Relation between Perimeter and hypotenuse of triangle

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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)

Answer is B

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)

I stuck here :)

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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

silly me :D

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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+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
Since it is an isoceles right angled triangle its a 45-45-90 triangle with the sides in the ratio 1:1:√2

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Perimeter = x + x + x√2 = 2x + x√2
But the perimeter is 16+16√2

Equating:
2x + x√2 = 16+16√2
x√2 (√2 + 1) = 16 (1 + √2)
x√2 = 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 :D

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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 + 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

Answer is B
An important point here 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 left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from 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 [email protected] » Wed Nov 06, 2013 6:52 am
amirhakimi wrote: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

Answer is B
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 x√2 for some positive value of x. The hypotenuse is x√2

So, if x is the length of one leg of the right triangle, the perimeter = x + x + x√2

Start with answer choice D.
If the hypotenuse is 8√2, then x = 8 [since the hypotenuse = x√2]
So, the perimeter = 8 + 8 + 8√2 = 16 + 8√2
NO GOOD! We're told that the perimeter is 16 + 16√2, so we need the triangle to be LARGER THAN 8√2.
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.

Let's try answer choice B.
If the hypotenuse is 16, then we know that x√2 = 16
So, x = 16/√2 = 8√2
So, in this instance, the perimeter is 8√2 + 8√2 + 16 = 16√2 + 16
PERFECT! We're told that the perimeter is 16 + 16√2

Answer: B

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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 :)!!