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2013 AMC 8 Problems/Problem 23: Difference between revisions

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==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>8.5π</math>. The arc length of a semicircle can be calculated using the formula  
πr, where  
<math>πr</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>2πr</math>. Since the semicircle is half of a circle, its arc length is  
πr, which was given as  
<math>πr</math>, which was given as  
8.5π. Solving for  
<math>8.5π</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>2x8.5=17</math>
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 as mentioned above by SavannahSolver
the answer is <math>\boxed{\textbf{(B)}\ 7.5}</math>
the answer is <math>\boxed{\textbf{(B)}\ 7.5}</math>
Line 54: Line 54:


. - TheNerdWhoIsNerdy.
. - TheNerdWhoIsNerdy.
Minor edits by -Coin1


==Solution 2==
==Solution 2==

Revision as of 20:09, 9 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 $8.5π$ (Error compiling LaTeX. Unknown error_msg). The arc length of a semicircle can be calculated using the formula $πr$ (Error compiling LaTeX. Unknown error_msg), where $r$ is the radius. let’s use the full circumference formula for a circle, which is $2πr$ (Error compiling LaTeX. Unknown error_msg). Since the semicircle is half of a circle, its arc length is $πr$ (Error compiling LaTeX. Unknown error_msg), which was given as $8.5π$ (Error compiling LaTeX. Unknown error_msg). Solving for $r$, we get $𝑟=8.5$ (Error compiling LaTeX. Unknown error_msg) . 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. 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

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