Frustum Surface Area 067E54

1. **State the problem:** We have a frustum formed by cutting a smaller square-based pyramid from a larger similar pyramid. The frustum has trapezium-shaped faces all the same size. 2. **Given data:** - Height of the frustum (vertical distance between the two square bases) is $17 + 17 = 34$ mm. - Side length of the larger square base is $32$ mm. - The smaller square base side length is unknown; we will find it using similarity. 3. **Find the side length of the smaller square base:** Since the pyramids are similar, the ratio of corresponding lengths is the same. Let the height of the larger pyramid be $H$ and the smaller pyramid be $h$. The frustum height is $H - h = 34$ mm. The side length of the larger base is $32$ mm. Let the smaller base side length be $x$ mm. Because the slant edge is split into two equal parts of $17$ mm each, the height ratio is: $$\frac{h}{H} = \frac{17}{34 + 17} = \frac{17}{51} = \frac{1}{3}$$ Therefore, the smaller pyramid height $h = \frac{1}{3}H$. 4. **Using similarity for side lengths:** Side lengths scale with height, so: $$\frac{x}{32} = \frac{h}{H} = \frac{1}{3}$$ Thus, $$x = \frac{32}{3} \approx 10.67 \text{ mm}$$ 5. **Calculate the surface area of the frustum:** The surface area includes: - The area of the larger square base: $$A_{large} = 32^2 = 1024 \text{ mm}^2$$ - The area of the smaller square top: $$A_{small} = \left(\frac{32}{3}\right)^2 = \frac{1024}{9} \approx 113.78 \text{ mm}^2$$ - The area of the four trapezium-shaped faces. 6. **Find the slant height of the frustum:** The slant height $l$ is the length of the trapezium side (the slant edge of the frustum). Using Pythagoras theorem on the trapezium face: The difference in half side lengths of bases: $$\frac{32 - \frac{32}{3}}{2} = \frac{32 \times \frac{2}{3}}{2} = \frac{32}{3} \approx 10.67 \text{ mm}$$ The vertical height of the frustum is $34$ mm. So, $$l = \sqrt{34^2 + 10.67^2} = \sqrt{1156 + 113.78} = \sqrt{1269.78} \approx 35.63 \text{ mm}$$ 7. **Calculate the area of one trapezium face:** The trapezium has parallel sides of lengths $32$ mm and $\frac{32}{3}$ mm, and height (slant height) $l$. Area of trapezium: $$A_{trap} = \frac{(32 + \frac{32}{3})}{2} \times 35.63 = \frac{(32 + 10.67)}{2} \times 35.63 = \frac{42.67}{2} \times 35.63 = 21.33 \times 35.63 \approx 760.1 \text{ mm}^2$$ 8. **Total lateral surface area:** There are 4 trapezium faces, so: $$4 \times 760.1 = 3040.4 \text{ mm}^2$$ 9. **Total surface area of the frustum:** $$A_{total} = A_{large} + A_{small} + 4 \times A_{trap} = 1024 + 113.78 + 3040.4 = 4178.18 \text{ mm}^2$$ **Final answer:** $$\boxed{4178.18 \text{ mm}^2}$$