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

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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>
 
<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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== Solution ==
 
== Solution ==

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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