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

 

 Feb 22, 2020
 #1
avatar+531 
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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  indecision

 Feb 22, 2020
edited by Dragan  Feb 22, 2020
 #2
avatar+109450 
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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  !!!!!

 

cool cool cool

 Feb 23, 2020
edited by CPhill  Feb 23, 2020

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