Since ABC is isosceles.....then angle ABC = 45°
PQ is parallel to BA
Therefore...angle QPC = angle ABC =45°
And angle MPQ = 60°
So angle MPB =180 - 45 -60 = 75°
Using the Law of Sines....we can find MP = the side of MPQ thusly
MB / sin MPB = MP /sin ABC
36 / sin (75) = MP / sin 45
36 sin 45 / [(sin (30 + 45) = MP
36 sin 45 / [ sin 45cos 30 + cos 45 sin 30] {sin 45 = cos 45 }
36sin 45 / [ sin 45 * ( √3/2 + 1/2) ] =
2 * 36 /[ √3 + 1 ] =
72 / [ √3 + 1] =
72 [ √3 -1] / [ 2 ] =
36 [√3 -1 ] = MP
So.....the area of triangle MPQ =
(1/2 ) (MP*^2 * sin (60) =
(1/2) [ 36 * [√3 - 1] ]^2 * (√3 / 2 ) ≈
300.74 units^2
Just as Dragan found !!!!!