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Question
In a right-angled triangle PQR, right-angled at Q, S and T are points on PQ and QR respectively such as PT = SR = 13 cm, QT = 5 cm and PS = TR. Find the length of PQ and PS.
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Solution
In ΔPQT, ∠Q = 90°
∴ PT2 = PQ2 + QT2 ....(By Pythagoras Theorem)
⇒ PQ2 = PT2 - QT2
⇒ PQ2 = PT2 - QT2
= 132 - 52
= 169 - 25
= 144
⇒ PQ = 12cm
Now, PS = TR = a (say)
In ΔSQR, ∠Q = 90°
∴ SR2 = QS2 + QR2 ....(By Pythagoras Theorem)
⇒ SR2 = (PQ - PS)2 + (QT + TR)2
⇒ SR2 = (PQ - PS)2 + (QT + PS)2
⇒ SR2 = PQ2 - 2 x PQ x PS + PS2 + QT2 + 2 x QT x PS + PS2
⇒ 132 = 122 - 2 x 12 x a + a2 + 52 + 2 x 5 x a + a2
⇒ 169 - 144 - 24a + a2 + 25 + 10a + a2
⇒ 169 = 169 - 14a + 2a2
⇒ 2a2 = 14a
⇒ a = 7
Hence, PS = 7cm.
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