Semicircle inception

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$\quad \quad In\quad \triangle DEF,\quad \\ \quad \quad { R }^{ 2 }={ 8 }^{ 2 }+r^{ 2 }\\ \Rightarrow { R }^{ 2 }-r^{ 2 }=64\\ ar(shaded\quad part)=\frac { \pi }{ 2 } [{ R }^{ 2 }-r^{ 2 }]\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \Rightarrow \frac { \pi }{ 2 } [64]=32\pi \\ so\\ x=32$

Let R be the radius of the big circle and r be the small circle's radius,now CD=16,and half of CD is 8.Now R^2-r^2=8^2=64.the shaded region is (pi R^2)/2-(pi r^2)/2=pi/2(R^2-r^2)=pi/2 (8)^2=pi 32=32*pi.So x=32

We assume that this problem holds true for all pairs (R, r) which satisfy the above conditions, where R is the radius of the large semicircle, and r is the radius of the small semicircle. We then choose (R, r) = (16/2, 0), which is clearly a solution. The area of the resulting semicircle is 1/2 pi 8^2 = 32pi.

Is this okay?

Let r be the radius of the big semicircle and j be the radius of the small semicircle. By power of a point, we see that (r-j)(r+j) = 8*8, so $r^2 - j^2 = 64$ . Note that we want x such that $x\pi = \pi/2(r^2-j^2)$ , so the answer is $x = 32$ .