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Difference between revisions of "2013 AMC 8 Problems/Problem 23"

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(→‎Solution 2)
 
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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));
−
</asy>
+
dot(A);
 +
dot(B);
 +
dot(C);
 +
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 16pi. Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagorean theorem says that the other side has length 15, so the radius is <math>\boxed{(B) 7.5}</math>.
+
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>
 +
 
 +
~Edited by Theraccoon to correct typos.
 +
 
 +
==Brief Explanation==
 +
SavannahSolver got a diameter of 17 because the given arc length of the semicircle was
 +
8.5π. The arc length of a semicircle can be calculated using the formula
 +
πr, where
 +
r is the radius. let’s use the full circumference formula for a circle, which is
 +
2πr. Since the semicircle is half of a circle, its arc length is
 +
πr, which was given as
 +
8.5π. Solving for
 +
r, we get
 +
𝑟=8.5
 +
. Therefore, the diameter, which is
 +
2r, is
 +
2x8.5=17
 +
Then, the other steps to solve the problem will be the same as mentioned above by SavannahSolver
 +
the answer is <math>\boxed{\textbf{(B)}\ 7.5}</math>
 +
 
 +
 
 +
. - TheNerdWhoIsNerdy.
 +
 
 +
==Solution 2==
 +
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>.
 +
 
 +
~Note by Theraccoon: The person who posted this did not include their name.
 +
 
 +
==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 17:47, 27 September 2024

Problem

Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$? [asy] 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((0,0)..(7.5,-7.5)..(15,0)--(0,0)); real theta = aTan(8/15); draw(arc((15/2,4),17/2,-theta,180-theta)); draw((0,8)--(15,0)); dot(A); dot(B); dot(C); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE);[/asy]

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9$

Video Solution

https://youtu.be/crR3uNwKjk0 ~savannahsolver

Solution 1

If the semicircle on $\overline{AB}$ were a full circle, the area would be $16\pi$.

$\pi r^2=16 \pi \Rightarrow r^2=16 \Rightarrow r=+4$, therefore the diameter of the first circle is $8$.

The arc of the largest semicircle is $8.5 \pi$, so if it were a full circle, the circumference would be $17 \pi$. So the $\text{diameter}=17$.


By the Pythagorean theorem, the other side has length $15$, so the radius is $\boxed{\textbf{(B)}\ 7.5}$

~Edited by Theraccoon to correct typos.

Brief Explanation

SavannahSolver got a diameter of 17 because the given arc length of the semicircle was 8.5π. The arc length of a semicircle can be calculated using the formula πr, where r is the radius. let’s use the full circumference formula for a circle, which is 2πr. Since the semicircle is half of a circle, its arc length is πr, which was given as 8.5π. Solving for r, we get 𝑟=8.5 . Therefore, the diameter, which is 2r, is 2x8.5=17 Then, the other steps to solve the problem will be the same as mentioned above by SavannahSolver the answer is $\boxed{\textbf{(B)}\ 7.5}$


. - TheNerdWhoIsNerdy.

Solution 2

We go as in Solution 1, finding the diameter of the circle on $\overline{AC}$ and $\overline{AB}$. 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 $\frac{289\pi}{8}$, and the middle one is $\frac{289\pi}{8}-\frac{64\pi}{8}=\frac{225\pi}{8}$, so the radius is $\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}$.

~Note by Theraccoon: The person who posted this did not include their name.

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. AMC logo.png

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