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

Hint: We will use the fact that $ABCD$ is a parallelogram and quadrilateral $PQRS$ is formed due to the intersections of the angle bisectors of the angles of parallelogram $ABCD$. We will use the congruence of opposite angles and the properties of adjacent angles of a parallelogram. We will look at $\Delta DRC$, $\Delta APQ$, $\Delta BQZ$ and $\Delta ASD$ and the sum of the angles of these triangles.

Complete step-by-step answer:
We know that the adjacent angles of a parallelogram are supplementary. Therefore, we have
$\begin{align}
  & 2x+2y=180{}^\circ \\
 & 2y+2z=180{}^\circ \\
 & 2z+2w=180{}^\circ \\
 & 2w+2x=180{}^\circ \\
\end{align}$
Dividing by 2 on both sides of all four equations, we get the following set of equations,
$\begin{align}
  & x+y=90{}^\circ \\
 & y+z=90{}^\circ \\
 & z+w=90{}^\circ \\
 & w+x=90{}^\circ \\
\end{align}$
Now, consider $\Delta APB$. The sum of the angles of a triangle is $180{}^\circ $. Therefore, we have
$x+y+\angle APB=180{}^\circ $.
We have already noted that $x+y=90{}^\circ $. This implies that $\angle APB=90{}^\circ $.
Similarly, in $\Delta DRC$, $w+z+\angle DRC=180{}^\circ $. Substituting $w+z=90{}^\circ $, we get $\angle DRC=90{}^\circ $

Now, we know that opposite angles are congruent. Hence, $\angle ASD=\angle PSR$ and $\angle BQC=\angle PQR$.

Let us consider $\Delta ASD$. Taking the sum of all the angles of this triangle, we get $x+w+\angle ASD=180{}^\circ $. Again, we already know that $x+w=90{}^\circ $. Hence, we have $\angle ASD=90{}^\circ $ and therefore, $\angle PSR=90{}^\circ $.

Now, considering $\Delta BQC$ and adding all the angles of this triangle, we get $y+z+\angle BQC=90{}^\circ $.

As $y+z=90{}^\circ $ , we get $\angle BQC=90{}^\circ $. And because $\angle BQC$ is the opposite angle of $\angle PQR$, we have $\angle PQR=90{}^\circ $.

Now, in quadrilateral $PQRS$, all four angles are $90{}^\circ $. Hence, it is a rectangle.

So, the correct answer is “Option C”.

Note: From the given information in the question, we are able to conclude that all the angles of the quadrilateral $PQRS$ are right angles. But we have no information to comment on the sides of this quadrilateral. Since, we cannot claim that all sides of this quadrilateral $PQRS$ are equal, we cannot say that it is a square.