PQRS is a trapezium in which $ QR \parallel PS $ and $ PR = RS $. If $\angle PQR = 96^{\circ}$ and $\angle QPR

Question

PQRS is a trapezium in which $ QR \parallel PS $ and $ PR = RS $. If $\angle PQR = 96^{\circ}$ and $\angle QPR = 36^{\circ}$, then what is the measure of $\angle PRS $?

Examoogle · Accepted Answer

The correct answer is Option C: 84^{\circ}. Use the alternate interior angle property for parallel lines and the triangle sum property. In $\Delta PQR$, find $\angle PRQ = 180-96-36 = 48^{\circ}$. By parallel lines, $\angle RPS = \angle PRQ = 48^{\circ}$. In isosceles $\Delta PRS$ ($PR=RS$), base angles are equal ($\angle RPS = \angle RSP = 48^{\circ}$). Finally, $\angle PRS = 180 - (48+48) = 84^{\circ}$.

Step-by-Step Solution & Concepts:
Shortcut Trick: Use the alternate interior angle property for parallel lines and the triangle sum property. In $\Delta PQR$, find $\angle PRQ = 180-96-36 = 48^{\circ}$. By parallel lines, $\angle RPS = \angle PRQ = 48^{\circ}$. In isosceles $\Delta PRS$ ($PR=RS$), base angles are equal ($\angle RPS = \angle RSP = 48^{\circ}$). Finally, $\angle PRS = 180 - (48+48) = 84^{\circ}$.

Formula Used: $$\angle QRP = 180^{\circ} - (\angle PQR + \angle QPR)$$, $$\angle RPS = \angle QRP$$, $$\angle PRS = 180^{\circ} - 2 	imes \angle PRS$$, $$\angle PRS = \frac{180^{\circ} - \angle RSP}{2}$$

Given: PQRS is a trapezium with QR || PS. PR = RS. ∠PQR = 96°, ∠QPR = 36°.

Steps: Calculate angle in triangle PQR: In $\Delta PQR$, the sum of angles is $180^{\circ}$. Given $\angle PQR = 96^{\circ}$ and $\angle QPR = 36^{\circ}$.. Use alternate interior angles: Since $QR || PS$, the alternate interior angle $\angle RPS$ is equal to $\angle PRQ$.. Use isosceles triangle property: In $\Delta PRS$, it is given that $PR = RS$. Therefore, the base angles opposite to these sides are equal: $\angle RSP = \angle RPS = 48^{\circ}$.. Calculate the final angle: Sum of angles in $\Delta PRS$ is $180^{\circ}$.

Answer: C is correct because the calculated value of $\angle PRS$ is $84^{\circ}$.

Exam Tip: This problem connects geometry of parallel lines with triangle properties, which is frequently tested in exams under properties of quadrilaterals.

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Discussion (0)

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)