Question

Pick the option which matches the two columns.

• A. $\left(A\right)\to q;\left(B\right)\to s;\left(C\right)\to p;\left(D\right)\to r$
• B. $\left(A\right)\to q;\left(B\right)\to p;\left(C\right)\to r;\left(D\right)\to s$
• C. $\left(A\right)\to s;\left(B\right)\to p;\left(C\right)\to q;\left(D\right)\to r$
• D. $\left(A\right)\to s;\left(B\right)\to r;\left(C\right)\to p;\left(D\right)\to q$

Answer 1

The correct option is A

$\left(A\right)\to q;\left(B\right)\to s;\left(C\right)\to p;\left(D\right)\to r$

$\left(A\right)\to q;\left(B\right)\to s;\left(C\right)\to p;\left(D\right)\to r$
(A) In $\Delta PQS$ and $\Delta PRS$
PQ = PR = 6 cm [given]
PS = PS [common]
$\angle PSQ=\angle PSR={90}^{\circ }$
$\Rightarrow \Delta PQS\cong \Delta PRS$ (by R.H.S congruency)

(B) In $\Delta POQ$ and $\Delta ROS$,
$PQ=RS=6cm$
$\angle QPO=\angle SRO={30}^{\circ }$
$\angle POQ=\angle ROS$ (vertically opposite angles)
$\Rightarrow \Delta POQ\cong \Delta ROS$ (by AAS congruency)

(C) Similarly, we can prove
$\Delta POR\cong \Delta QOS$ (by SAS congruency)

(D) In$\Delta PMQand\Delta PMR$,
$\begin{array}{c}PQ=PR\left(given\right)\\ QM=RM\left(given\right)\\ PM=PM\left(common\right)\end{array}$
$\Rightarrow \Delta PMQ\cong \Delta PMR$ (by SSS congruency)