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.

Answer 1

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.