Question

In the figure, seg QS perpendicular side PR, side RT perpendicular side PQ, P-S-R and P-T-Q. Show that triangle PQS ~ PRT and QS ×TP = PS ×RT

Answer 1

Answer:

Given:  $QS\perp RT$

Also,  $RT\perp PQ$

We have to prove that:  $\triangle PQS\sim \triangle PRT$

And, QS ×TP = PS ×RT

Proof:

In triangle, PQS and PRT,

$\angle PSQ\cong \angle PTR$        ( Right angles )

$\angle QPS\cong \angle RPS$

Thus, By AA similarity postulates,

$\triangle PQS\sim \triangle PRT$,

Thus, by the property of similar triangle,

$\frac{PS}{PT}=\frac{QS}{RT}$

$\implies PS\times RT=QS\times PT$

Hence, proved.

Answer 2

Given : In figure,

Seg QS _|_ seg PR

Seg RT _|_ seg PQ

To prove : 1) PQS ~ PRT

2) QS × TP = PS × RT

Proof : In PQS and PRT

Angle PSQ = 90° (I)

Angle PTR = 90° (II)

Angle PSQ = Angle PTR....... from (I) and (II)

Angle SPQ = Angle TPR......... common angle

PQS ~ PRT ....... AA test

Now,

PQS ~ PRT by c. s. s. t.

$\frac{pq}{pr} = \frac{qs}{rt} = \frac{ps}{tp} \\ \frac{qs}{rt} = \frac{ps}{tp} \\ \\ qs \: \times \: tp \: = \: ps \: \times \: rt$

QS × TP = PS × RT

HOPE IT HELPS.

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