Difference between revisions of "2006 AMC 10B Problems/Problem 8"
Ragnarok23 (talk | contribs) |
(Added problem and solution) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? | ||
+ | |||
+ | [[Image:2006amc10b08.gif]] | ||
+ | |||
+ | <math> \mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi </math> | ||
+ | |||
== Solution == | == Solution == | ||
+ | Since the area of the square is <math>40</math> the length of the side is <math>\sqrt{40}=2\sqrt{10}</math> | ||
+ | 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 <math>\sqrt{10}</math> | ||
+ | |||
+ | Using the Pythagorean Theorem to find the square of radius: | ||
+ | |||
+ | <math>(2\sqrt{10})^2 + (\sqrt{10})^2 = r^2 </math> | ||
+ | |||
+ | <math>50=r^2</math> | ||
+ | |||
+ | So the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Rightarrow B </math> | ||
+ | |||
== See Also == | == See Also == | ||
*[[2006 AMC 10B Problems]] | *[[2006 AMC 10B Problems]] |
Revision as of 20:51, 13 July 2006
Problem
A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?
Solution
Since the area of the square is the length of the side is
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
Using the Pythagorean Theorem to find the square of radius:
So the area of the semicircle is