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hypotenus of an isoceles right triangle

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by lukaswelker » Tue Apr 08, 2014 10:05 am
Hey Guys, here is the question

The perimeter of a certain isosceles triangle is, 16+ 16√2. What is the length of the hypotenuse of the triangle?

8; 16 ; 4√2; 8√2; 16√2

In order to resolve this question I wrote down the following;
y= hypothenus
x= one of the two equal sides.

y+2x=16+16√2 --> y=8+8√2-y/2
y2(exponent)=2x2(exponent)

Then I would try the function with the given answers for y, but the calculations were long and prompt to errors. Any other way to resolve this?
Many thanks
Lukas
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by Brent@GMATPrepNow » Tue Apr 08, 2014 10:09 am
When you transcribed the question, you missed an important part. I've posted the original question below.
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
Last edited by Brent@GMATPrepNow on Thu Apr 10, 2014 3:43 pm, edited 1 time in total.
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by lukaswelker » Wed Apr 09, 2014 5:15 am
I knew there was an easier way. Thanks for showing it to me Brent.

Have a lovely day
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by GMATGuruNY » Wed Apr 09, 2014 8:27 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 proportioned: 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 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 susahntgupta402 » Wed Apr 09, 2014 3:40 pm
Mitch and Brent is the reason for assuming right isosceles triangle occurrence of 2^1/2?
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by susahntgupta402 » Wed Apr 09, 2014 3:41 pm
Mitch and Brent is the reason for assuming right isosceles triangle occurrence of 2^1/2?
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by Brent@GMATPrepNow » Wed Apr 09, 2014 4:07 pm
susahntgupta402 wrote:Mitch and Brent is the reason for assuming right isosceles triangle occurrence of 2^1/2?
The question tells us that we have a right isosceles triangle.

Cheers,
Brent
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by susahntgupta402 » Thu Apr 10, 2014 3:04 pm
I couldn't see it in the question. Did I miss something?
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by Brent@GMATPrepNow » Thu Apr 10, 2014 3:41 pm
Oops, sorry.
Looks like the person who posted the question transcribed it incorrectly.
The original question states that it's an isosceles right triangle.

I've since noted this in my response above.

Cheers,
Brent
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