1. **Problem Statement:** Calculate the perimeter and area of each composite figure given. --- ### a) Composite polygon with sides 5 m, 6 m, 9 m, 3 m, and an unknown side 4 m (calculated from the figure). 2. **Perimeter Calculation:** The perimeter $P$ is the sum of all side lengths: $$P = 5 + 6 + 4 + 3 + 9 = 27\text{ m}$$ 3. **Area Calculation:** We can divide the figure into a rectangle and a triangle. - Rectangle area: length $= 5$ m, width $= 3$ m $$A_{rect} = 5 \times 3 = 15\text{ m}^2$$ - Triangle area: base $= 4$ m, height $= 6 - 3 = 3$ m $$A_{tri} = \frac{1}{2} \times 4 \times 3 = 6\text{ m}^2$$ 4. **Total area:** $$A = A_{rect} + A_{tri} = 15 + 6 = 21\text{ m}^2$$ --- ### b) Composite figure: semicircle on top of an inverted isosceles triangle 5. **Given:** - Diameter of semicircle $d = 8$ cm, so radius $r = \frac{8}{2} = 4$ cm - Two equal sides of triangle $= 10$ cm each 6. **Perimeter Calculation:** - Semicircle perimeter (half circumference plus diameter): $$P_{semi} = \pi r + d = \pi \times 4 + 8 = 4\pi + 8$$ - Triangle base is the diameter $8$ cm, so the triangle perimeter excluding base is $10 + 10 = 20$ cm - Total perimeter: $$P = 20 + 4\pi + 8 = 28 + 4\pi \approx 28 + 12.57 = 40.6\text{ cm}$$ 7. **Area Calculation:** - Area of semicircle: $$A_{semi} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \times 4^2 = 8\pi \approx 25.1\text{ cm}^2$$ - Height of triangle using Pythagoras: $$h = \sqrt{10^2 - 4^2} = \sqrt{100 - 16} = \sqrt{84} \approx 9.17\text{ cm}$$ - Area of triangle: $$A_{tri} = \frac{1}{2} \times 8 \times 9.17 = 36.7\text{ cm}^2$$ - Total area: $$A = A_{semi} + A_{tri} = 25.1 + 36.7 = 61.8\text{ cm}^2$$ --- **Final answers:** - a) Perimeter $= 27$ m, Area $= 21$ m$^2$ - b) Perimeter $\approx 40.6$ cm, Area $\approx 61.8$ cm$^2$