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In the given figure. PQ = PS, P =R = 90°. RS = 20 cm and QR = 21 cm. Find the length of PQ correct to two decimal places.

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#### Solution

In ΔSRQ, ∠R = 90°

∴ QS^{2} = RS^{2} + QR^{2} ....(Pythagoras Theorem)

= 20^{2 }+ 21^{2}

= 440 + 441

= 841

Now,

In ΔQSP, ∠P = 90°

∴ QS^{2} = PQ^{2} + PS^{2}

⇒ QS^{2} = PQ^{2 }+ PQ^{2} ....(Pythagoras Theorem)

⇒ QS^{2} = 2PQ^{2} ....(Given PQ = PS)

⇒ PQ2 = `"QS"^2/(2) = (841)/(2)` = 420.5

⇒ PQ = 20.50cm.

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