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

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== Problem ==
 
== Problem ==
A [[square]] of area 40 is [[inscribe]]d 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?
<!-- [[Image:2006amc10b08.gif]] -->
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<asy>
 
<asy>
 
defaultpen(linewidth(0.8)); size(100);
 
defaultpen(linewidth(0.8)); size(100);
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draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle);
 
draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle);
 
</asy>
 
</asy>
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<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>
  

Revision as of 23:24, 18 October 2020

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 (ProblemsAnswer KeyResources)
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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