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	<title>Comments on: An isosceles triangle inscribed in a circle. How should this be accomplished if triangle&#8217;s area is maximized?!?</title>
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		<title>By: goober</title>
		<link>http://www.thearealist.com/triangle-area/an-isosceles-triangle-inscribed-in-a-circle-how-should-this-be-accomplished-if-triangles-area-is-maximized/comment-page-1#comment-1846</link>
		<dc:creator>goober</dc:creator>
		<pubDate>Mon, 15 Feb 2010 02:57:59 +0000</pubDate>
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		<description>Draw a triangle in a circle with a height greater than the radius.  The height is r+x.
The 1/2 of the base is sqrt(r^2 - x^2).
The area is A = (1/2) b h =  (r+x)*sqrt(r^2 - x^2)

The max area occurs when dA/dx =0

Compute the derivative and solve for the value of x as a fraction of r or simply let r=1.&lt;br&gt;&lt;b&gt;References : &lt;/b&gt;&lt;br&gt;</description>
		<content:encoded><![CDATA[<p>Draw a triangle in a circle with a height greater than the radius.  The height is r+x.<br />
The 1/2 of the base is sqrt(r^2 &#8211; x^2).<br />
The area is A = (1/2) b h =  (r+x)*sqrt(r^2 &#8211; x^2)</p>
<p>The max area occurs when dA/dx =0</p>
<p>Compute the derivative and solve for the value of x as a fraction of r or simply let r=1.<br /><b>References : </b></p>
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