Questions: Determine the surface area of this composite object, which is a right cylinder and a hemisphere, to the nearest tenth of a square metre. HINT: One end of the cylinder is covered, so should not be included in the final answer.

Solution Steps

The radius of the hemisphere is \(r = 2.5 \div 2 = 1.25\) m. The surface area of a sphere is \(4\pi r^2\), so the surface area of the hemisphere is half of that, plus the circular base:

\( \text{Area}_\text{hemisphere} = \frac{1}{2} \times 4\pi r^2 + \pi r^2 = 2\pi r^2 + \pi r^2 = 3\pi r^2 = 3\pi (1.25)^2 = 3\pi(1.5625) \approx 14.73 \text{ m}^2\)

The height of the cylinder is \(h = 6.5\) m, and the radius is \(r = 1.25\) m. The lateral surface area of the cylinder is given by the formula \(2\pi rh\). One end of the cylinder is covered by the hemisphere, so it's not included in the surface area calculation. Thus, \( \text{Area}_\text{cylinder} = 2\pi rh = 2\pi(1.25)(6.5) = 16.25\pi \approx 51.05 \text{ m}^2 \)

The total surface area is the sum of the hemisphere's surface area and the lateral surface area of the cylinder: \( \text{Total Area} = \text{Area}_\text{hemisphere} + \text{Area}_\text{cylinder} \approx 14.73 + 51.05 = 65.78 \text{ m}^2\)

Final Answer