rsrighosh
took me ages to calculate every option. There has to be some shorter process to find out.

Perimeter of 80. Eliminate option E
try units digit since a^2+b^2=c^2
A) _4+_5=_9 can work
B)_5+_1=_6 can work
C)_6+_0=_6 can work
D)_1+_6=_5 cannot
leaves ABC, answer choice A has the largest side so highest chance of obtuse angle (smallest sum of the squares of shorter legs). both BC have "hypotenuse" of 34. between the two, test the triangle (15-31-34) with the shorter legs, it have the largest difference so your squares would be the largest. Remember, largest area with a given perimeter is when sides are equal.
15^2+31^2=34^2?
225+961=1156? NO it is >, so with this anchor, you choose the other option (16-30-34) C
Or if you caught that the 16-30-34 triangle has a common factor, it can be reduced to an 8-15-17 triangle (it is a right triangle!) or have simpler math if you need to check it out.