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Perimeter of a Isosceles Triangle

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by Brent@GMATPrepNow » Sun Nov 29, 2015 4:21 pm
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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by DavidG@VeritasPrep » Sun Nov 29, 2015 4:35 pm
vrn2vw wrote:The perimeter of a certain isosceles triangle is 16 + 16(sqrt(2)). What is the length of the hypotenuse?

A)8
B)16
C)4(sqrt(2))
D)8(sqrt(2))
E)16(sqrt(2))

OA is B
Or trying back-solving. If we start with answer choice D, the hypotenuse would be 8√2, and the sides of the triangle would be 8, giving us a Perimeter of 8 + 8 + 8√2 = 16 + 8√2. Well, the actual perimeter is 16 + 16√2, so that's too small, and the hypotenuse must be larger than 8√2. The only answer choices that offer a larger value are B and E.

So try E. Now the hypotenuse would be 16√2, and the sides of the triangle would be 16, giving us a Perimeter of 16 + 16 + 16√2 = 32 + 16√2. Well, the actual perimeter is 16 + 16√2, so that's too big.

The only option remaining is B.
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by [email protected] » Sun Nov 29, 2015 11:00 pm
Hi vrn2vw,

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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