In given figure, PQR is a right triangle, right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, then find QS, RS and QR.

Given,

∆PQR in which ∠Q = 90°,

QS ⊥ PR and PQ = 6cm.

PS = 4cm

In ∆SQP and ∆SRQ,

∠PSQ = ∠RSQ [each equal to 90°]

∠SPQ = ∠SQR [each equal to 90°-∠R]

∴ ∆SQP ∼ ∆SRQ

Then,

= SQ/PS = SR/SQ

⇒ SQ² = PSXSR …..(i)

In right angled ∆PSQ,

PQ² = PS² + QS² [by Pythagoras theorem]

= (6)² = (4)² + QS²

= 36 = 16 + QS²

= QS² = 36-16 = 20

∴ QS = √20 = 2√5 cm

On putting the value of QS in Eq. (i),

we get,

In right angled ∆QSR,

QR² = QS² + SR²

= QR² = (2√5)² + (5)²

= QR² = 20 + 25

∴ QR = √45 = 3√5cm

Hence,

QS = 2√5 cm, RS = 5 cm and QR = 3√5 cm.