Question

In the given figure, $PT=QRandPT||QR$ and the three sides $PQ=RS=TS$. If $\angle QPT=108°$, then $\angle QRS$ is equal to

• A. ${122}^0$
• B. ${148}^0$
• C. ${168}^0$
• D. ${180}^0$

Answer 1

The correct option is C ${168}^0$

Given that $PT=QR$ and $PT||QR$.
Thus, $PQRT$ is a parallelogram.
$\Rightarrow TR=PQ$ and $TR||PQ$
$\Rightarrow TR=RS=TS$
($\because PQ=RS=TS$)
Therefore, $\Delta RST$ is an equilateral triangle.
$\Rightarrow \angle TRS=60°$
Also, in a parallelogram opposite angles are equal.
$\therefore \angle QRT=\angle QPT=108°$
$\therefore \angle QRS=\angle QRT+\angle TRS$
$=108°+60°$
$=168°$
Hence, the correct answer is option (c).