Question

In a parallelogram $PQRS,$ $PQ=12cm$ and $PS=9cm.$ The bisector of $\angle P$ meets $SR$ in $M.$ $PM$ and $QR$ both when produced meet at $T.$ Find the length of $RT.$

Answer 1

Given: In parallelogram $PQRS,PQ=12cm$ and $PS=9cm.$ The bisector of $\angle SPQ$ meets $SR$ at $M.$
Let $\angle SPQ=2x.$
⇒ $\angle SRQ=2x$ [Opposite angles of a parallelogram]

and, $\angle TPQ=x.$ [$PM$ is the angle bisector]
Also, $PQ\parallel SR$
⇒ $\angle TMR=\angle TPQ=x.$ [Corresponding angles]

In $△TMR$, $\angle SRQ$ is an exterior angle.
$\Rightarrow \angle SRQ=\angle TMR+\angle MTR$
$\Rightarrow 2x=x+\angle MTR$
$\Rightarrow \angle MTR=x$
$\Rightarrow △TPQ$ is an isosceles triangle. $[\because \angle MTR=\angle TPQ=x]$

$\Rightarrow TQ=PQ=12cm$ [sides opposite to equal angles ]
Now,
$RT=TQ-QR$
$=TQ-PS$ [(\because PQRS\) is a parallelogram, opposite sides are equal, $QR=PS$]
$=12-9$
$=3cm$