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Triangle

Expert replies
by eitijan » Thu May 12, 2016 2:03 am
It took me a lot of time in calculations for this question. Is there a shorter way of doing it?
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Source: — Problem Solving |

by GMATGuruNY » Thu May 12, 2016 2:15 am
The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse?
a. 8
b. 16
c. 4√2
d. 8√2
e. 16√2
The sides of an isosceles right triangle are in the following ratio: s : s : s√2.
Thus:
If s=side and h=hypotenuse, then h = s√2 and s = h/√2.

We can PLUG IN THE ANSWERS, which represent the length of the hypotenuse.
Because h = s√2, the average test-taker will assume that the correct answer must include √2.
Don't be an average test-taker.
Start with an answer choice that does NOT include √2.

Answer choice B: h = 16
In this case:
s = 16/√2 = (16*√2)/(√2*√2) = (16√2)/2 = 8√2.
Thus:
p = 8√2 + 8√2 + 16 = 16 + 16√2.
Success!

The correct answer is B.
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by Brent@GMATPrepNow » Thu May 12, 2016 9:37 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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Thu May 12, 2016 9:55 am
Hi eitijan,

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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