Let us assume that we have a triangle ABC in which we know the lengths of AB and BC.
Now, let us assume BC is the base of the triangle as shown in the following diagram.

In the above diagram ABC, A'BC, and A"BC are three triangles each with the same base BC and AB = A'B = A"B
Now, are of a triangle is given by (length of base)*(length of height)/2

In triangle ABC, AB is perpendicular on BC.
So, area of ABC = (AB)*(BC)/2

Now in triangles A'BC and A"BC, the heights of the triangles (Green and red) with respect base BC are clearly less than AB.

This proves that maximum area of the triangle whose two sides are given is when the two given sides are at right angle to each other.

This can be proved using trigonometry also but that is beyond the scope of GMAT.

Blue_Skies wrote: An important assertion given in the Solution that i saw is that maximum area of the triangle whose two sides are given is when the two given sides are at right angle to each other. Can experts comment on this.

Thanks!

I posted an explanation here:

https://www.beatthegmat.com/area-when-si ... 86637.html