Find hypotenuse of Isoscelese Right angled triangle

[hiteshdalal]

I got the question from GMAT prep exam download available at http://www.mba.com site.

The question is :

The perimeter of a certain isosceles right triangle is 16 + 16 * sqrt(2). What is the length of the hypotenuse of the triangle?

Options :
(a) 8 (b) 16 (c) 4 sqrt(2) (d) 8 sqrt(2) (e) 16 sqrt(2)

As per the course material of MGMAT, for a isosceles triangle with side x, the hypotenuse shall be x*sqrt(2), for a 45-90-45 triangle.

So if the side is 16, perimeter shall be 16+16+16*sqrt(2) = 32+16 * sqrt(2). This answer is missing in the options..

Is the understanding correct ?

[jnelson0612]

hitesh,
Your understanding is correct that an isosceles right triangle has sides x, x, and x(sqrt2), thus the perimeter is the sum of those sides or 2x + x(sqrt2). The problem tells us this perimeter is actually 16 + 16(sqrt2).

i handled this problem by backsolving, or testing the answer choices. I always start with answer choice C when I do this, so I know if I need to try larger or smaller answers if C does not work out. Usually the answer choices are organized in ascending order.

We are trying to find the hypotenuse of the triangle. C says that the hypotenuse would be 4(sqrt2). In that case, the sides would be 4 each, and the total perimeter would be 8 + 8(sqrt2). This does not match the perimeter given of 16 + 16(sqrt2).

Knowing I need to start with a larger value, I'm going to move to answer choice B, 16. If the hypotenuse is 16, I know that equals x(sqrt2). I need to divide 16 by the square root of 2. I can't leave a radical in the denominator, so I need to multiply the top and bottom by the square root of 2. I get 16(sqrt2) on top, divided by 2 on the bottom. This gives me a result of 8(sqrt2) for x.

Aha! Notice that if x is 8(sqrt2), that 2x + x(sqrt2) would be 16sqrt 2 + 16! Bingo, we have a match. The answer is B.

Thus, hitesh, your only mistake was assuming that the hypotenuse itself would include a square root of two. In this case, the hypotenuse was an integer, and the two sides contained the square root of 2.

Good question--thanks for posting!

[hiteshdalal]

Many Thanks for clearing the point...

Regards,

[RonPurewal]

see here:

post36393.html#p36393