Question on hypotenuse

[poonamchiK]

Perimetre of a isoceles triangle is 16+16 (square root of 2). WHAT IS THE LENGTH OF THE HYPOTENUSE OF THE TRIANGLE?
answer choices
8
16
4(square root 2)
8(square root of 2)
16 (square root of 2).

thx
poonam

[RB_51273]

If there is question on hypotenuse then it must be right angled triangle(pls check again).If it is RA isosceles triangle then IMO:D - 8(square root of 2)

Equate
a(2+sqrt2)=16+16sqrt(2)

a=8*sqrt(2)

thanks

ram

[poonamchiK]

thx Ram,
correct Answer is 16

I also got the same answer as you! bt its wrong!

:-(
If this question comes in GMAT i am sure to get it wrong.
this is from the test prep software btw.

Poonam

[RB_51273]

If 16 is the answer,then it looks like based on the logic only.
If we divide 16 in two parts ,Then triangle will not be formed as sum of the two side will be less than third.Hence only 16sqrt(2) will have to be divided in two parts two represent the two sides of isosceles triangle,and 16 will represent the third side.
But it is based on purely on logic,I am not too convinced with this logic.May be Gmat work like this.
wait for MGMAT staff to respond

Ron/Stacey solution please

Thanks

Ram

[nelloriumrao]

correct me if i am wrong

We need S.sqrt2 = ? instead

2S+S.sqrt2=16+16.sqrt2
S.sqrt2(sqrt2+1)=16(1+sqrt2)
S.sqrt2=16

that what we need

[rchitta]

The answer is 16.

What the problem meant is that the triangle is a right isosceles triangle.

So, if the 2 sides are 'x' and 'x' the hypotenuse would be x * sqrt(2) [use the pythagoras theorem to get that]

perimeter of the right isosceles triangle would be 2x + sqrt(2) * x = 16 + 16 * sqrt(2) [given]

so x would be 16 * ( 1 + sqrt(2)) / (2 + sqrt(2))

and hypotenuse would be sqrt(2) * x = sqrt(2) * 16 * ( 1 + sqrt(2)) / (2 + sqrt(2))

=> hypotenuse = 16

[RonPurewal]

this is a gmat prep problem.

please post GMATPREP PROBLEMS in the DEDICATED GMATPREP FOLDER. do not post those problems in the general folder.

also, this problem has been answered (several times, actually) in the gmatprep folder. here's one of the better threads:
isosceles-right-triangle-t2694.html