In ΔPQR
∠PQR = 90∘ [Given]
QS⊥PR [From vertex Q to hypotenuse PR]
∴QS2 = PS × SR (i) [By theorem]
Now , in ΔPSQ we have
QS2 = PQ2 − PS2 [By Pythagoras theorem]
= 62 − 42
= 36 − 16
QS2 = 20
⇒ QS = 2√5 cm
QS2 = PS × SR (i)
⇒ (2√5)2 = 4 × SR
⇒ 20/4 = SR
⇒ SR = 5cm
Now , QS⊥PR
∴ ∠QSR = 90∘
⇒ QR2 = QS2 + SR2 [By Pythagoras theorem]
= (2√5)2 + 52
= 20 + 25
⇒ QR2 = 45
⇒ QR = 3√5 cm
Hence, QS = 2√5 cm, RS = 5 cm and QR = 3√5cm.