Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AMC 10B Problems/Problem 13"

(Created page with "==Problem== Two circles of radius <math>2</math> are centered at <math>(2,0)</math> and at <math>(0,2).</math> What is the area of the intersection of the interiors of the two c...")
 
(undo previous edits)
 
(6 intermediate revisions by 3 users not shown)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is a quarter of the circle with radius <math>2,</math> and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is
+
You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is one-fourth of the area of the circle with radius <math>2,</math> and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is
 
<cmath>\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2</cmath>
 
<cmath>\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2</cmath>
 
That means the area of the whole intersection is <math>\boxed{\mathrm{(D) \ } 2(\pi-2)}</math>
 
That means the area of the whole intersection is <math>\boxed{\mathrm{(D) \ } 2(\pi-2)}</math>
Line 14: Line 14:
  
 
{{AMC10 box|year=2007|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2007|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 20:42, 15 February 2024

Problem

Two circles of radius $2$ are centered at $(2,0)$ and at $(0,2).$ What is the area of the intersection of the interiors of the two circles?

$\textbf{(A) } \pi -2 \qquad\textbf{(B) } \frac{\pi}{2} \qquad\textbf{(C) } \frac{\pi \sqrt{3}}{3} \qquad\textbf{(D) } 2(\pi -2) \qquad\textbf{(E) } \pi$

Solution

You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is one-fourth of the area of the circle with radius $2,$ and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is \[\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2\] That means the area of the whole intersection is $\boxed{\mathrm{(D) \ } 2(\pi-2)}$

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems/Problem_13&oldid=215557"