1. **State the problem:** We have rectangle ABCD with perimeter 26. A half circle with diameter AD is inside the rectangle, and its area is given as $8\pi$. We need to find the perimeter of the part of the figure that is not shaded. 2. **Identify variables:** Let the length of AD be $d$ and the length of AB be $h$. 3. **Use the perimeter of the rectangle:** Perimeter $P = 2(d + h) = 26$. So, $$d + h = 13.$$ 4. **Use the area of the half circle:** Area of half circle = $\frac{1}{2} \pi r^2 = 8\pi$. Since diameter $d = 2r$, radius $r = \frac{d}{2}$. Substitute radius: $$\frac{1}{2} \pi \left(\frac{d}{2}\right)^2 = 8\pi$$ $$\frac{1}{2} \pi \frac{d^2}{4} = 8\pi$$ $$\frac{\pi d^2}{8} = 8\pi$$ Divide both sides by $\pi$: $$\frac{d^2}{8} = 8$$ Multiply both sides by 8: $$d^2 = 64$$ Take positive root (length): $$d = 8$$ 5. **Find $h$ using $d + h = 13$:** $$8 + h = 13$$ $$h = 5$$ 6. **Find the perimeter of the non-shaded part:** The total perimeter is $26$. The shaded part is the half circle arc with length: $$\text{arc length} = \pi r = \pi \times 4 = 4\pi$$ The non-shaded perimeter consists of the three sides of the rectangle excluding AD plus the straight line AD (diameter) replaced by the half circle arc. So, non-shaded perimeter = perimeter of rectangle - arc length + length of AD $$= 26 - 4\pi + 8$$ $$= (26 + 8) - 4\pi = 34 - 4\pi$$ But this counts AD twice (once as straight line and once as arc). Actually, the perimeter of the figure is the sum of the three sides (AB, BC, CD) plus the arc length of the half circle. Sum of three sides: $$AB + BC + CD = h + d + h = 5 + 8 + 5 = 18$$ Add the arc length: $$18 - d + 4\pi = 18 - 8 + 4\pi = 10 + 4\pi$$ This is incorrect because BC = h, CD = d, AB = h, so three sides excluding AD are $h + h + d = 5 + 5 + 8 = 18$. The perimeter of the non-shaded part is the sum of these three sides plus the arc length replacing AD: $$18 + 4\pi$$ 7. **Check options:** None of the options match $18 + 4\pi$ exactly, but option D is $18 + 4\pi$. **Final answer:** $$\boxed{18 + 4\pi}$$