In the diagram, line segment $P S$ has length 4. Points $Q$ and $R$ are on line segment $P S$. Four semi-circles are drawn on the same side of $P S$. The diameters of these semi-circles are $P S, P Q$, $Q R$, and $R S$. The region inside the largest semicircle and outside the three smaller semi-circles is shaded. What is the area of a square whose perimeter equals the perimeter of the shaded region?

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-058.jpg?height=317&width=572&top_left_y=2078&top_left_x=1172)
(A) 4
(B) $\pi$
(C) $\pi^{2}$
(D) $2 \pi^{2}$
(E) $\frac{\pi^{2}}{4}$

Part C: Each correct answer is worth 8.
The perimeter of the shaded region consists of four pieces: a semi-circle with diameter $P S$, a semi-circle with diameter $P Q$, a semi-circle with diameter $Q R$, and a semi-circle with diameter $R S$.

We note that since a circle with diameter $d$ has circumference equal to $\pi d$, then, not including the diameter itself, the length of a semi-circle with diameter $d$ is $\frac{1}{2} \pi d$.

Suppose that the diameter of semi-circle $P Q$ is $x$, the diameter of semi-circle $Q R$ is $y$, and the diameter of semi-circle $R S$ is $z$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-100.jpg?height=306&width=563&top_left_y=2297&top_left_x=846)

We are given that the diameter of the semi-circle $P S$ is 4 .

Since $P Q+Q R+R S=P S$, then $x+y+z=4$.

Thus, the perimeter of the shaded region is

$\frac{1}{2} \pi(P Q)+\frac{1}{2} \pi(Q R)+\frac{1}{2} \pi(R S)+\frac{1}{2} \pi(P S)=\frac{1}{2} \pi x+\frac{1}{2} \pi y+\frac{1}{2} \pi z+\frac{1}{2} \pi(4)=\frac{1}{2} \pi(x+y+z+4)=\frac{1}{2} \pi(4+4)=4 \pi$

We want to determine the area of the square whose perimeter equals this perimeter (that is, whose perimeter is $4 \pi$ ).

If a square has perimeter $4 \pi$, then its side length is $\frac{1}{4}(4 \pi)=\pi$, and so its area is $\pi^{2}$.

Because this problem is multiple choice, then we should get the same answer regardless of the actual lengths of $P Q, Q R$ and $R S$, since we are not told what these lengths are. Therefore, we can assign to these lengths arbitrary values that satisfy the condition $P Q+Q R+R S=4$. For example, if $P Q=Q R=1$ and $R S=2$, then we can calculate the perimeter of the shaded region to be

$\frac{1}{2} \pi(P Q)+\frac{1}{2} \pi(Q R)+\frac{1}{2} \pi(R S)+\frac{1}{2} \pi(P S)=\frac{1}{2} \pi(1)+\frac{1}{2} \pi(1)+\frac{1}{2} \pi(2)+\frac{1}{2} \pi(4)=\frac{1}{2} \pi(1+1+2+4)=4 \pi$

and obtain the same answer as above.

ANSWER: (C)

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