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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)

 

 

 

 

 

Posted by

seema garhwal

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