In a quadrilateral ^@PQRS,^@ if ^@PQ \parallel RS, \angle S = 2 | Math Question and Answer

Answer:

$p + q$

Step by Step Explanation:

Let us first draw the quadrilateral $P Q R S$ $P Q R S$ .
Let $∠ Q = x$ $∠ Q = x$
So, $∠ S = 2 x$ $∠ S = 2 x$
Let us join $R$ $R$ to a point $E$ $E$ on the side $P Q$ $P Q$ such that $P E R S$ $P E R S$ is a parallelogram.
We know that opposite sides of a parallelogram are equal.
So, $\angle PSR = \angle PER = 2x \ldots (i)$
Also, $\begin{align} & \angle PER + \angle REQ = 180^\circ &&[ \text{ Angles on a straight line } ] \\ \implies & \angle REQ = 180^ \circ - \angle PER \\ \implies & \angle REQ = 180^ \circ - 2x && [\text{From } (i)] \\ \end{align}$
The sum of angles of a triangle is $180^ \circ.$
In $\triangle ERQ$ $\begin{align} & \angle REQ + \angle EQR + \angle QRE = 180^ \circ \\ \implies & 180^\circ - 2x + x + \angle QRE = 180^\circ \\ \implies & \angle QRE = x \end{align}$
In $\triangle ERQ,$ $\begin{align} & \angle QRE = \angle EQR \\ \implies & ER = EQ \ldots (ii) && [\text{ Sides opposite to equal angles are equal. }] \\ \end{align}$
We are given that $RS = p$ and $PS = q.$
As, opposite sides of a parallelogram are equal,
$RS = PE = p$
and $PS = ER = q$
$\implies EQ = q \ldots [\text{From } (ii)]$
We can see that $PQ = PE + EQ = p + q$
Thus, $PQ = p + q$

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