1. **State the problem:** We have a frustum formed by removing a smaller square pyramid from a larger similar square pyramid. The base edge length of the frustum is 36 mm, and the height is divided into two equal segments of 41 mm each, so the total height of the original larger pyramid is $41 + 41 = 82$ mm. 2. **Identify given data:** - Smaller pyramid height $h_s = 41$ mm - Larger pyramid height $h_l = 82$ mm - Base edge length of frustum (bottom square) $a = 36$ mm - The top square edge length of the frustum is unknown, call it $b$ 3. **Use similarity of pyramids:** Since the pyramids are similar, the ratio of corresponding edges equals the ratio of their heights. $$\frac{b}{a} = \frac{h_s}{h_l} = \frac{41}{82} = \frac{1}{2}$$ So, $$b = \frac{a}{2} = \frac{36}{2} = 18 \text{ mm}$$ 4. **Calculate the slant height of the frustum:** The frustum height $h_f = h_l - h_s = 82 - 41 = 41$ mm. The difference in base edges is $a - b = 36 - 18 = 18$ mm. Half the difference is $\frac{18}{2} = 9$ mm. The slant height $l$ of the trapezium face is found by Pythagoras theorem: $$l = \sqrt{h_f^2 + 9^2} = \sqrt{41^2 + 9^2} = \sqrt{1681 + 81} = \sqrt{1762} \approx 41.97 \text{ mm}$$ 5. **Calculate the lateral surface area:** Each trapezium face has parallel sides $a = 36$ mm and $b = 18$ mm, and height (slant height) $l \approx 41.97$ mm. Area of one trapezium face: $$A_{face} = \frac{(a + b)}{2} \times l = \frac{36 + 18}{2} \times 41.97 = 27 \times 41.97 \approx 1133.19 \text{ mm}^2$$ There are 4 such faces, so total lateral surface area: $$A_{lateral} = 4 \times 1133.19 = 4532.76 \text{ mm}^2$$ 6. **Calculate the areas of the two square bases:** Bottom base area: $$A_{bottom} = a^2 = 36^2 = 1296 \text{ mm}^2$$ Top base area: $$A_{top} = b^2 = 18^2 = 324 \text{ mm}^2$$ 7. **Calculate total surface area of the frustum:** $$A_{total} = A_{lateral} + A_{bottom} + A_{top} = 4532.76 + 1296 + 324 = 6152.76 \text{ mm}^2$$ **Final answer:** The surface area of the frustum is approximately **6153 mm\textsuperscript{2}**.