Question

The bisectors of the angles of a parallelogram enclose a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Answer 1

Let $ABCD$ be a parallelogram as given below:

Since, sum of adjacent angles in a parallelogram is supplementary,
So, $\angle A+\angle B={180}^{\circ }$
$\Rightarrow \frac{\angle A}{2}+\frac{\angle B}{2}=\frac{{180}^{\circ }}{2}$
Since $AR$ and $BR$ are bisectors of $\angle A$ and $\angle B$ respectively.
So, $\angle RAB+\angle RBA={90}^{\circ }$....(i)
Since, In a triangle $\Delta ARB$,
We have $\angle ARB+\angle RAB+\angle RBA={180}^{\circ }$
$\Rightarrow \angle ARB+90=180$ [from equation (i)]
$\therefore \angle ARB=\angle QRS={90}^{\circ }$
Similar way, we can prove $\angle QPS={90}^{\circ }$
Since, $\angle A+\angle D={180}^{\circ }$
$\Rightarrow \frac{\angle A}{2}+\frac{\angle D}{2}=\frac{{180}^{\circ }}{2}$
$\Rightarrow \angle DAQ+\angle ADQ={90}^{\circ }$
In a triangle $AQD$, we have $\angle AQD+\angle DAQ+\angle ADQ=180$
$\Rightarrow \angle AQD+90=180$
$\therefore \angle AQD=\angle PQR={90}^{\circ }$ [As vertically opposite angles]
Similarly, we can prove $\angle PSR={90}^{\circ }$
Therefore, $PQRS$ is a rectangle.
Hence, the correct option is C (Rectangle).