GMATPREP question #2

Consider two of the sides to be x and the hypothenuse y.

p=16+16sqrt2 = 2x+y
y^2=2x^2, since it is a right iscoscele triangle.
So, x=y/sqrt2
16+16sqrt2=(2y/sqrt2)+y

Solve and u will get y=16

So, it is B.

This may be a simple question - but how do you simplify that down to 16?

I am having some trouble with this. Thanks in advance!

we know one of the ratios of the right triangle is (1:1: root 2); when angle are 45, 45, 90

Here, two values are given 16, 16 root 2.

So, the third value must be 16

I think this is the easiest method

I am not sure that last method works... wouldn't that mean that one side is the same length as c? (a^2+b^2=c^2)

I may be wrong, but I think c must always be the longest?

I know I should know this one.. ack.

The hypotenuse of the triange cannot be 16. If it was the other side would be 16/ sq root of 2 and the perimeter would be 16 + 32/ sq root of 2

can you please confirm that the perimeter is indeed 16+16 sq root of 2 ? and not 16 + 8 sq root of 2?

thanks

I took this test this morning and it is indeed 16 + 16 sqrt. 2

Still pretty stumped by this - any help anyone?

I believe the qs itself is having some problem! It cannot be 16! But I am not being able to find out any option specified here!

Hi
Just started solving all these prbs and came to this one.

If we take x =8sqrt2 then dont we get the perimeter as 16+16sqrt2?
Am not too good with this so may be wrong but 8 sqrt2 +8sqrt2 + 8 sqrt2 *rt2= 16sqrt2 + 16(as rt2 *rt2 =2 Thus 8 * 2=16)
Hope i am correct

Thanks
Preeti

Ok. think of it this way.

16√ 2 = 8√ 2 + 8√ 2

these two are the two equal sides of the isocelse triangle.

and 16 is the hypotenuse.

you can verify this by using the a^2 + b^2 = c^2 formuale

where (8√ 2)^2 +(8√ 2)^2 = 256
which is also 16^2 ( square of 16 = 256)

think strategically, dont assume that the sides of the triangle have to be 8 and 8. and add up to 16.. thats the joe bloggs generic response.

CRACK THE GMAT NOW

To reiterate what a few have said.

We know that a right isosceles triangle has sides in the ratio of:

x : x : xroot2

So, if we want to try to write the 3 sides in that ratio.

Here, we know that a + a + hypotenuse = 16 + 16root2

and that the hypotenuse is greater than the length of either side.

Using the "numbers have to work out nicely" principle for the GMAT, let's rewrite our equation as:

2a + h = 16root2 + 16

And assume that a = 8root2 and h = 16.

Now to double check:

(8root2)^2 + (8root2)^2 = 16^2

62*2 + 64*2 = 256
128 + 128 = 256

It works, therefore hyp = 16.

Lets name the triangle as ABC.

Its an Isoceles triangle with an hypotenuse so AB^2+Bc^2=AC^2(Pythogoras)

or 2AB^2=AC^2

AC is the hypotenuse so when AC=16 then 2AB^2=256 or AB=8 sq.root 2

Now the perimeter would be 8rt2+8rt2+16=16rt2+16