Problem Solving

Hello GMAT Beaters. First, thanks in advance for the response. Second, on the GMAT software I came upon a geometry question that goes as follows:

The perimeter of a certain isosceles triangle is 16 + 16(square root of 2). What is the length of the hypotenuse of the triangle?

I can't seem to figure out how they came up with the answer. Please help. Thanks.

The perimeter of a right angled isosceles triangle is: 2a + b where 2a are the 2 equal sides and b is the unequal side...

we have 16 + 16 (sq root 2).

Now I used the POE:

if 2a=16 a=8...then b = 16(sq root 2) doesnt make sense

if 2a= 16(sq root 2) a=8(sq root 2)...then b=hyp=16 makes sense.
you can get it from the 1:1:sq root 2 property of an isosceles right triangle or u can use Pythagoras theorem to solve to check.

Not sure if this is the standard approach.

crossingfingers wrote:The perimeter of a right angled isosceles triangle is: 2a + b where 2a are the 2 equal sides and b is the unequal side...

we have 16 + 16 (sq root 2).

Now I used the POE:

if 2a=16 a=8...then b = 16(sq root 2) doesnt make sense

if 2a= 16(sq root 2) a=8(sq root 2)...then b=hyp=16 makes sense.
you can get it from the 1:1:sq root 2 property of an isosceles right triangle or u can use Pythagoras theorem to solve to check.

Not sure if this is the standard approach.

Thanks crossing fingers but first what is POE? Second, why did you think that the first choice didn't make sense? Is it cause you tried the properties of isoscelese and it didn't work?

Also does anyone know how to do this algebraically cause I'm sure this will not be the exact question on the test and I want to understand the concept that I'm missing. Thanks.

oops...sorry...POE - process of elimination (am pretty much doing an either or)

yes...for the first choice applying Pythagoras theorem a^2 + a^2 = b^2...gives us sq root of (64 + 64) = 8(sq root of 2) which is not what's expected - 16(sq root of 2). Hence proceed to 2

what is the OA?

Does anyone know how to do this algebraically cause I'm sure this will not be the exact question on the test and I want to understand the concept that I'm missing. Thanks.

crossingfingers wrote:The perimeter of a right angled isosceles triangle is: 2a + b where 2a are the 2 equal sides and b is the unequal side...

we have 16 + 16 (sq root 2).

Now I used the POE:

if 2a=16 a=8...then b = 16(sq root 2) doesnt make sense

if 2a= 16(sq root 2) a=8(sq root 2)...then b=hyp=16 makes sense.
you can get it from the 1:1:sq root 2 property of an isosceles right triangle or u can use Pythagoras theorem to solve to check.

Not sure if this is the standard approach.

But where it is mentioned that the triangle is right angle triangle

farooq wrote:

crossingfingers wrote:The perimeter of a right angled isosceles triangle is: 2a + b where 2a are the 2 equal sides and b is the unequal side...

we have 16 + 16 (sq root 2).

Now I used the POE:

if 2a=16 a=8...then b = 16(sq root 2) doesnt make sense

if 2a= 16(sq root 2) a=8(sq root 2)...then b=hyp=16 makes sense.
you can get it from the 1:1:sq root 2 property of an isosceles right triangle or u can use Pythagoras theorem to solve to check.

Not sure if this is the standard approach.

But where it is mentioned that the triangle is right angle triangle

Sorry about that guys. I neglected to write that it was a right triangle. Crossingfingers was correct both in the answer and in that it was a right triangle.

What was the answer? You can put it in a spoiler