Hi Mike! Digital SAT #3… Can you show a simple way to manage Question #25 on Math Module 2? Thanks!

Sure, and sorry it took me a few days to get to this!

The question says you have an isosceles right triangle (AKA a 45-45-90 triangle) with a perimeter of 94+94\sqrt{2}. We know that any 45-45-90 triangle with legs of x has a hypotenuse of x\sqrt{2}. The question asks us for the length of one leg of the triangle so we want to solve for x in this equation:

x+x+x\sqrt 2=94+94 \sqrt{2}\\2x+x\sqrt{2}=94+94 \sqrt{2}

Hmm…at first maybe not so obvious!

But wait, this is a multiple choice question! Let’s see if the choices give us any clues. Look at choices B and D. They have \sqrt{2} in them. If there’s a \sqrt{2} in the length of a leg already, then we’re suddenly in better shape.

So let’s see if choice B works. If x=47\sqrt{2}, then the perimeter of x+x+\sqrt{2} would be:

47\sqrt{2}+47\sqrt{2}+47\sqrt{2}\sqrt{2}\\=94\sqrt{2}+47(2)\\=94\sqrt{2}+94

So B works, and backsolving got us there.

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