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# In Fig. 6.36, QR by QS equals QT by PR and angle 1 equals angle 2. Show that triangle PQS ~ triangle TQR.

Q4 In Fig. 6.36, $\frac{QR }{QS } = \frac{QT}{PR}$  and $\angle 1 = \angle 2$ . Show that $\Delta PQS \sim \Delta TQR$

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Given :  $\frac{QR }{QS } = \frac{QT}{PR}$  and $\angle 1 = \angle 2$

To prove : $\Delta PQS \sim \Delta TQR$

In $\triangle PQR$ , $\angle PQR=\angle PRQ$

$\therefore PQ=PR$

$\frac{QR }{QS } = \frac{QT}{PR}$                 (Given)

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

In $\Delta PQS\, and\, \Delta TQR$,

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

$\angle Q=\angle Q$        (Common)

$\Delta PQS \sim \Delta TQR$   ( By SAS)

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