1. **Calculate the distance around the running track.** The running track is a rectangle with two semicircles at the ends. 2. **Formula for perimeter of running track:** $$\text{Perimeter} = 2 \times \text{length} + 2 \times \text{radius} \times \pi$$ Here, the length is the straight part (80 m), and the radius is half the width (15 m). 3. **Calculate radius:** $$r = 15\,m$$ 4. **Calculate perimeter:** $$\text{Perimeter} = 2 \times 80 + 2 \times 15 \times \pi = 160 + 30\pi$$ 5. **Approximate value:** $$30\pi \approx 30 \times 3.1416 = 94.248$$ 6. **Total distance:** $$160 + 94.248 = 254.248\,m$$ **Answer:** The distance around the running track is approximately $254.25$ meters. --- 2. **Calculate the perimeter of the four-lobed shape.** The shape is complex, but given the problem, we assume the perimeter is the sum of the outer edges. Since the shape is not fully described, we cannot calculate the exact perimeter without more information. --- 3. **Calculate the length of gold trim Jamie needs.** Jamie has a rectangular piece with a semicircle cut out. The gold trim goes around the entire frame and the picture (the semicircle). 4. **Calculate perimeter of rectangle:** $$P_{rect} = 2 \times (14 + 8) = 44\,cm$$ 5. **Calculate circumference of full circle with diameter 10 cm:** $$C = \pi \times d = \pi \times 10 = 10\pi$$ 6. **Calculate length of semicircle (half circumference):** $$L_{semi} = \frac{10\pi}{2} = 5\pi \approx 15.708\,cm$$ 7. **Total gold trim length:** $$44 + 15.708 = 59.708\,cm$$ **Answer:** Jamie needs approximately $59.71$ cm of gold trim. --- 4. **Calculate $x$ given a semicircle perimeter of 80 cm.** 5. **Formula for semicircle perimeter:** $$P = \pi r + 2r$$ Here, $x$ is the diameter, so radius $r = \frac{x}{2}$. 6. **Substitute $r$ into formula:** $$80 = \pi \times \frac{x}{2} + 2 \times \frac{x}{2} = \frac{\pi x}{2} + x$$ 7. **Simplify:** $$80 = x \left( \frac{\pi}{2} + 1 \right)$$ 8. **Solve for $x$:** $$x = \frac{80}{\frac{\pi}{2} + 1} = \frac{80}{\frac{\pi + 2}{2}} = \frac{80 \times 2}{\pi + 2} = \frac{160}{\pi + 2}$$ 9. **Approximate value:** $$\pi + 2 \approx 3.1416 + 2 = 5.1416$$ $$x \approx \frac{160}{5.1416} = 31.11\,cm$$ **Answer:** $x$ is approximately $31.11$ cm.