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2013 AMC 8 Problems/Problem 23: Difference between revisions

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draw((0,8)--(15,0));
draw((0,8)--(15,0));
dot(A);
dot(A);
dot(B);
dot(C);
dot(C);
label("$A$", A, NW);
label("$A$", A, NW);
Line 33: Line 32:




y the Pythagorean theorem, the other side has length <math>15</math>, so the radius is <math>\boxed{\textbf{(B)}\ 7.5}</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==
==Brief Explanation==
SavannahSolver got a diameter of 17 because the given arc length of the semicircle was  
SavannahSolver got a diameter of <math>17</math> because the given arc length of the semicircle was  
8.5π. The arc length of a semicircle can be calculated using the formula  
<math>8.5\pi</math>. The arc length of a semicircle can be calculated using the formula  
πr, where  
<math>\pi r</math>, where  
r is the radius. let’s use the full circumference formula for a circle, which is  
<math>r</math> 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  
<math>2\pi r</math>. Since the semicircle is half of a circle, its arc length is  
πr, which was given as  
<math>\pi r</math>, which was given as  
8.5π. Solving for  
<math>8.5\pi</math>. Solving for  
r, we get  
<math>r</math>, we get  
𝑟=8.5
<math>r=8.5</math>
. Therefore, the diameter, which is  
. Therefore, the diameter, which is  
2r, is  
<math>2r</math>, is  
2x8.5=17
<math>2\cdot8.5=17</math>
\boxed{\textbf{(B)}\ 7.5}$.
Then, the other steps to solve the problem will be the same as mentioned above by SavannahSolver
Then, the other steps to solve the problem will be the same.
the answer is <math>\boxed{\textbf{(B)}\ 7.5}</math>
 
 
. - TheNerdWhoIsNerdy.
. - TheNerdWhoIsNerdy.
Minor edits by -Coin1


==Solution 2==
==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>.
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==
==Video Solution by OmegaLearn==
Line 60: Line 66:
~ pi_is_3.14
~ pi_is_3.14


===Answer (B) 7.5===
~ Mia Wang the Author
~skibidi


==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 14:06, 24 August 2025

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(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 $8.5\pi$. The arc length of a semicircle can be calculated using the formula $\pi r$, where $r$ is the radius. let’s use the full circumference formula for a circle, which is $2\pi r$. Since the semicircle is half of a circle, its arc length is $\pi r$, which was given as $8.5\pi$. Solving for $r$, we get $r=8.5$ . Therefore, the diameter, which is $2r$, is $2\cdot8.5=17$ 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. Minor edits by -Coin1

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


Answer (B) 7.5

~ Mia Wang the Author ~skibidi

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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