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1959 AHSME Problems/Problem 31

Problem

A square, with an area of $40$, is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius, is: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 120\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 200$

Solution

[asy] import geometry; point O=(0,0); real r=5*sqrt(2); point A=(-r/sqrt(5),2r/sqrt(5)); point B=(-r/sqrt(5),0); point C=(5,5); point D=-C; markscalefactor=0.1; dot(O); label("O",O,S); // Circle with diameter draw(circle(O,r)); draw((-r,0)--(r,0)); // Small square draw((-r/sqrt(5),2r/sqrt(5))--(-r/sqrt(5),0)--(r/sqrt(5),0)--(r/sqrt(5),2r/sqrt(5))--(-r/sqrt(5),2r/sqrt(5))); dot(A); label("A",A,NW); dot(B); label("B",B,SW); draw(O--A); draw(rightanglemark(A,B,O)); // Big Square draw((5,5)--(5,-5)--(-5,-5)--(-5,5)--(5,5)); dot(C); label("C",C,NE); dot(D); label("D",D,SW); draw(C--D); [/asy]

Let the points be labeled as in the diagram, with $O$ being the center of the circle. Because we know that the small square has an area of $40$, it must have a side length of $\sqrt{40}=2\sqrt{10}$. It is simple to prove that $O$ is the midpoint of the bottom side of the small square, so $OB=\frac{2\sqrt{10}}{2}=\sqrt{10}$. By the Pythagorean Theorem, $AO=\sqrt{50}=5\sqrt{2}$, which is the radius of the circle. Thus, $OC=5\sqrt{2}$, and so the diagonal $\overline{CD}$ of the big square has length $10\sqrt{2}$. Thus, the big square has side length $\frac{10\sqrt{2}}{\sqrt{2}}=10$, and, consequently, it has an area of $\boxed{\textbf{(B) }100}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 30 		Followed by
Problem 32
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All AHSME Problems and Solutions

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