Question

In triangle PQR measure angle PQR is 135 degree prove that PR^2=PQ^2+QR^2+4(area of triangle PQR)
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Answer 1

See the full explanation below

Step-by-step explanation:

cos(∠PQR) = $\frac{PQ^{2} + QR^{2} - PR^{2} }{2 *PQ *QR}$

cos(135°) = $\frac{-1}{\sqrt{2} }$ = $\frac{PQ^{2} + QR^{2} - PR^{2} }{2 *PQ *QR}$

${PQ^{2} + QR^{2} - PR^{2} }$ =  - 2* PQ* QR/$\sqrt{2}$

$- PQ^{2} - QR^{2} + PR^{2}$ = 2 * 2 * ($\frac{1}{2} * PQcos(45) * QR$)

$- PQ^{2} - QR^{2} + PR^{2}$ = 4*  ( Area of triangle PQR)

$PR^{2}$ = $PQ^{2} + QR^{2}$ + 4*  ( Area of triangle PQR)

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Answer 2

The area of the triangle is PQR = PR^2

Step-by-step explanation:

In ∆ pqr, construct PN perpendicular to QR

In triangle PNQ, angle Q=45°

So QN = PN

PQ^2+QR^2-4 x area PQR = PN^2 + QN^2 + (QN + NR)2-4 x 0.5 x QR x PN

=2PN^2+PN^2+NR^2+2 x PN x NR-2 x QR x PN

=2PN^2+PR^2+2PN.NR-2(PN+NR)*PN

=2PN^2-2PN^2+2PN x NR-2PN x NR + PR^2 =PR^2

Thus the area of the triangle is PQR = PR^2

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