Question

6. In the given figure, PS is the median then angle QPS
a) 40°
c) 80°
b) 50°
d) 90°​

Answer 1

$Given \: in\: \triangle PQR , PQ = PR$

$\angle {PQR} = 40\degree ,$

$PS \: is \:a \:median .$

$\underline{\green{ Solution: }}$

$in\: \triangle PQR , PQ = PR$

$\angle {PQR} = \angle {PRQ} = 40\degree$

$\blue {( Angles \: opposite \:to \: equal \:sides}$

$\blue{ equal. )}$

$\angle {QPR} + \angle {PQR} + \angle {PRQ}= 180\degree$

$\orange { ( Angle \:Sum \: Property )}$

$\implies \angle {QPR} + 40\degree + 40 \:degree = 180\degree$

$\implies \angle {QPR} + 80\degree = 180\degree$

$\implies \angle {QPR} = 180\degree - 80\degree$

$\implies \angle {QPR} = 100\degree$

$ii) Now, in\: \triangle QPS \:and \:in\: \triangle RPS$

$PQ = PR \: (given)$

$\angle {PQR} = \angle {PRQ}$

$QS = SR \: ( \because PS \:is \:a \: median )$

$\therefore \triangle QPS \cong \triangle RPS$

$\blue{ ( S .A.S \: congruence \:rule )}$

$\angle {QPS} = \angle {RPS} \: ( C.P.CT )$

$\implies \angle {QPS} = \frac{1}{2} \times \angle {QPR}$

$= \frac{1}{2} \times 100$

$= 50\degree$

Therefore.,

$Option \: \pink { ( b ) } \:is \: correct.$

•••♪

Answer 2

Answer:

50 is your answer buddy