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 $\mathrm{\Delta }DRC$, $\mathrm{\Delta }APQ$, $\mathrm{\Delta }BQZ$ and $\mathrm{\Delta }ASD$ 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
$\begin{array}{rl}& 2x+2y=180{}^{\circ }\\ & 2y+2z=180{}^{\circ }\\ & 2z+2w=180{}^{\circ }\\ & 2w+2x=180{}^{\circ }\end{array}$
Dividing by 2 on both sides of all four equations, we get the following set of equations,
$\begin{array}{rl}& x+y=90{}^{\circ }\\ & y+z=90{}^{\circ }\\ & z+w=90{}^{\circ }\\ & w+x=90{}^{\circ }\end{array}$
Now, consider $\mathrm{\Delta }APB$. The sum of the angles of a triangle is $180{}^{\circ }$. Therefore, we have
$x+y+\mathrm{\angle }APB=180{}^{\circ }$.
We have already noted that $x+y=90{}^{\circ }$. This implies that $\mathrm{\angle }APB=90{}^{\circ }$.
Similarly, in $\mathrm{\Delta }DRC$, $w+z+\mathrm{\angle }DRC=180{}^{\circ }$. Substituting $w+z=90{}^{\circ }$, we get $\mathrm{\angle }DRC=90{}^{\circ }$

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

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

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

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

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.