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Originally Posted by arcade79 Is this the answer? |
Hats off !!! You're right !
My approach was a bit different... Would like to share with you
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Originally Posted by arcade79 Consider a triangle ABC
let D be a point such that AD=BD=10 and CD=18.
Let CD cut AB at O.
Now focus on Triangles AOC and BOC.. It shouldnt be difficult to prove that they both are congruent triangles....
Thus angle AOC=angle BOC=90 degrees |
Precisely... that was it!
Now, CO is the altitude of triangle ABC and OD is the median ( and altitude ) of the triangle ABD.
If a is the side of the equilateral triangle, then
CO = alt = [sqrt(3)/2] * a
OD = 18 - alt
In triangle, ABD,
AD^2 + BD^2 = 2 (OD^2 + OB^2)
OB=a/2 (since equilateral)
This gives the solution a=9*sqrt(3) (+/-) sqrt(19) which comes to 19.9 / 11.2
Regards,
Bharathi.