In triangle ABC, it is known that AB = 9 and BC/AC = 40/41.
Find the maximum possible area of the triangle ABC.
Let A = (-c,0), B = (c,0), and C = (x,y)
where c = 9/2.
|BC| sqrt[(x - c)^2 + y^2]
------ = ----------------------- = k = 40/41.
|AC| sqrt[(x + c)^2 + y^2]
or
(x - c)^2 + y^2 = k^2*[(x + c)^2 + y^2] (1)
taking the derivative of (1) with respect
to x and setting dy/dx = 0 gets
x = c*(1 + k^2)/(1 - k^2).
Plugging x into (1) and solving for y:
y = 2*c*k/(1 - k^2)
max. area of triangle ABC = c*y
= 2*c^2*k/(1 - k^2)
= 820
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Posted by Bractals
on 2014-10-24 18:28:09 |
Solution
