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.

#### Solution

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]

∴ ∠SQP ∼ ∠SRQ

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

⇒ `SQ^2 = PS xx SR` .....(i)\

In right angled ΔPSQ,

`PQ^2 = PS^2 + QS^2 ......[By Pythagoras theorem]

⇒ `(6)^2 = (4)^2 + QS^2`

⇒ 36 = `16 + QS^2`

⇒ `QS^2 = 36 - 16 = 20`

∴ `QS = sqrt(20) = 2sqrt(5)` cm

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

`(2sqrt(5))^2 = 4 xx SR`

⇒ `SR = (4 xx 5)/4` = 5 cm

In right angled ΔQSR,

`QR^2 = QS^2 + SR^2`

⇒ `QR^2 = (2sqrt(5))^2 + (5)^2`

⇒ `QR^2 = 20 + 25`

∴ `QR = sqrt(45) = 3sqrt(5)` cm

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