Question 10 In given figure PQR is a right triangle, right angled at Q and QS ⊥ PR. If PQ = 6cm and PS = 4cm, then find QS, RS and QR.
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Solution
Given, ΔPQR in which ∠Q=90∘, QS⊥PR and PQ = 6cm, PS = 4cm.
In ΔSQP and ΔSRQ, ∠PSQ=∠RSQ [each equal to 90∘ ] ∠SPQ=∠SQR [each equal to 90∘−∠R] ∴ΔSQP∼ΔSRQ [ by AAA similarity criterion] Then, SQPS=SRSQ ⇒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=√20=2√5cm On putting the value of QS in Eq..(i), we get (2√5)2=4×SR ⇒SR=4×54=5cm
In right angled ΔQSR, QR2=QS2+SR2 ⇒QR2=(2√5)2+(5)2 ⇒QR2=20+25 ∴QR=√45=3√5cm Hence, QS=2√5cm,RS=5cm and QR=3√5cm