A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$. A smaller semicircle with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below. What is the area of the resulting figure, shown shaded? $\textbf{(A) } 16\pi \qquad$ $\textbf{(B) } 24\pi \qquad$ $\textbf{(C) } 32\pi \qquad$ $\textbf{(D) } 48\pi \qquad$ $\textbf{(E) } 64\pi$
Solution
Let $R$ be the radius of the large semicircle (center $O$) and $h$ the distance from $\overline{CD}$ to $\overline{AB}$.
The small semicircle has its diameter on $\overline{AB}$ and is tangent to $\overline{CD}$, so its radius is $h$.
The midpoint of $\overline{CD}$ is at distance $h$ above $O$ and half the chord is $8$, so $R^2=8^2+h^2$.
The area is $\frac12\pi R^2-\frac12\pi h^2=\frac12\pi(R^2-h^2)=\frac12\pi\cdot64=32\pi$.
A semicircle has diameter $\overline{AB}$. Two congruent smaller semicircles have their diameters on $\overline{AB}$; they are tangent to each other and each is internally tangent to the large semicircle (so their diameters together fill $\overline{AB}$). A chord $\overline{CD}$ of the large semicircle, of length $20$, is parallel to $\overline{AB}$ and tangent to both small semicircles. The two small semicircles are cut away from the large one. What is the area of the resulting figure?
What is the area of the resulting figure, shown shaded? $\textbf{(A) } 16\pi \qquad$ $\textbf{(B) } 24\pi \qquad$ $\textbf{(C) } 32\pi \qquad$ $\textbf{(D) } 48\pi \qquad$ $\textbf{(E) } 64\pi$ 
