## Relation between Perimeter and hypotenuse of triangle

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### Relation between Perimeter and hypotenuse of triangle

by amirhakimi » Wed Nov 06, 2013 5:21 am
The Perimeter of a certain isosceles right triangle is 16+16*sqrt(2). What is the length of the hypotenuse of the triangle?

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

Is this question categorized as a tough one?

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by amirhakimi » Wed Nov 06, 2013 5:30 am
Here is what I've done:
Since its a right isosceles triangle, the perimeter is 2x+x*sqrt(2)
2x+x*sqrt(2)=16+16*sqrt(2)

x(2+sqrt(2))=16(1+sqrt(2)) -----> X=[16(1+sqrt(2))]/[2+sqrt(2)]
using FOIL to simplify denominator ---> X= [(16*sqrt(2))]/2 ---> X=8*sqrt(2)

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by amirhakimi » Wed Nov 06, 2013 5:36 am
Ohh, X is one of the sides not hypotenuse, hypotenuse will be X*sqrt(2) and will be 8*sqrt(2) * sqrt(2) which is 16

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by mevicks » Wed Nov 06, 2013 6:27 am
amirhakimi wrote:The Perimeter of a certain isosceles right triangle is 16+16âˆš2. What is the length of the hypotenuse of the triangle?
A)8
B)16
C)4âˆš2
D)8âˆš2
E)16âˆš2
Since it is an isoceles right angled triangle its a 45-45-90 triangle with the sides in the ratio 1:1:âˆš2 Perimeter = x + x + xâˆš2 = 2x + xâˆš2
But the perimeter is 16+16âˆš2

Equating:
2x + xâˆš2 = 16+16âˆš2
xâˆš2 (âˆš2 + 1) = 16 (1 + âˆš2)
xâˆš2 = 16

From the figure this is the hyp. thus the answer is B

P.S: Wont call it a tough problem; the apt word would be tricky, especially in time crunch situations ### GMAT/MBA Expert

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by [email protected] » Wed Nov 06, 2013 6:42 am
amirhakimi wrote:The Perimeter of a certain isosceles right triangle is 16 + 16âˆš2. What is the length of the hypotenuse of the triangle?

A)8
B)16
C)4âˆš2
D)8âˆš2
E)16âˆš2

An important point here is that, in any isosceles right triangle, the sides have length x, x, and xâˆš2 for some positive value of x.

Note: xâˆš2 is the length of the hypotenuse, so our goal is to find the value of xâˆš2

From here, we can see that the perimeter will be x + x + xâˆš2

In the question, the perimeter is 16 + 16âˆš2, so we can create the following equation:
x + x + xâˆš2 = 16 + 16âˆš2,
Simplify: 2x + xâˆš2 = 16 + 16âˆš2
IMPORTANT: Factor xâˆš2 from left side to get : xâˆš2(âˆš2 + 1) = 16 + 16âˆš2
Now factor 16 from right side to get: xâˆš2(âˆš2 + 1) = 16(1 + âˆš2)
Divide both sides by (1 + âˆš2) to get: xâˆš2 = 16

Cheers,
Brent

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by [email protected] » Wed Nov 06, 2013 6:52 am
amirhakimi wrote:The Perimeter of a certain isosceles right triangle is 16 + 16âˆš2. What is the length of the hypotenuse of the triangle?

A)8
B)16
C)4âˆš2
D)8âˆš2
E)16âˆš2

Another option is to test the answer choices.

We'll use the fact that, for any isosceles right triangle, the sides have length x, x, and xâˆš2 for some positive value of x. The hypotenuse is xâˆš2

So, if x is the length of one leg of the right triangle, the perimeter = x + x + xâˆš2

If the hypotenuse is 8âˆš2, then x = 8 [since the hypotenuse = xâˆš2]
So, the perimeter = 8 + 8 + 8âˆš2 = 16 + 8âˆš2
NO GOOD! We're told that the perimeter is 16 + 16âˆš2, so we need the triangle to be LARGER THAN 8âˆš2.
This means we can ELIMINATE A, C and D

IMPORTANT: At this point, we can test either B or E. It doesn't matter which one we test. If B works, we're done. If B doesn't work, then we'll automatically choose E.

If the hypotenuse is 16, then we know that xâˆš2 = 16
So, x = 16/âˆš2 = 8âˆš2
So, in this instance, the perimeter is 8âˆš2 + 8âˆš2 + 16 = 16âˆš2 + 16
PERFECT! We're told that the perimeter is 16 + 16âˆš2

Cheers,
Brent

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by gmatclubmember » Thu Nov 07, 2013 8:13 am
amirhakimi wrote:Here is what I've done:
Since its a right isosceles triangle, the perimeter is 2x+x*sqrt(2)
2x+x*sqrt(2)=16+16*sqrt(2)

x(2+sqrt(2))=16(1+sqrt(2)) -----> X=[16(1+sqrt(2))]/[2+sqrt(2)]
using FOIL to simplify denominator ---> X= [(16*sqrt(2))]/2 ---> X=8*sqrt(2)

I stuck here You were on track until 2x+x*sqrt(2)=16+16*sqrt(2).
After this you could have tried this way:
x(2+sqrt(2)) = 8sqrt(2) (sqrt(2)+2) => x=8sqrt(2)
a lil' Thank note goes a long way !!

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