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galactus · Dec 20, 2010, 02:49 PM

Here is a fun little geometry problem. Give it a go if you so desire.

Look at the graph. The radius of the larger circle is 2r and the radius of the smaller circle is r. Find the area of the triangle.

There are many ways to go about it. I used simialr triangles, but perhaps you have a better way.

I just left r=1 in order to graph it. But consider r to be some radius r, not 2 as I have in the graph. The area of the triangle should be expressed in terms of r alone.

ebaines · Dec 20, 2010, 03:21 PM

I get:

$<br /> Area = 8 \sqrt{8} r^2<br />$

Similar triangles seems to be the best way.

galactus · Dec 20, 2010, 03:33 PM

Yep. That's it.

Except, I expressed it as $16\sqrt{2}r^{2}$ .

Same thing.

Cool, ebaines.

This reminds me of the infinite circles inside the triangle, except it stops at 2 circles. :)

harum · Dec 20, 2010, 04:46 PM

Because FE/GD = 2, CG = FG. I.e. CH = 8r. BH = CH * tan(BCH) = CH * (2/sqrt(32)) = CH/sqrt(8) = r*sqrt(8). S = CH*BH = 16*sqrt(8)*r^2.

harum · Dec 20, 2010, 09:08 PM

Is there a way to edit a typo? 16*sqrt(2)*r^2

galactus · Dec 21, 2010, 02:03 AM

$16sqrt{2}r^{2}$ .

As long as no one else has posted afterwards, I have had no trouble editing a post.

You meant $16\sqrt{2}r^{2}$

harum · Dec 21, 2010, 02:02 PM

Thanks! 16\sqrt{2}r^{2}

galactus · Dec 21, 2010, 02:21 PM

You have to place the tags around your LaTex code.

Math in brackets at the front and /math in brackets when closing

Unknown008 · Dec 22, 2010, 12:39 PM

And it may sound weird... but the site has got several 'skins' On the one you are on, harum, you can't edit your own post. It's the 'newer skin' and I personally don't like it. You might try the skin made before that, by looking for your settings and look for something about 'Take me back to the old AMHD' or something along those lines. It's been a while I haven't been there myself.