2013 AMC 8 Problems/Problem 23

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 for Problems 21-25

https://youtu.be/-mi3qziCuec

Video Solution

https://youtu.be/crR3uNwKjk0 ~savannahsolver

Solution 1

If the semicircle on 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 $\boxed{\textbf{(B)}\ 7.5}$

Solution 2

We go as in Solution 1, finding the diameter of the circle on AC and 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 $\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}$ .

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=2584

~ pi_is_3.14

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: Unable to save thumbnail to destination

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