Difference between revisions of "2010 AMC 8 Problems/Problem 19"
m (→Solution) |
|||
(4 intermediate revisions by 4 users not shown) | |||
Line 5: | Line 5: | ||
unitsize(45); | unitsize(45); | ||
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); | import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); | ||
− | draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); | + | draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); |
dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); | dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); | ||
</asy> | </asy> | ||
+ | |||
<math> \textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi </math> | <math> \textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi </math> | ||
Line 16: | Line 17: | ||
Note: The length <math>AC</math> is necessary information, as this tells us the radius of the larger circle. The area of the annulus is <math>\pi(AC^2-BC^2)=\pi AB^2=64\pi</math>. | Note: The length <math>AC</math> is necessary information, as this tells us the radius of the larger circle. The area of the annulus is <math>\pi(AC^2-BC^2)=\pi AB^2=64\pi</math>. | ||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/Q6rnoQChiyU. Soo, DRMS, NM | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/51K3uCzntWs?t=3206 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | |||
+ | ==Video by MathTalks== | ||
+ | |||
+ | https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/yjhitUYSAI0 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=18|num-a=20}} | {{AMC8 box|year=2010|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:04, 10 March 2023
Contents
Problem
The two circles pictured have the same center . Chord
is tangent to the inner circle at
,
is
, and chord
has length
. What is the area between the two circles?
Solution
Since is isosceles,
bisects
. Thus
. From the Pythagorean Theorem,
. Thus the area between the two circles is
Note: The length is necessary information, as this tells us the radius of the larger circle. The area of the annulus is
.
Video Solution
https://youtu.be/Q6rnoQChiyU. Soo, DRMS, NM
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=3206
~ pi_is_3.14
Video by MathTalks
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
Video Solution by WhyMath
~savannahsolver
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.