Question

PQR is a triangle right angled at Q and QS is perpendicular to PR.If PQ=6cm qnd PS=4cm,find RS,QS and QR​

Answer 1

Answer:

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 …..(1)

In right angled ∆PSQ,

PQ2 = PS2 + QS^2 [by Pythagoras theorem]

= (6)2 = (4)2 + QS^2

= 36 = 16 + QS^2

= QS^2 = 36-16 = 20

∴ QS = √20 = 2√5 cm

On putting the value of QS in Equation...(1)

we get,

(2√5)^2=4×SR

SR=4×5/4=5cm

In right angled ∆QSR,

QR^2 = QS^2 + SR^2

= QR^2 = (2√5)^2+ (5)^2

= QR^2 = 20 + 25

∴ QR = √45 = 3√5cm

Hence,

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