Bunuel wrote:

Official Solution:

(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), from which it follows that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral.

Let's find what this area would be: \(Area=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even equilateral triangle with perimeter of 30 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.

Answer: D

I got my answer. Sorry to bother you. Now I know the reason why we used the equilateral triangle.

But anyways, I tried using both equilateral (the triangle is supposed to have the smallest perimeter for a given area) as well as isosceles triangle. See below:

1/2 ab= 50

or ab=100

a and b can be any of the following combinations or many more

1X 100 Now the third side "c" can be anything between 99-101. Hence, the smallest perimeter solution would be 1+100+100

2X 50 48-52. Hence, the smallest perimeter solution would be 2+49+50

4X 25 21-29. Hence, the smallest perimeter solution would be 4+22+25

5X 20 15-25. Hence, the smallest perimeter solution would be 5+16+20=41

Thus, in every case, the answer would be more than 30.

Similarly, in equilateral triangle with area 50, each side will come out to be 10.84