A) AP = QP
B) AP > AQ
C) \[\angle APS>\angle APQ\]
D) AR>AQ
Correct Answer: D
Solution :
In\[\Delta APQ\] and\[\Delta APS,\] we have \[PQ=PS\] [Given] \[AP=AP\] [Common] \[\angle APQ=\angle APS\] [Each\[{{90}^{o}}\]] \[\therefore \]\[\Delta \Alpha PQ=\Delta APS\] [By SAS congruency] \[\therefore \]\[\angle AQP=\angle ASP\] [By C.P.C.T.] or \[\angle AQS=\angle ASQ\] ?(1) But, \[\angle ASQ>\angle ARS\] \[\therefore \]\[\angle AQS>\angle ARS\] [From (1)] \[\Rightarrow \]\[\angle AQR>\angle ARQ\] \[\therefore \] \[AR>AQ.\] [Side opposite to greater angle is longer]
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