Right Triangle

by **Stockmoose16** Wed Oct 22, 2008 4:39 pm

I was doing an MGMAT challenge question, and I came across the following statement:

The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length).

This doesn't make sense. A 30-60-90 right triangle would have an area of (X^2 * sqrt 3)/2, which is clearly greater than an isoceles right triangle (45-45-90), which would have an area of just (x^2)/2

Can anyone explain?

by **Saurav** Sat Oct 25, 2008 3:53 pm

A 30-60-90 right triangle would have an area of (X^2 * sqrt 3)/2, which is clearly greater than an isoceles right triangle (45-45-90), which would have an area of just (x^2)/2

What is X in both the expressions ?

[/quote]

by **JonathanSchneider** Thu Nov 13, 2008 1:30 am

Saurav's question is apt. The x must represent the long side. Imagine, for instance, if we could make the triangle have degrees of 90/89/1. This would be a very tall triangle, if we were to put the angle of 1 degree a the top. Now, if we were to say that the side opposite that smallest angle were x, then clearly we would have a very large area in compared to x.

The question hinges on the fact that x is thus the longest side possible.