Question

In △PQR, PQ = PR and QR = 18 in. The altitude PN = 12 in and S is a point on the extension of QR so that PS = 20 in. Prove that ∠QPS is a right angle.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Thus, $\angle QPS$ is right angle.

Hence Proved.

Step-by-step explanation:

Given,

In $\triangle PQR$, $PQ=PR$,$QR=18in.$,altitude $PN=12in.$ and $PS=20in.$

From figure,

$QN=NR=\frac{QR}{2}=\frac{18}{2}=9in.$ (∵ $PQ=PR$ )

From $\triangle PQN$,

$PQ^{2}=PR^{2} =PN^{2} +QN^{2}$

⇒$PQ^{2}=PR^{2} =12^{2} +9^{2}=144+81$

⇒$PQ=PR=\sqrt{225}$

⇒$PQ=PR=15in.$

From $\triangle PNS$,

$NS^{2} =PS^{2} -PN^{2}$

⇒$NS^{2} =20^{2} -12^{2}$

⇒$NS=\sqrt{256}=16in.$

∴$QS=QN+NS=9+16$

⇒$QS=25$

From $\triangle PQS$,

$PQ^{2}+PS^{2} =15^{2} +20^{2}$

⇒$PQ^{2}+PS^{2} =225+400$

⇒$PQ^{2}+PS^{2} =625$

⇒$PQ^{2}+PS^{2} =25^{2}$

⇒$PQ^{2}+PS^{2} =QS^{2}$

Which is satisfied Pythagorean Theorem.

So, $\angle QPS$ is right angle.

Hence Proved.