Triangle pqr is right angled at q. qs is altitude. pq is 2root13 cm and ps 8 cm. find length of sr.
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PQ=4√29 ,PS=8, SR=.........
IN RIGHT PQS
PQ^2=QS^2+PS^2
(29*16)=(QS^2)+64
√(29*16)-64=QS
QS=20
IN PQR
(4√29)^2+(QR) ^2=(PR) ^2
464+(QR) ^2=(PS+SR)^2
IN QSR
(QR) ^2=(QS) ^2 +((SR) ^2
substitute in above eqn
464+400+SR^.2=64+SR^2+16SR
800=16SR
SR=50
IN RIGHT PQS
PQ^2=QS^2+PS^2
(29*16)=(QS^2)+64
√(29*16)-64=QS
QS=20
IN PQR
(4√29)^2+(QR) ^2=(PR) ^2
464+(QR) ^2=(PS+SR)^2
IN QSR
(QR) ^2=(QS) ^2 +((SR) ^2
substitute in above eqn
464+400+SR^.2=64+SR^2+16SR
800=16SR
SR=50
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