help

In the diagram below, triangle ABC is isosceles, and triangle MPQ is equilateral. Find the area, in cm2, of triangle MPQ.

BM = 36 cm

AC = BC

PC = QC

Let an F to be the foot of altitude from M to BC

∠FMP = 15°

FM = sqrt [( 36² )/2] = 25.455844

MP = FM / cos(15°) = 26.353829

PQ/2 = MP/2 = 13.1769145

The height of ΔMPQ is H = tan(60°) * (PQ/2) = 22.8230854

Area of  Δ MPQ = (PQ/2) * H = 300.7378467 cm²  

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  !!!!!