A frustum is formed by removing a small square-based pyramid from a similar, larger one, as shown below. Each trapezium-shaped face of the frustum is the same size. Work out the surface area of the frustum.

The frustum is formed from similar pyramids. Let $s_1$ be the side length of the top base, and $s_2$ be the side length of the bottom base.
From the image, the slant height of the removed small pyramid is $L_1 = 41 \text{ mm}$.
The slant height of the frustum's lateral face is $L_2 = 41 \text{ mm}$.
The total slant height of the original large pyramid is $L_{total} = L_1 + L_2 = 41 \text{ mm} + 41 \text{ mm} = 82 \text{ mm}$.
The side length of the bottom base is $s_2 = 36 \text{ mm}$.
Due to similarity, the ratio of corresponding lengths is equal:

$$\frac{s_1}{s_2} = \frac{L_1}{L_{total}}$$ $$\frac{s_1}{36 \text{ mm}} = \frac{41 \text{ mm}}{82 \text{ mm}}$$ $$\frac{s_1}{36 \text{ mm}} = \frac{1}{2}$$ $$s_1 = \frac{1}{2} \times 36 \text{ mm} = 18 \text{ mm}$$

Area of top base ($A_{top}$) = $s_1^2$

$$A_{top} = (18 \text{ mm})^2 = 324 \text{ mm}^2$$

Area of bottom base ($A_{bottom}$) = $s_2^2$

$$A_{bottom} = (36 \text{ mm})^2 = 1296 \text{ mm}^2$$

Each lateral face is a trapezoid with parallel sides $s_1 = 18 \text{ mm}$ and $s_2 = 36 \text{ mm}$, and height (slant height of the frustum) $L_2 = 41 \text{ mm}$.
The area of one trapezoid ($A_{trap}$) is given by the formula $\frac{1}{2}(a+b)h$.

$$A_{trap} = \frac{1}{2}(s_1 + s_2)L_2$$ $$A_{trap} = \frac{1}{2}(18 \text{ mm} + 36 \text{ mm})(41 \text{ mm})$$ $$A_{trap} = \frac{1}{2}(54 \text{ mm})(41 \text{ mm})$$ $$A_{trap} = 27 \text{ mm} \times 41 \text{ mm} = 1107 \text{ mm}^2$$

Since there are four identical lateral faces:

$$A_{lateral} = 4 \times A_{trap}$$ $$A_{lateral} = 4 \times 1107 \text{ mm}^2 = 4428 \text{ mm}^2$$

The total surface area ($A_{total}$) is the sum of the areas of the top base, bottom base, and the four lateral faces.

$$A_{total} = A_{top} + A_{bottom} + A_{lateral}$$ $$A_{total} = 324 \text{ mm}^2 + 1296 \text{ mm}^2 + 4428 \text{ mm}^2$$ $$A_{total} = 1620 \text{ mm}^2 + 4428 \text{ mm}^2$$ $$A_{total} = 6048 \text{ mm}^2$$

The surface area of the frustum is $\text{6048 mm}^2$.