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Hypotenuse of Triangle?

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Hypotenuse of Triangle?

by bml1105 » Thu Jun 05, 2014 1:40 pm
The perimeter of a certain isosceles right triangle is 16 + 16(sqrroot 2). What is the length of the hypotenuse of the triangle?

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

OA: B
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by Brent@GMATPrepNow » Thu Jun 05, 2014 1:43 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 GMATGuruNY » Thu Jun 05, 2014 1:54 pm
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.
So 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.

Answer choice C: h = 4√2
s = (4√2)/√2 = 4.
p = 4 + 4 + 4√2 = 8 + 4√2.
Eliminate C. The perimeter needs to be quite a bit larger.

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

The correct answer is B.
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by [email protected] » Thu Jun 05, 2014 3:48 pm
Hi bml1105,

Mitch's approach to TEST THE ANSWERS is perfect for this scenario. There's a Number Property about Isosceles triangles that will help you to avoid most of the wrong answers...

The ratio of sides in an Isosceles Triangle is: x, x, and x√2

If, for example, one of the "short sides" was 5, then we'd have 5, 5, 5√2 and the perimeter would be 10 + 5√2. Notice that the 10 and 5 are different numbers....That will ALWAYS happen IF the short side is an integer and the hypotenuse has a √2 attached to it....

We're given a perimeter of 16 + 16√2; here, both numbers are 16, so we are NOT dealing with the above scenario. The "short sides" CANNOT be integers and the hypotenuse CANNOT have a √2 attached to it. Eliminate C, D and E.

From there, we can TEST A and B to find the match.

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by bml1105 » Thu Jun 05, 2014 5:59 pm
Brent@GMATPrepNow wrote:
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
Hi Brent,

I'm a little confused. How can we factor out x√2 from 2x + x√2 and get x√2(√2 + 1) when the 2x doesn't have an x√2. If anything wouldn't it be 2x/x√2 (√2/√2) = (2√2)/2?
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by bml1105 » Thu Jun 05, 2014 6:01 pm
Thanks Rich, I was definitely confused because I was trying to use x:x:x√2 and I couldn't understand how it could work so I panicked!
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by Brent@GMATPrepNow » Thu Jun 05, 2014 6:21 pm
bml1105 wrote:
Hi Brent,

I'm a little confused. How can we factor out x√2 from 2x + x√2 and get x√2(√2 + 1) when the 2x doesn't have an x√2. If anything wouldn't it be 2x/x√2 (√2/√2) = (2√2)/2?
Good question.
Notice that 2 = (√2)(√2)
So, let's start at the point where we have: 2x + x√2 = 16 + 16√2
We can rewrite this as (√2)(√2)x + x√2 = 16 + 16√2
So, on the left side, both terms have an x and a √2
So we can factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2

I hope that helps.

Cheers,
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by garmahis » Tue Jun 10, 2014 6:10 am
I'm so disappointed of myself. I tried to answer the problem but couldn't get the answer from the choices. Hmmmmn. Maybe I should start reviewing on how to get the hypotenuse of a triangle. Should have been an easy question since it's been taught all over from high school to college. Thanks for the explanations you've provided here. I am enlightened as well. Thanks for posting the question too. It was mind-boggling and tickled the brain cells in me. Best of luck to all test takers. Especially on the Math section. This section proved to be the most challenging for me.

-Garmahis
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by Brent@GMATPrepNow » Tue Jun 10, 2014 7:15 am
Don't be too hard on yourself, Garmahis. This is a tricky question. In fact, I'd put it in the 700 range (difficulty-wise).

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