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Question
In the following figure, QS ⊥ PR, RT ⊥ PQ and QS = RT.
- Is ∆QSR = ∆RTO? Give reasons.
- Is ∠PQR = ∠PRQ? Give reasons.
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Solution
i. In ∆QSR and ∆RTQ,
∠QSR = ∠RTQ = 90° ...(∵ QS ⊥ PR and RT ⊥ PQ (given))
QS = RT ...(Given)
QR = RQ ...(Common hypotenuse)
∴ ∆QSR ≅ ∆RTQ ...(RHS criterion)
ii. Yes, by using (i) part, we get
∠TQR = ∠SRQ ...(By C.P.C.T)
⇒ ∠PQR = PRQ
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