1. **Problem Statement:** We have several problems involving volumes and areas of prisms, cylinders, and tanks. --- ### Problem 1: Swimming Pool Cross-Section and Volume 2. **Given:** - Depth at AE = 2.1 m - Depth at BC = 1 m - AB = 6.5 m - BC = 1 m - ED = 3 m - Width of pool = 4.6 m 3. **(i) Find length of DC:** - Since AE and BC are vertical depths, and ED is vertical 3 m, DC is horizontal. - Using the polygon properties and right angles, length DC = AB - BC = 6.5 - 1 = 5.5 m. 4. **(ii) Find area of uniform cross-section:** - The cross-section is a polygon composed of rectangles and triangles. - Area = sum of areas of rectangles and triangles. - Calculate area of trapezium or polygon using coordinates or decomposition. - Using average depth: average depth = (2.1 + 1)/2 = 1.55 m - Area = length AB * average depth = 6.5 * 1.55 = 10.075 m² 5. **(b) Find volume of pool:** - Volume = cross-sectional area * width - Volume = 10.075 * 4.6 = 46.345 m³ --- ### Problem 2: Volume of Solid with Semicircular Top 6. **Given:** - Base length = 22 mm - Semicircle radius = 17 mm 7. **Volume formula:** - Volume = base area * height - Base area = rectangle area + semicircle area - Rectangle area = length * width (width not given, assume uniform cross-section) - Semicircle area = \( \frac{1}{2} \pi r^2 \) 8. **Calculate semicircle area:** - \( \frac{1}{2} \pi (17)^2 = \frac{1}{2} \pi 289 = 144.5\pi \approx 454.0 \text{ mm}^2 \) 9. **Total cross-section area:** - Assuming width = 22 mm (same as length), total area = rectangle + semicircle = 22 * 22 + 454 = 484 + 454 = 938 mm² 10. **Volume:** - Volume = cross-section area * height (height not given, problem incomplete) --- ### Problem 3: Cylindrical Can Fuel Volume 11. **Given:** - Diameter = 6 cm - Height = 12 cm - Volume of liquid = 320 cm³ 12. **(a)(i) Calculate volume of can:** - Radius = 3 cm - Volume = \( \pi r^2 h = \pi (3)^2 (12) = 108\pi \approx 339.29 \text{ cm}^3 \) 13. **(a)(ii) Volume of empty space:** - Empty space = total volume - liquid volume = 339.29 - 320 = 19.29 cm³ 14. **(a)(iii) Height of gap:** - Volume of liquid = \( \pi r^2 h_{liquid} = 320 \) - \( h_{liquid} = \frac{320}{\pi (3)^2} = \frac{320}{28.274} \approx 11.31 \text{ cm} \) - Height of gap = total height - liquid height = 12 - 11.31 = 0.69 cm 15. **(b) Number of cans to fill tank:** - Tank dimensions: 22 cm x 12 cm x 17 cm - Tank volume = 22 * 12 * 17 = 4488 cm³ - Number of cans = \( \left\lfloor \frac{4488}{339.29} \right\rfloor = 13 \) cans --- ### Problem 4: Rain Volume on Roof 16. **Given:** - Roof area = 20 m * 8 m = 160 m² - Rain depth = 1 cm = 0.01 m 17. **Volume of water:** - Volume = area * depth = 160 * 0.01 = 1.6 m³ --- **Final answers:** - 1(i) Length DC = 5.5 m - 1(ii) Area cross-section = 10.075 m² - 1(b) Volume pool = 46.345 m³ - 3(a)(i) Volume can = 339.29 cm³ - 3(a)(ii) Empty space = 19.29 cm³ - 3(a)(iii) Height gap = 0.69 cm - 3(b) Number of cans = 13 - 4 Volume rain = 1.6 m³