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In an acute angled triangle PQR,angle PQR=45 degree.Prove that (PR) 2 =(PQ) 2 +(QR) 2 -4ar(trianglePQR).

In an acute angled triangle PQR,angle PQR=45 degree.Prove that (PR)2=(PQ)2+(QR)2-4ar(trianglePQR).

Grade:10

1 Answers

ATHARV DEO
11 Points
6 years ago
construct PN perpendicular to QR
in triangle PNQ, angle Q=45
so QN=PN
PQ^2+QR^2-4*areaPQR=PN^2+QN^2+(QN+NR)2-4*0.5*QR*PN
=2PN^2+PN^2+NR^2+2*PN*NR-2*QR*PN
=2PN^2+PR^2+2PN.NR-2(PN+NR)*PN
=2PN^2-2PN^2+2PN*NR-2PN*NR+PR^2

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