The diagram below shows a bucket with top diameter 30cm and bottom diameter 20cm. The height of the bucket is 28cm. Find (a) The capacity of the bucket in litres. (b) The area of the metal sheet required to make 100 such buckets, taking 10% extra overlapping and wastage.

The formula for the volume of a frustum is $V = \frac{1}{3}\pi h (R_1^2 + R_1 R_2 + R_2^2)$.

$$V = \frac{1}{3}\pi (28) (15^2 + 15 \times 10 + 10^2)$$ $$V = \frac{28}{3}\pi (225 + 150 + 100)$$ $$V = \frac{28}{3}\pi (475)$$ $$V = \frac{13300}{3}\pi \text{ cm}^3$$

Since $1 \text{ litre} = 1000 \text{ cm}^3$:

$$V_{litres} = \frac{13300\pi}{3 \times 1000}$$ $$V_{litres} = \frac{133\pi}{30} \text{ litres}$$

Using $\pi \approx 3.14159$:

$$V_{litres} \approx \frac{133 \times 3.14159}{30} \approx 13.92968 \text{ litres}$$

Rounding to three significant figures:
The capacity of the bucket is $\boxed{13.9 \text{ litres}}$.

The formula for the slant height is $l = \sqrt{h^2 + (R_1 - R_2)^2}$.

$$l = \sqrt{28^2 + (15 - 10)^2}$$ $$l = \sqrt{28^2 + 5^2}$$ $$l = \sqrt{784 + 25}$$ $$l = \sqrt{809} \text{ cm}$$

The formula for the lateral surface area of a frustum is $A_{lateral} = \pi (R_1 + R_2) l$.

$$A_{lateral} = \pi (15 + 10) \sqrt{809}$$ $$A_{lateral} = 25\pi \sqrt{809} \text{ cm}^2$$

The bottom base is a circle with radius $R_2$.

$$A_{base} = \pi R_2^2 = \pi (10)^2$$ $$A_{base} = 100\pi \text{ cm}^2$$

$$A_{bucket} = A_{lateral} + A_{base}$$ $$A_{bucket} = 25\pi \sqrt{809} + 100\pi$$ $$A_{bucket} = \pi (25\sqrt{809} + 100) \text{ cm}^2$$

$$A_{100\_buckets} = 100 \times A_{bucket}$$ $$A_{100\_buckets} = 100\pi (25\sqrt{809} + 100) \text{ cm}^2$$

$$A_{required} = A_{100\_buckets} \times (1 + 0.10)$$ $$A_{required} = 100\pi (25\sqrt{809} + 100) \times 1.10$$ $$A_{required} = 110\pi (25\sqrt{809} + 100) \text{ cm}^2$$

Now, calculate the numerical value:

$$A_{required} \approx 110 \times 3.14159 \times (25 \times \sqrt{809} + 100)$$ $$A_{required} \approx 110 \times 3.14159 \times (25 \times 28.442923 + 100)$$ $$A_{required} \approx 110 \times 3.14159 \times (711.073075 + 100)$$ $$A_{required} \approx 110 \times 3.14159 \times 811.073075$$ $$A_{required} \approx 280319.64 \text{ cm}^2$$

Converting to square meters ($1 \text{ m}^2 = 10000 \text{ cm}^2$):

$$A_{required} \approx \frac{280319.64}{10000} \text{ m}^2 \approx 28.031964 \text{ m}^2$$

Rounding to three significant figures:
The area of the metal sheet required is $\boxed{28.0 \text{ m}^2}$.