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zqueez · Apr 11, 2007, 11:00 AM

What is the hypotenuse of an isosceles triangle if the perimeter is 16 + 16 radical 2?

galactus · Apr 11, 2007, 11:15 AM

The perimeter is $2a+b=16+16\sqrt{2}$

The hypoteneuse is $\sqrt{a^{2}+a^{2}}=\sqrt{2}a=b$

$2a+\sqrt{2}a=16(1+\sqrt{2})$

Solve for a.

zqueez · Apr 11, 2007, 11:28 AM

Is that the final answer or does it have to be simplified?

galactus · Apr 11, 2007, 11:30 AM

I said "solve for a".

zqueez · Apr 11, 2007, 11:40 AM

Oh ok, sorry I didn't see that sentence. Thank you for your help.

galactus · Apr 11, 2007, 11:52 AM

You're so welcome. Few that post on this site bother to even drop a line and say thanks.

galactus · Apr 11, 2007, 12:08 PM

Originally Posted by galactus

The perimeter is $2a+b=16+16\sqrt{2}$

The hypoteneuse is $\sqrt{a^{2}+a^{2}}=\sqrt{2}a=b$

$2a+\sqrt{2}a=16(1+\sqrt{2})$

Solve for a.

Factor out 'a' and divide through.

$a(2+\sqrt{2})=16(1+\sqrt{2})$

Now, it's easy. Ain't it? Just be sure to simplify it down.

zqueez · Apr 11, 2007, 12:40 PM

Got it. Thank you very much.

zqueez · Apr 11, 2007, 04:31 PM

May I ask you if the answer to this is 16(1-radical2)?

galactus · Apr 11, 2007, 04:51 PM

Poor kid, you're sure having a time with basic algebra.

$a=\frac{16(1+\sqrt{2})}{2(1+\frac{\sqrt{2}}{2})}$

Practice simplfying this. Rationalize the denominator.

It reduces to $8\sqrt{2}$

That's a side. Now, use that to find the length of the hypoteneuse. Should be easy.

There ya' go. ;)

ironsferri · Jun 15, 2010, 04:19 PM

Dude, relax - these people in the post are trying to get some help - no need to show off or have this rude attitude...

Unknown008 · Jun 16, 2010, 08:59 AM

Originally Posted by ironsferri

Dude, relax - these people in the post are trying to get some help - no need to show off or have this rude attitude...

And this thread is 3 years old :rolleyes:

Just try to have a look at the post date before posting next time. Apparently, 'reviving' threads can slow the site down.

ironsferri · Jun 16, 2010, 10:12 AM

Sorry man, just couldn't stand to write it... will look at the date first next time...
Thanks

Originally Posted by Unknown008

And this thread is 3 years old :rolleyes:

Just try to have a look at the post date before posting next time. Apparently, 'reviving' threads can slow the site down.

Unknown008 · Jun 16, 2010, 10:13 AM

It's all right, I didn't know either of that when I joined, and it's through mistakes that we learn, right? :D

I just wanted to point that fact out. Thank you! :)

ironsferri · Jun 16, 2010, 10:19 AM

You're truly right, learning by mistake :)

Thanks bud - glad to see there are still nice people around in this blog...

C-

Originally Posted by Unknown008

It's all right, I didn't know either of that when I joined, and it's through mistakes that we learn, right? :D

I just wanted to point that fact out. Thank you! :)

loveall_things · Jan 7, 2011, 04:00 PM

How do you go from the square root of a squared plus a squared to the square root of two times a. Please respond asap if at all possible.

Unknown008 · Jan 8, 2011, 11:48 PM

how do you go from the square root of a squared plus a squared to the square root of two times a. Please respond asap if at all possible.

This:

$\sqrt{a^2 + a^2}$

becomes this:

$\sqrt{2a^2}$

Which then is the same thing as:

$\sqrt2 \sqrt{a^2}$

and $\sqrt{a^2} = a$

Therefore:

$\sqrt{2a^2} = \sqrt2 a$