1. **State the problem:** We need to find the perimeter of a combined shape consisting of a semicircle with diameter 13 m and a trapezium with parallel sides 18 m and 27 m, and height 12 m. 2. **Calculate the perimeter of the semicircle:** - Radius $r$ of the semicircle is half the diameter: $r=\frac{13}{2}=6.5$ m. - Circumference of a full circle is $2\pi r$, so the semicircle arc length is half of that: $\pi r=\pi \times 6.5=6.5\pi$ m. - Semicircle perimeter includes the diameter line, which for the combined shape is shared with the trapezium and not counted separately. 3. **Calculate the sides of trapezium:** - Given parallel sides (bases) $a=18$ m and $b=27$ m. - Height $h=12$ m. - Calculate non-parallel sides (legs) using the Pythagorean theorem. - The difference of the bases is $27-18=9$ m. - Each leg length $=\sqrt{12^2 + \left(\frac{9}{2}\right)^2}=\sqrt{144 + 20.25}=\sqrt{164.25}$. - Approximate $\sqrt{164.25} \approx 12.82$ m. 4. **Combine perimeter parts:** - The diameter of the semicircle (13 m) lies along one base of trapezium, so it is shared. - Assume the diameter aligns with side 18 m of trapezium, so that side is replaced by semicircle arc. - Perimeter = semicircle arc length $+ $ other base $+ $ two legs. $$\text{Perimeter} = 6.5\pi + 27 + 2 \times 12.82 = 6.5\pi + 27 + 25.64$$ 5. **Calculate numerical value:** - Use $\pi \approx 3.1416$. - $6.5 \times 3.1416 = 20.42$. - Sum: $20.42 + 27 + 25.64 = 73.06$ m. **Final answer:** The perimeter of the combined shape is approximately $73.06$ meters.