The correct option is A (A)→q;(B)→s;(C)→p;(D)→r
(A)→q;(B)→s;(C)→p;(D)→r
(A) InΔPQS and ΔPRS
PQ = PR = 6 cm [Given]
PS = PS [Common]
∠PSQ=∠PSR=90∘
⇒ΔPQS≅ΔPRS (by R.H.S congruency)
(B) InΔPOQ and ΔROS,
PQ=RS=6cm
∠QPO=∠SRO=30∘
∠POQ=∠ROS
(vertically opposite angles)
⇒ΔPOQ≅ΔROS (by AAS congruency)
(C) InΔPOR and ΔQOS,
PO=QO (Given)
∠POR=∠QOS
(vertically opposite angles)
RO=SO (Given)
ΔPOR≅ΔQOS (by SAS congruency)
(D) InΔPMQ and ΔPMR,
PQ=PR (given)QM=RM (given)PM=PM (common)
⇒ΔPMQ≅ΔPMR (by SSS congruency)