Question : In the given figure, $\Delta QPS \cong \Delta SRQ $. Find the measure of $\angle PSR$.
Option 1: 64°
Option 2: 74°
Option 3: 52°
Option 4: 82°
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Correct Answer: 74°
Solution :
$\triangle$QPS = $\triangle$SRQ
$\angle$QPS = $\angle$SRQ (Congruent part of congruent triangles)
So, 106° = 2x + 12°
⇒ 106° – 12° = 2x
⇒ 94° = 2x
$\therefore$ x = 47°
$\angle$QPS = $\angle$SRQ = 106°
So, PQRS is a parallelogram.
$\angle$QSR = 180° – (42° + 106°) = 180° – 148° = 32°
$\angle$PQS = 32° (alternate interior angles)
$\angle$SQR =$\angle$PSQ = 42° (alternate interior angles)
$\therefore\angle$PSR = $\angle$QSR + $\angle$PSQ = 32° + 42° = 74°
Hence, the correct answer is 74°.
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