The perimeter of a certain isosceles right triangle is 16 +

[email2sharad]

For an isoscles triangle - 2 sides are of equal size and the third size is sqrt(2) * (size of sides of equal length).

Therefore if x is one of the equal length sides, the perimeter of the isosceles triangle is x + x + sqrt(2) * x.

so, 2(x) + sqrt(2) * x = 16 + sqrt(2) * 16
=> x(1 + sqrt (2)) = 16 (1 + sqrt(2))
=> x = 16.

-Sharad

[jnelson0612]

For an isoscles triangle - 2 sides are of equal size and the third size is sqrt(2) * (size of sides of equal length).

Therefore if x is one of the equal length sides, the perimeter of the isosceles triangle is x + x + sqrt(2) * x.

so, 2(x) + sqrt(2) * x = 16 + sqrt(2) * 16
=> x(1 + sqrt (2)) = 16 (1 + sqrt(2))
=> x = 16.

-Sharad

Hi Sharad,
Check your math again. You determined that a length of one of the sides, x, is 16. If x is 16 the perimeter would be 16 + 16 + 16sqrt2, or 32 + 16sqrt2. However, the problem tells me that the perimeter is 16 + 16sqrt2. Can you figure out how to fix this? :-) Let us know if you need more help.

[email2sharad]

Hi Jamie, Thanks for point out the mistake -

if x is the length of the equal sides, perimeter is -

x + x + sqrt(2)*x = 16 + 16*sqrt (2)
=> x(2 + sqrt (2)) = 16 (1 + sqrt(2))
=> x = 16 (1 + sqrt(2)) / (2 + sqrt(2))

Now, 2 + sqrt(2) can also be written as sqrt(2) * (sqrt(2) + 1).
Thus,
x = 16 (1 + sqrt(2)) / sqrt(2) * (sqrt(2) + 1)
or x = 16/sqrt(2).

We know that in isosceles triangle, length of hypotenuse = sqrt(2) * x (x being length of the 2 equal sides).

So, hypotenuse = 16/sqrt (2) * sqrt (2) = 16.

now that length of all 3 sides are known, lets see the perimeter (reverse check).

perimeter p = 16/sqrt(2) + 16/sqrt(2) + 16

=> p = sqrt (2) * 16 + 16, this is what the question stem said.

-Sharad

[RonPurewal]

make sure you've also noticed that you can solve this problem by backsolving (plugging in the answer choices):
see here
post36393.html#p36393

[KevinF866]

Sorry to bump up an old thread, but I had a question about one approach to this that I'm not quite getting. This was an alternative approach on the 9/25 Thursday with Ron on Misc. Geometry. This approach lets you set up your equation for x√2 which is what you are trying to solve for the hypotenuse.

If you have x = (16+16√2)/(2+√2), you can multiply both sides by √2 to get:

x√2 = (16√2 + 32)/(2+√2)

This simplifies to 16. How do you get 16 from this? I keep getting 32, or other numbers. I'm having a mental block with the remaining algebra here. Thanks!

[RonPurewal]

Try replacing "√2" with, say, "y".

If you can get (16y + 32)/(y + 2) to simplify to 16, then you can do this, too. It's the same thing.