQRB$.

Answer 1

In $△$ABC, PQ || BC ($\because$ In a rhombus opposite sides are parallel)

$\therefore \frac{AP}{AB}=\frac{PQ}{BC}$ ($\because$ Corollary of Thales theorem)

$\frac{12-x}{12}=\frac{x}{6}$

$12x=72-6x$

$18x=72$

$x=4cm$

The sides of the rhombus is $4$ cm each.