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प्रश्न
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.
उत्तर
In ΔSRQ, ∠R = 90°
∴ QS2 = RS2 + QR2 ....(Pythagoras Theorem)
= 202 + 212
= 440 + 441
= 841
Now,
In ΔQSP, ∠P = 90°
∴ QS2 = PQ2 + PS2
⇒ QS2 = PQ2 + PQ2 ....(Pythagoras Theorem)
⇒ QS2 = 2PQ2 ....(Given PQ = PS)
⇒ PQ2 = `"QS"^2/(2) = (841)/(2)` = 420.5
⇒ PQ = 20.50cm.
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