1. **State the problem:** We have a rectangle with width 17 cm and height 8 cm. On the left side, there is a quarter circle arc with radius 8 cm (the height), and on the right side, there is a semicircle arc whose diameter equals 8 cm. We need to find the perimeter of this composite figure. 2. **Analyze the components:** - Rectangle sides: The rectangle normally has two widths and two heights, but parts of these sides are replaced by arcs. - Left side arc: a quarter circle with radius 8 cm. - Right side arc: a semicircle with diameter 8 cm, hence radius 4 cm. 3. **Calculate lengths of arcs:** - Quarter circle arc length = $\frac{1}{4} \times 2\pi r = \frac{1}{4} \times 2\pi \times 8 = \frac{1}{4} \times 16\pi = 4\pi$ cm. - Semicircle arc length = $\frac{1}{2} \times 2\pi r = \pi r = \pi \times 4 = 4\pi$ cm. 4. **Calculate the straight edges of the perimeter:** - Top horizontal side: full width = 17 cm. - Bottom horizontal side: full width = 17 cm. - Vertical sides are partially arcs, so exclude those lengths replaced by arcs. 5. **Determine the total perimeter:** The perimeter consists of: - Top side: 17 cm - Right vertical straight side below semicircle: Since the semicircle replaces half the height (radius 4 cm), the remaining vertical segment on the right is $8 - 8 = 0$ cm since the semicircle covers the entire height vertically. Actually, the diameter equals the full 8 cm height, so no vertical segment remains. - Bottom side: length under semicircle is marked as "a cm" which is 17 cm minus the base length under the quarter circle on the left? But bottom side is continuous at 17 cm since arcs are vertical. - Left vertical straight side replaced by quarter circle arc, so vertical sides are replaced by arcs only. Hence perimeter = top side + bottom side + left quarter circle arc + right semicircle arc $$P = 17 + 17 + 4\pi + 4\pi = 34 + 8\pi$$ 6. **Final answer:** $$\boxed{34 + 8\pi \text{ cm}}$$ This is the total perimeter of the figure composed of the rectangle's horizontal sides and the arcs replacing the vertical sides.