Question

In the given figure, $ABCD$ is a parallelogram. The quadrilateral $PQRS$ is exactly

• A. a square
• B. a parallelogram
• C. a rectangle
• D. a rhombus

Answer 1

The correct option is C a rectangle

In parallelogram $ABCD$, $\angle A+\angle D={180}^{\circ }$
$\Rightarrow$ In $\Delta ASD,\angle SDA+\angle SAD=\frac{{180}^{\circ }}{2}={90}^{\circ }$
($\because DS$ and AS bisects $\angle sD$ and A respectively)
In parallelogram $\angle ASD={180}^{\circ }-(\angle ADS+\angle SAD)$
$={180}^{\circ }-{90}^{\circ }={90}^{\circ }$
$\Rightarrow \angle PSR=\angle ASD={90}^{\circ }$ (vert. opp. $\angle s$)
Similarly $\angle SPQ=\angle PQR$
$=\angle QRS={90}^{\circ }$.
Therefore, quadrilateral $$PQRS is a rectangle.