Difference between revisions of "1951 AHSME Problems/Problem 48"

Revision as of 11:27, 5 July 2013

Problem

The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

$\textbf{(A)}\ 1: 2\qquad\textbf{(B)}\ 2: 3\qquad\textbf{(C)}\ 2: 5\qquad\textbf{(D)}\ 3: 4\qquad\textbf{(E)}\ 3: 5$

Solution

Let the radius of the circle be $r$ . Let $s_1$ be the side length of the square inscribed in the semicircle and $s_2$ be the side length of the square inscribed in the entire circle. For the square in the semicircle, we have $s_1^2 + (\frac{s_1}{2})^2 = r^2 \Rightarrow s_1^2 = \frac{4}{5}r^2$ . For the square in the circle, we have $s_2 \sqrt{2} = 2r \Rightarrow s_2^2 = 2r^2$ .

Therefore, $s_1^2 : s_2^2 = \frac{4}{5}r^2 : 2r^2 = \boxed{\textbf{(C)}\ 2: 5}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 47	Followed by Problem 48
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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Difference between revisions of "1951 AHSME Problems/Problem 48"

Revision as of 11:27, 5 July 2013

Problem

Solution

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