https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_10&feed=atom&action=history 1964 AHSME Problems/Problem 10 - Revision history 2021-01-28T05:35:12Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_10&diff=107821&oldid=prev Talkinaway: Created page with "== Problem== Given a square side of length $s$. On a diagonal as base a triangle with three unequal sides is constructed so that its area equals that of the square..." 2019-07-23T16:26:48Z <p>Created page with &quot;== Problem== Given a square side of length &lt;math&gt;s&lt;/math&gt;. On a diagonal as base a triangle with three unequal sides is constructed so that its area equals that of the square...&quot;</p> <p><b>New page</b></p><div>== Problem==<br /> <br /> Given a square side of length &lt;math&gt;s&lt;/math&gt;. On a diagonal as base a triangle with three unequal sides is constructed so that its area<br /> equals that of the square. The length of the altitude drawn to the base is:<br /> <br /> &lt;math&gt;\textbf{(A)}\ s\sqrt{2} \qquad<br /> \textbf{(B)}\ s/\sqrt{2} \qquad<br /> \textbf{(C)}\ 2s \qquad<br /> \textbf{(D)}\ 2\sqrt{s} \qquad<br /> \textbf{(E)}\ 2/\sqrt{s}&lt;/math&gt;<br /> <br /> == Solution==<br /> <br /> The area of the square is &lt;math&gt;s^2&lt;/math&gt;. The diagonal of a square with side &lt;math&gt;s&lt;/math&gt; bisects the square into two &lt;math&gt;45-45-90&lt;/math&gt; right triangles, so the diagonal has length &lt;math&gt;s\sqrt{2}&lt;/math&gt;.<br /> <br /> The area of the triangle is &lt;math&gt;\frac{1}{2}bh&lt;/math&gt;. The base &lt;math&gt;b&lt;/math&gt; of the triangle is the diagonal of the square, which is &lt;math&gt;b = s\sqrt{2}&lt;/math&gt;. If the area of the triangle is equal to the area of the square, we have:<br /> <br /> &lt;math&gt;s^2 = \frac{1}{2}bh&lt;/math&gt;<br /> <br /> &lt;math&gt;s^2 = \frac{1}{2}s\sqrt{2}\cdot h&lt;/math&gt;<br /> <br /> &lt;math&gt;s = \frac{\sqrt{2}}{2}h&lt;/math&gt;<br /> <br /> &lt;math&gt;h = \frac{2}{\sqrt{2}}s&lt;/math&gt;<br /> <br /> &lt;math&gt;h = s\sqrt{2}&lt;/math&gt;<br /> <br /> This is option &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AHSME 40p box|year=1964|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> <br /> {{MAA Notice}}</div> Talkinaway