perimeter of a certain isosceles

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by Brent@GMATPrepNow » Mon Sep 07, 2015 7:39 am
The Perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?

A) 8
B) 16
C) 4√2
D) 8√2
E) 16√2
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

Answer = B

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Brent
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by Brent@GMATPrepNow » Mon Sep 07, 2015 7:39 am
oops - double post
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by [email protected] » Mon Sep 07, 2015 8:41 am
Hi anant03,

Here's a larger discussion on this question, including other ways to solve it and some noteworthy aspects about how the prompt is written:

https://www.beatthegmat.com/hypotenuse-o ... 76892.html

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by Matt@VeritasPrep » Tue Sep 08, 2015 2:42 am
Here's a very fast (<30 seconds) approach for test day:

x + x + x√2 = 16 + 16√2

or

2x + x√2 = 16 + 16√2

Clearly x can't equal 8, or the sides would have different values. So we must have x√2 = 16 and 2x = 16√2. This gives us x = 8√2. From there, x√2 = (8√2)√2, or 16, so the hypotenuse is 16.