Given: triangle PQR, area of triangle PQR=50, m

Construct OQ
Consider triangle PQO, it is constructed by two radii, thus this triangle is isosceles, m∠PQO must also have 36°, then m∠POQ is 108°.
Consider triangle QOR, it is also constructed by two radii, thus this triangle is also isosceles, m∠OQR must be 54°, then m∠QOR is 72°.
Find PQ and QR in terms of r (radius) with the law of cosines. [tex]PQ^2=r^2+r^2-2*r*r*Cos108\\PQ=\sqrt{2r^2-2r^2*Cos108}\\QR^2=r^2+r^2-2*r*r*Cos72\\QR=\sqrt{2r^2-2r^2*Cos72}[/tex]
Set up an equation for the area of the triangle. [tex]\frac{(\sqrt{2r^2-2r^2*Cos108})(\sqrt{2r^2-2r^2*Cos72})}{2} =50\\[/tex]
With the help from a calculator, we get two solutions: [tex]x=7.25073177..., -7.25073177...[/tex]
Since lengths cannot be negative, the positive solution is our only solution. Multiply it by 2 to get PR.

[tex]PR=14.501...[/tex]

Now let me receive treatment for my headache...