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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, QSPR and PQ = 6cm, PS = 4cm.

In ΔSQP and ΔSRQ,
PSQ=RSQ [each equal to 90 ]
SPQ=SQR [each equal to 90R]
ΔSQPΔSRQ [ 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=3616=20
QS=20=25cm
On putting the value of QS in Eq..(i), we get
(25)2=4×SR
SR=4×54=5cm

In right angled ΔQSR,
QR2=QS2+SR2
QR2=(25)2+(5)2
QR2=20+25
QR=45=35cm
Hence, QS=25cm,RS=5cm and QR=35cm

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