## The perimeter of a certain isosceles right triangle is $$16+16\sqrt2.$$ What is the length of the hypotenuse of the tria

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### The perimeter of a certain isosceles right triangle is $$16+16\sqrt2.$$ What is the length of the hypotenuse of the tria

by Gmat_mission » Sun Apr 11, 2021 2:53 am

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The perimeter of a certain isosceles right triangle is $$16+16\sqrt2.$$ What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt2$$

(D) $$8\sqrt2$$

(E) $$16\sqrt2$$

Source: GMAT Prep

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### Re: The perimeter of a certain isosceles right triangle is $$16+16\sqrt2.$$ What is the length of the hypotenuse of the

by [email protected] » Tue Apr 13, 2021 7:22 am
Gmat_mission wrote:
Sun Apr 11, 2021 2:53 am
The perimeter of a certain isosceles right triangle is $$16+16\sqrt2.$$ What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt2$$

(D) $$8\sqrt2$$

(E) $$16\sqrt2$$

Source: GMAT Prep
An IMPORTANT point to remember is that, in any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x.

Note: x√2 is the length of the hypotenuse, so our goal is to find the value of x√2

From here, we can see that the perimeter will be x + x + x√2

In the question, the perimeter is 16 + 16√2, so we can create the following equation:
x + x + x√2 = 16 + 16√2,
Simplify: 2x + x√2 = 16 + 16√2
IMPORTANT: Factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from the right side to get: x√2(√2 + 1) = 16(1 + √2)
Divide both sides by (1 + √2) to get: x√2 = 16