Difference between revisions of "2006 AMC 10B Problems/Problem 8"
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== Problem == | == Problem == | ||
− | A square of area 40 is | + | A [[square]] of area 40 is [[inscribe]]d in a [[semicircle]] as shown. What is the area of the semicircle? |
− | + | <!-- [[Image:2006amc10b08.gif]] --> | |
− | [[Image:2006amc10b08.gif]] | + | <asy> |
− | + | defaultpen(linewidth(0.8)); size(100); | |
+ | real r=sqrt(50), s=sqrt(10); | ||
+ | draw(Arc(origin, r, 0, 180)); | ||
+ | draw((r,0)--(-r,0), dashed); | ||
+ | draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); | ||
+ | </asy> | ||
<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> | <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>. | + | 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>. |
− | 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>. | ||
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− | <math>(2\sqrt{10})^2 + (\sqrt{10})^2 = | + | Using the [[Pythagorean Theorem]] to find the square of radius, <math>r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50</math>. So, the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Longrightarrow \boxed{\text{(B)}}</math>. |
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− | So, the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \ | ||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=7|num-a=9}} | |
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] | ||
+ | [[Category:Circle Problems]] |
Revision as of 23:23, 20 August 2011
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 .
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |