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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.
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]
∴ 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.
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