Question

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

Answer 1

Given,

ΔPQR in which ∠Q = 90°,

QS ⊥ PR and PQ = 6 cm,

PS = 4 cm

In ΔSQP and ΔSRQ,

∠PSQ = ∠RSQ  ...[Each equal to 90°]

∠SPQ = ∠SQR   ...[Each equal to 90° – ∠R]

∴ ΔSQP ∼ ΔSRQ   ...[By AA similarity criterion]

Then, `("SQ")/("PS") = ("SR")/("SQ")`

⇒ SQ² = PS × SR   ...(i)

In right angled ΔPSQ,

PQ² = PS² + QS²   ...[By pythagoras theorem]

⇒ (6)² = (4)² + QS²

⇒ 36 = 16 + QS²

⇒ QS² = 36 – 16 = 20

∴ QS = `sqrt(20) = 2sqrt(5)` cm

On putting the value of QS in equation (i), we get

`(2sqrt(5))^2` = 4 × SR

⇒ SR = `(4 xx 5)/4` = 5 cm

In right angled ΔQSR,

QR² = QS² + SR²

⇒ QR² = `(2sqrt(5))^2 + (5)^2`

⇒ QR² = 20 + 25

∴ QR = `sqrt(45) = 3sqrt(5)` cm

Hence, QS = `2sqrt(5)` cm, RS = 5 cm and QR = `3sqrt(5)` cm.