https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sanchitb&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-19T07:09:18Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=37211 2010 AMC 10B Problems/Problem 16 2011-03-01T08:12:32Z <p>Sanchitb: </p> <hr /> <div>Radius of circle = &lt;math&gt;\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} = 0.577&lt;/math&gt;<br /> <br /> Half the diagonal of the square = &lt;math&gt;\sqrt{.5^2 + .5^2} = \frac{\sqrt{2}}{2} = 0.707&lt;/math&gt;<br /> <br /> Therefore the picture will look something like this:<br /> <br /> [[Image:squarecircle.png]]<br /> <br /> Then you proceed to find: 4 * (area of the sector - area of the triangle) to get answer (B)</div> Sanchitb https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=37210 2010 AMC 10B Problems/Problem 16 2011-03-01T08:12:13Z <p>Sanchitb: </p> <hr /> <div>Radius of circle = &lt;math&gt;\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} = 0.577&lt;/math&gt;<br /> <br /> Half the diagonal of the square = &lt;math&gt;\sqrt{.5^2 + .5^2} = \frac{\sqrt{2}}{2} = 0.707&lt;/math&gt;<br /> <br /> Therefore the picture will look something like this:<br /> <br /> [[Image:squarecircle.png]]<br /> <br /> Then you proceed to find the 4 * (area of the sector - are of the triangle) to get answer (B)</div> Sanchitb https://artofproblemsolving.com/wiki/index.php?title=File:Squarecircle.png&diff=37209 File:Squarecircle.png 2011-03-01T08:06:20Z <p>Sanchitb: A Square and a Circle where the radius of the circle is more than half the length of a side but diameter is less than the diagonal of the square.</p> <hr /> <div>A Square and a Circle where the radius of the circle is more than half the length of a side but diameter is less than the diagonal of the square.</div> Sanchitb https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=37208 2010 AMC 10B Problems/Problem 16 2011-03-01T06:56:02Z <p>Sanchitb: Undo revision 37207 by Sanchitb (Talk)</p> <hr /> <div></div> Sanchitb https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=37207 2010 AMC 10B Problems/Problem 16 2011-03-01T06:49:39Z <p>Sanchitb: </p> <hr /> <div>Area of Circle = &lt;math&gt;\pi r^2 = \frac{\pi}{3}&lt;/math&gt;<br /> <br /> Area of Square = &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> Therefore answer = (A) &lt;math&gt;\frac{\pi}{3} - 1&lt;/math&gt;</div> Sanchitb https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=37206 2010 AMC 10B Problems/Problem 16 2011-03-01T06:47:05Z <p>Sanchitb: </p> <hr /> <div>Area of Circle = &lt;math&gt;&amp;pi r^2 = &amp;pi/3&lt;/math&gt;<br /> <br /> Area of Square = &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> Therefore answer = (A) &lt;math&gt;&amp;pi/3 - 1&lt;/math&gt;</div> Sanchitb