Difference between revisions of "2004 AMC 12B Problems/Problem 7"

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{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #7]] and [[2004 AMC 10B Problems|2004 AMC 10B #9]]}}
 
{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #7]] and [[2004 AMC 10B Problems|2004 AMC 10B #9]]}}
  
== Problem 7 ==
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== Problem ==
A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?
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A square has sides of length <math>10</math>, and a circle centered at one of its vertices has radius <math>10</math>. What is the area of the union of the regions enclosed by the square and the circle?
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<math>\mathrm{(A)\ }200+25\pi\quad\mathrm{(B)\ }100+75\pi\quad\mathrm{(C)\ }75+100\pi\quad\mathrm{(D)\ }100+100\pi\quad\mathrm{(E)\ }100+125\pi</math>
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== Video Solution 1==
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https://youtu.be/IGN4XxJIbE0
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~Education, the Study of Everything
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<math>(\mathrm {A}) 200+25\pi \quad (\mathrm {B}) 100+75\pi \quad (\mathrm {C}) 75+100\pi \quad (\mathrm {D}) 100+100\pi \quad (\mathrm {E}) 100+125\pi</math>
 
  
 
== Solution ==
 
== Solution ==
  
The area of the circle is <math>S_{\bigcirc}=100\pi</math>, the area of the square is <math>S_{\square}=100</math>.
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The area of the circle is <math>S_{\bigcirc}=100\pi</math>; the area of the square is <math>S_{\square}=100</math>.
  
Exactly <math>1/4</math> of the circle lies inside the square. Thus the total area is <math>\dfrac34 S_{\bigcirc} + S_{\square} = \boxed{100+75\pi} \Longrightarrow \mathrm{(B)}</math>.
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Exactly <math>\frac{1}{4}</math> of the circle lies inside the square. Thus the total area is <math>\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}</math>.
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<asy>
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Draw(Circle((0,0),10));
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Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0));
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label("$10$",(5,0),S);
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label("$10$",(0,5),W);
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dot((0,0));
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</asy>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2004|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2004|ab=B|num-b=6|num-a=8}}
{{AMC12 box|year=2004|ab=B|num-b=8|num-a=10}}
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{{AMC10 box|year=2004|ab=B|num-b=8|num-a=10}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Area Problems]]
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{{MAA Notice}}

Latest revision as of 18:23, 22 October 2022

The following problem is from both the 2004 AMC 12B #7 and 2004 AMC 10B #9, so both problems redirect to this page.

Problem

A square has sides of length $10$, and a circle centered at one of its vertices has radius $10$. What is the area of the union of the regions enclosed by the square and the circle?

$\mathrm{(A)\ }200+25\pi\quad\mathrm{(B)\ }100+75\pi\quad\mathrm{(C)\ }75+100\pi\quad\mathrm{(D)\ }100+100\pi\quad\mathrm{(E)\ }100+125\pi$

Video Solution 1

https://youtu.be/IGN4XxJIbE0

~Education, the Study of Everything


Solution

The area of the circle is $S_{\bigcirc}=100\pi$; the area of the square is $S_{\square}=100$.

Exactly $\frac{1}{4}$ of the circle lies inside the square. Thus the total area is $\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}$.

[asy] Draw(Circle((0,0),10)); Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); label("$10$",(5,0),S); label("$10$",(0,5),W); dot((0,0)); [/asy]

See Also

2004 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 12 Problems and Solutions
2004 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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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