1. **Problem 5(i): Area of shaded section in a square with quarter circle** - Given: Square side = 7 cm, quarter circle radius = 7 cm. - Formula for area of quarter circle: $$\text{Area} = \frac{1}{4} \pi r^2$$ - Area of square: $$7 \times 7 = 49$$ cm² - Area of quarter circle: $$\frac{1}{4} \pi (7)^2 = \frac{49\pi}{4}$$ cm² - Shaded area = Area of square - Area of quarter circle = $$49 - \frac{49\pi}{4} = 49\left(1 - \frac{\pi}{4}\right)$$ cm² 2. **Problem 5(ii): Area of shaded section in trapezoid with semicircle** - Given: Trapezoid base = 20 cm, semicircle diameter = 7 cm, radius = 3.5 cm. - Area of semicircle: $$\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3.5)^2 = \frac{1}{2} \pi \times 12.25 = 6.125\pi$$ cm² - Area of trapezoid is not fully given, so shaded area is semicircle area = $$6.125\pi$$ cm² 3. **Problem 5(iii): Area of two semicircles with diameters a and a/2** - Radius of large semicircle: $$\frac{a}{2}$$ - Radius of small semicircle: $$\frac{a}{4}$$ - Area of large semicircle: $$\frac{1}{2} \pi \left(\frac{a}{2}\right)^2 = \frac{1}{2} \pi \frac{a^2}{4} = \frac{\pi a^2}{8}$$ - Area of small semicircle: $$\frac{1}{2} \pi \left(\frac{a}{4}\right)^2 = \frac{1}{2} \pi \frac{a^2}{16} = \frac{\pi a^2}{32}$$ - Total shaded area = $$\frac{\pi a^2}{8} + \frac{\pi a^2}{32} = \frac{4\pi a^2}{32} + \frac{\pi a^2}{32} = \frac{5\pi a^2}{32}$$ 4. **Problem 6(i): Arc length of the semicircle** - Semicircle radius = half of side of square (assumed side length not given, so let side = s) - Arc length of semicircle: $$\pi r$$ 5. **Problem 6(ii): Perimeter of the badge** - Perimeter = sum of square sides + arc length of sector + arc length of semicircle - Square perimeter = $$4s$$ - Arc length of quarter circle sector: $$\frac{1}{4} \times 2\pi r = \frac{\pi r}{2}$$ - Arc length of semicircle: $$\pi r$$ - Total perimeter = $$4s + \frac{\pi r}{2} + \pi r = 4s + \frac{3\pi r}{2}$$ 6. **Problem 6(iii): Ratio of area of ABC sector to area of badge** - Area of sector ABC (quarter circle): $$\frac{1}{4} \pi r^2$$ - Area of badge = Area of square + Area of semicircle - overlapping parts (assumed negligible) - Area of square: $$s^2$$ - Area of semicircle: $$\frac{1}{2} \pi r^2$$ - Ratio = $$\frac{\frac{1}{4} \pi r^2}{s^2 + \frac{1}{2} \pi r^2}$$ 7. **Problem 6(iv): Sketching triangle with area equal to semicircle and base DE** - Area of semicircle: $$\frac{1}{2} \pi r^2$$ - Triangle base DE = length of side of square = s - Triangle area formula: $$\frac{1}{2} \times \text{base} \times \text{height}$$ - Set equal: $$\frac{1}{2} s h = \frac{1}{2} \pi r^2 \Rightarrow h = \frac{\pi r^2}{s}$$ - Sketch triangle with base DE = s and height = $$\frac{\pi r^2}{s}$$ **Final answers:** - 5(i) Shaded area = $$49\left(1 - \frac{\pi}{4}\right)$$ cm² - 5(ii) Shaded area = $$6.125\pi$$ cm² - 5(iii) Shaded area = $$\frac{5\pi a^2}{32}$$ - 6(i) Arc length semicircle = $$\pi r$$ - 6(ii) Perimeter badge = $$4s + \frac{3\pi r}{2}$$ - 6(iii) Ratio area sector to badge = $$\frac{\frac{1}{4} \pi r^2}{s^2 + \frac{1}{2} \pi r^2}$$ - 6(iv) Triangle height = $$\frac{\pi r^2}{s}$$ with base DE = s