A Problem From My Exam In the image, a semicircle with diameter $AD$ has smaller semicircles with diameters $AB$, $BC$, $CD$, all next to each other inside it. The area between them is shaded.
The perimeter of the painted area is $24\pi$, what is the area?

 A: Ratio of diameters
$$
\begin{align}
AB : BC : CD = 1 : 2 : 3 =
r_{1} : r_{2} : r_{3}
\end{align}
$$
Radius of perimeter segment: $\pi r = 24 \pi$.
$$ 
 r = 24
$$
Add up the sgements:
$$
 \begin{align}
  r &= r_{1} + r_{2} + r_{3} \\
    &= r_{1} + 2 r_{1} + 3 r_{1} = 6 r_{1} \\
  24 &= 6 r_{1} \\
   4 &= r_{1}
 \end{align}
$$

Shaded area
$$
 \begin{align}
%
A &= \frac{\pi}{2} \left( r^{2} - 
\left( 
  r^{2}_{1} + r^{2}_{2} + r^{2}_{3}
\right)
\right) \\
%
&=\frac{\pi}{2} \left( 24^{2} - 
\left( 
  4^{2} + 8^{2} + 12^{2}
\right)
\right) \\
&= 352 \frac{\pi}{2} = \boxed{176 \pi}
%
 \end{align}
$$
A: $$ 24 \pi =\text{perimeter} = \pi(AD + AB + BC + CD) $$
$$24 \pi = \pi (AD + AD)$$
$$(12)(2\pi)=2 \pi AD$$
Once you solved for $AD$, I believe you can recover $AB, BC, CD$ and solve for the desired area.
A: The perimeter of $|AD|$ is $24\pi$:
$$
P_{|AB|}=\pi |AD| = 24\pi
$$
If you solve for $|AD|$ you get:
$$
|AD|=24
$$

Now, you now that $|AB|$ is the sum of the other 3
$$
|AD| = |AB| + |BC| + |CD| = 24
$$
You also know that:
$$
|BC| = 2|AB|
$$
and
$$
|CD| = 3|AB|
$$
Therefore
$$
|AB| = \frac{24}{6} = 4 \quad , \quad |BC| = 8 \quad , \quad |CD|=12
$$

Now, the shaded area is the area of the $|AD|$ semicircle minus the areas of the other tree:
$$
A_{\text{shaded}} = A_{|AD|} - A_{|AB|} - A_{|BC|} - A_{|CD|}
$$
Using the formula for the area of the semicircle $A=\frac{\pi r^2}{2}$
$$
A_{\text{shaded}} = \frac{\pi}{2}(24^2 - 4^2 -8^2 - 12^2) = 176 \pi
$$

That's your final answer: $A_{\text{shaded}} = 176 \pi$

Let me know if you need me to explain any of the steps!
