Your quickest method to solve this algebra problem?

P = perimeter = b+h+t
b = base
h = height
t = hypoteneus

Area =(bh)/2 = 28 -> bh = 56 -> 2bh = 112
Pytagorean theorem -> b² + h² = 12² = 144

b² + 2bh + h² = 112+144
b² + 2bh + h² = 256
(b+h)² = 256
| b+h | = 16
b+h+t = 16 + 12 = 28 = p

Thank you MAAJ for your quick reply to the question. This is the same answer BTG uses in their video with the BTW Practice Questions.

The part that confuses me is how/why do you know to turn the area bh = 56 to 2bh = 112?
What am I missing that would quickly make us think "why dont we double the area"? (if we see this type question for the first time on GMAT)

Hopefully this is something easy and quick that I have missed which will help many people that read this.

It's something that comes with practice. At first I wasn't able deduct this step. But if you get used to all the quadratics formulas (in both ways, factored and unfactored), you'll see this patterns more easily.

Special products #1-> x²-y² = (x-y)(x+y)
Special product #2 -> x²+2xy+y² = (x+y)² = (x+y)(x+y)
Special product #3 -> x²-2xy+y² = (x-y)² = (x-y)(x-y)

The special thing about a right triangle is that it's base and height are both legs of the triangle.

So 1/2(b)(h) = 28 and b^2 + h^2 = 12

So b^2 + H^2 = 144 and b(h) = 56

From there I could solve this algebraically but it would be more helpful to have the answer choices available as you would on the actual GMAT - then you could use this info to apply different answer choices - subtract 12 from each of the answers and then see which one fits into the above categories.

You can use comon quadratics but you don't have to.

Best of Luck

For other readers, remember that "h" in this question represents the "height" and not the "hypotenuse". The equation "b² + h² = 12² = 144" can be misleading - where h represents "a" in the PT theorem b^2 + a^2 = c^2.

Cheers,