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Ajay wrote at 2011-05-25 05:38:42
The problem can also be solved via Pythagoras theorem.

We know that: given two sides of a triangle, the area is maximized if we form a right-angled triangle using those two sides (since we want to maximize height * width)

Hence the third side - hypotenuse = sqrt( (11)squared + (5)squared ) = 12.083


Eric wrote at 2013-05-08 01:47:20
I got a simple way. Please advise if it is not valid.



Let the triangle be ABC where AB=5 and AC=11.



The area of the triangle is given by

1/2 * (AB) * (AC) * sin(A)

= 1/2 * 5 * 11 * sin(A)



Because -1 <= sin(A) <= 1

In order to maximize it, we take sin(A) = 1

It means A = 90deg



Therefore,

BC = sqrt(5^2 + 11^2)   (pyth. theorem)

BC = sqrt(146)

BC = 12.08



The area is given by 5 * 11 / 2 = 27.5


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