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"

(→‎Solution)
(→‎Solution)
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 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
+
You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whosector. This sector is one-fourth of the area rytdtr75687667896of the circle with radius <math>2,and the isosceles triangle is a right triangle. Therefore, the area of half the intersectiongiyuyiuutyuiyt7t68 i
−
<cmath>\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2</cmath>
+
</math><math>\frac
−
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)}$
 +
Uhi987y8yy978
  
 
==See Also==
 
==See Also==

Revision as of 20:24, 13 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 whosector. This sector is one-fourth of the area rytdtr75687667896of the circle with radius $2,and the isosceles triangle is a right triangle. Therefore, the area of half the intersectiongiyuyiuutyuiyt7t68 i$$\frac That means the area of the whole intersection is$\boxed{\mathrm{(D) \ } 2(\pi-2)}$ Uhi987y8yy978

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=215456"