One response to “Triangle Problem”

1. 

   The Estonian Fox

   13 May 2025

   George,
   Not sure if this qualifies as expressing the Area of the triangle in terms of A1 & A2. I had to include r1 (radius of A1) & r2 (rad of A2) in the final answer.
   I broke the triangle into 4 separate parts. A top equilateral triangle (X), a lower trapezoid (Y), and 2 parallelograms (Z) on either side of the trap, defined by lines parallel to the triangle sides.
   Altitude of X is 3r1 (construct a small rt triangle with r1 as its base, hypotenuse of this small triangle is then 2r1). Base X is 2/sqrt(3) * 3r1. So base X = 2√3 r1. Area X =3√3 r1^2 = 3√3/π A1.
   Similarly, base Y =2√3 r2. Altitude of Y = 2 r2. Top Y = 2√3/3 r2. Area Y = 8√3/3 r2^2 = 8√3/3π A2.
   Altitude Z = 2 r2; base Z = (2√3 r1 – 2√3/3 r2)/2. So Area of both Zs = 4 r2 (√3 r1 – √3/3 r2).
   Area (triangle) = 3√3/π A1 + 8√3/3π A2 + 4 r2 (√3 r1 – √3/3 r2).
   Better way is to drop the areas, use just r1 and r2.
   Area (triangle) = 3√3 *r1^2 + 4√3/3 * r2^2 + 4√3 r1 r2.

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