2013 AMC 8 Problems/Problem 23: Difference between revisions

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

@@ Line 70: / Line 70: @@
 ~ Mia Wang the Author
 ~skibidi
-~Coin1
+==See Also==
+{{AMC8 box|year=2013|num-b=22|num-a=24}}
+{{MAA Notice}}

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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