Difference between revisions of "2013 AMC 8 Problems/Problem 23"
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<asy> | <asy> | ||
import graph; | import graph; | ||
+ | pair A,B,C; | ||
+ | A=(0,8); | ||
+ | B=(0,0); | ||
+ | C=(15,0); | ||
draw((0,8)..(-4,4)..(0,0)--(0,8)); | draw((0,8)..(-4,4)..(0,0)--(0,8)); | ||
draw((0,0)..(7.5,-7.5)..(15,0)--(0,0)); | draw((0,0)..(7.5,-7.5)..(15,0)--(0,0)); | ||
Line 8: | Line 12: | ||
draw(arc((15/2,4),17/2,-theta,180-theta)); | draw(arc((15/2,4),17/2,-theta,180-theta)); | ||
draw((0,8)--(15,0)); | draw((0,8)--(15,0)); | ||
− | label("$A$", | + | dot(A); |
− | label("$B$", | + | dot(B); |
− | label("$C$", | + | dot(C); |
− | </asy> | + | label("$A$", A, NW); |
+ | label("$B$", B, SW); | ||
+ | label("$C$", C, SE);</asy> | ||
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9</math> | <math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/crR3uNwKjk0 ~savannahsolver | ||
==Solution 1== | ==Solution 1== | ||
− | If the semicircle on AB were a full circle, the area would be | + | If the semicircle on <math>\overline{AB}</math> were a full circle, the area would be <math>16\pi</math>. |
+ | |||
+ | <math>\pi r^2=16 \pi \Rightarrow r^2=16 \Rightarrow r=+4</math>, therefore the diameter of the first circle is <math>8</math>. | ||
+ | |||
+ | The arc of the largest semicircle is <math>8.5 \pi</math>, so if it were a full circle, the circumference would be <math>17 \pi</math>. So the <math>\text{diameter}=17</math>. | ||
+ | |||
+ | By the Pythagorean theorem, the other side has length <math>15</math>, so the radius is <math>\boxed{\textbf{(B)}\ 7.5}</math> | ||
==Solution 2== | ==Solution 2== | ||
− | We go as in | + | We go as in Solution 1, finding the diameter of the circle on <math>\overline{AC}</math> and <math>\overline{AB}</math>. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is <math>\frac{289\pi}{8}</math>, and the middle one is <math>\frac{289\pi}{8}-\frac{64\pi}{8}=\frac{225\pi}{8}</math>, so the radius is <math>\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}</math>. |
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/abSgjn4Qs34?t=2584 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=22|num-a=24}} | {{AMC8 box|year=2013|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:56, 16 July 2024
Contents
Problem
Angle of
is a right angle. The sides of
are the diameters of semicircles as shown. The area of the semicircle on
equals
, and the arc of the semicircle on
has length
. What is the radius of the semicircle on
?
Video Solution
https://youtu.be/crR3uNwKjk0 ~savannahsolver
Solution 1
If the semicircle on were a full circle, the area would be
.
, therefore the diameter of the first circle is
.
The arc of the largest semicircle is , so if it were a full circle, the circumference would be
. So the
.
By the Pythagorean theorem, the other side has length , so the radius is
Solution 2
We go as in Solution 1, finding the diameter of the circle on and
. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is
, and the middle one is
, so the radius is
.
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=2584
~ pi_is_3.14
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.