Question is from MGMT

The lengths of the two shorter legs of a right triangle add up to 40 units. What is the maximum possible area of the triangle?

Answer is

MGMT set it up as a square then went from there by calculating it as half the area of a maximized square.

My approach was different and it yielded a wrong result, of course

, but I am sure that I had the right idea in mind and something went wrong.

Since we want to maximize a triangle,

the maximum area would be that of an Equalitral triangle which has 3 equal sides. Since s1+s2 = 40 (assuming we are dealing with an Equalitral) then S1 must be 20; therefore, s2 and s3 has to be each 20.

The area of an Equalitral triangle is s^2\sqrt{3}/4 which should give us

400 * \sqrt{3}/4 which is 100\sqrt{3}.

What did I do wrong there?

.

, so the the lengths of the two shorter legs must be equal, so each must be 20. Area=1/2*20*20=200.