1. **Problem statement:** (a) Calculate the volume of water from 1 cm of rain on a roof measuring 20 m by 8 m. (b) Determine if a tank with a given cross-section and height can hold all the rainwater. 2. **Formula and concepts:** - Volume of rainwater = area of roof × rainfall depth. - Volume of tank = cross-sectional area × height. - Cross-sectional area is composite: rectangle + 2 semicircles. 3. **Calculations for (a):** - Convert rainfall depth to meters: 1 cm = 0.01 m. - Roof area = $20 \times 8 = 160$ m$^2$. - Volume of rainwater = $160 \times 0.01 = 1.6$ m$^3$. 4. **Calculations for (b):** - Cross-section dimensions: - Diameter of semicircle = 1 m, so radius $r = \frac{1}{2} = 0.5$ m. - Rectangle height = 75 cm = 0.75 m. - Rectangle width = diameter of semicircle = 1 m. - Area of rectangle = width $\times$ height = $1 \times 0.75 = 0.75$ m$^2$. - Area of one semicircle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (0.5)^2 = \frac{1}{2} \pi \times 0.25 = 0.125\pi$ m$^2$. - Area of two semicircles = $2 \times 0.125\pi = 0.25\pi$ m$^2$. - Total cross-sectional area = rectangle area + semicircles area = $0.75 + 0.25\pi$ m$^2$. - Height of tank = 60 cm = 0.6 m. - Volume of tank = cross-sectional area $\times$ height = $(0.75 + 0.25\pi) \times 0.6$ m$^3$. 5. **Evaluate volume of tank:** - Approximate $\pi \approx 3.1416$. - $0.25\pi \approx 0.7854$. - Cross-sectional area $\approx 0.75 + 0.7854 = 1.5354$ m$^2$. - Volume of tank $\approx 1.5354 \times 0.6 = 0.92124$ m$^3$. 6. **Comparison and conclusion:** - Volume of rainwater = 1.6 m$^3$. - Volume of tank = 0.92124 m$^3$. - Since $1.6 > 0.92124$, the tank is not big enough to hold all the rainwater. **Final answer:** (a) Volume of rainwater = $1.6$ m$^3$. (b) The tank volume is approximately $0.92$ m$^3$, which is less than the rainwater volume, so the tank cannot hold all the rainwater.