AoPSWiki
Art of Problem Solving holds many free classes called Math Jams.
Click here for transcripts to past Math Jams.

2006 AMC 10B Problems/Problem 8

From AoPSWiki

Revision as of 01:24, 4 August 2006 by Xantos C. Guin (Talk | contribs)
(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

Problem

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?

Image:2006amc10b08.gif

\mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \...

Solution

Since the area of the square is 40, the length of the side is \sqrt{40}=2\sqrt{10}. The distance between the center of the semicircle and one of the bottom vertecies of the square is half the length of the side, which is \sqrt{10}.

Using the Pythagorean Theorem to find the square of radius:

(2\sqrt{10})^2 + (\sqrt{10})^2 = r^2

50=r^2

So, the area of the semicircle is \frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Rightarrow B

See Also

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us