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

Latest revision as of 21:47, 10 November 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 49
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

@@ Line 12: / Line 12: @@
 == See Also ==
-{{AHSME 50p box|year=1951|num-b=47|num-a=48}}
+{{AHSME 50p box|year=1951|num-b=47|num-a=49}}
 [[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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

Latest revision as of 21:47, 10 November 2013

Problem

Solution

See Also