https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_10&feed=atom&action=history1964 AHSME Problems/Problem 10 - Revision history2024-03-28T21:13:46ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_10&diff=107821&oldid=prevTalkinaway: Created page with "== Problem== Given a square side of length <math>s</math>. 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 "== Problem== Given a square side of length <math>s</math>. On a diagonal as base a triangle with three unequal sides is constructed so that its area equals that of the square..."</p>
<p><b>New page</b></p><div>== Problem==<br />
<br />
Given a square side of length <math>s</math>. 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 />
<math>\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}</math><br />
<br />
== Solution==<br />
<br />
The area of the square is <math>s^2</math>. The diagonal of a square with side <math>s</math> bisects the square into two <math>45-45-90</math> right triangles, so the diagonal has length <math>s\sqrt{2}</math>.<br />
<br />
The area of the triangle is <math>\frac{1}{2}bh</math>. The base <math>b</math> of the triangle is the diagonal of the square, which is <math>b = s\sqrt{2}</math>. If the area of the triangle is equal to the area of the square, we have:<br />
<br />
<math>s^2 = \frac{1}{2}bh</math><br />
<br />
<math>s^2 = \frac{1}{2}s\sqrt{2}\cdot h</math><br />
<br />
<math>s = \frac{\sqrt{2}}{2}h</math><br />
<br />
<math>h = \frac{2}{\sqrt{2}}s</math><br />
<br />
<math>h = s\sqrt{2}</math><br />
<br />
This is option <math>\boxed{\textbf{(A)}}</math><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