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Difference between revisions of "2006 AMC 10B Problems/Problem 8"

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== Problem ==
 
== Problem ==
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A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?  
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A [[square]] of area 40 is [[inscribe]]d in a [[semicircle]] as shown. What is the area of the semicircle?  
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<!-- [[Image:2006amc10b08.gif]] -->
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[[Image:2006amc10b08.gif]]
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<asy>
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defaultpen(linewidth(0.8)); size(100);
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real r=sqrt(50), s=sqrt(10);
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draw(Arc(origin, r, 0, 180));
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draw((r,0)--(-r,0), dashed);
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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 ==
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Since the area of the square is <math>40</math>, the length of the side is <math>\sqrt{40}=2\sqrt{10}</math>.
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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>.
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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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Using the Pythagorean Theorem to find the square of radius:
 
  
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<math>(2\sqrt{10})^2 + (\sqrt{10})^2 = r^2 </math>
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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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<math>50=r^2</math>
 
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So, the area of the semicircle is <math>\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Rightarrow B </math>
 
  
 
== See Also ==
 
== See Also ==
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*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=7|num-a=9}}
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*[[2006 AMC 10B Problems/Problem 7|Previous Problem]]
 
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*[[2006 AMC 10B Problems/Problem 9|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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[[Category:Area Problems]]
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[[Category:Circle Problems]]

Revision as of 00:23, 21 August 2011

Problem

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? [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] $\mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi$

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, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$. So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = 25\pi \Longrightarrow \boxed{\text{(B)}}$.

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
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