Difference between revisions of "2005 AMC 10B Problems/Problem 7"

Revision as of 21:48, 11 April 2013

Problem

A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?

$\mathrm{(A)} \frac{\pi}{16} \qquad \mathrm{(B)} \frac{\pi}{8} \qquad \mathrm{(C)} \frac{3\pi}{16} \qquad \mathrm{(D)} \frac{\pi}{4} \qquad \mathrm{(E)} \frac{\pi}{2}$

Solution

Let the side of the largest square be $x$ . It follows that the diameter of the inscribed circle is also $x$ . Therefore, the diagonal of the square inscribed inscribed in the circle is $x$ . The side length of the smaller square is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$ . Similarly, the diameter of the smaller inscribed circle is $\dfrac{x\sqrt{2}}{2}$ . Hence, its radius is $\dfrac{x\sqrt{2}}{4}$ . The area of this circle is $\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}$ , and the area of the largest square is $x^2$ . The ratio of the areas is $\dfrac{\dfrac{x^2\pi}{8}}{x^2}=\boxed{\mathrm{(B)}\ \dfrac{\pi}{8}}$ .

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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
All AMC 10 Problems and Solutions

@@ Line 7: / Line 7: @@
 == See Also ==
 {{AMC10 box|year=2005|ab=B|num-b=6|num-a=8}}
+[[Category:Introductory Geometry Problems]]
+[[Category:Area Ratio Problems]]

Page

Toolbox

Search

Difference between revisions of "2005 AMC 10B Problems/Problem 7"

Revision as of 21:48, 11 April 2013

Problem

Solution

See Also