1. **State the problem:** We have a pyramid VABCD with square base ABCD of side 20 cm. The angle between any sloping edge (e.g., VA) and the base plane ABCD is 55°. 2. **Calculate the slant height:** The sloping edges VA, VB, VC, and VD form an angle of 55° with the base plane. Each sloping edge length can be used to find the height of the pyramid. 3. **Find the height of the pyramid (V perpendicular to base):** The sloping edge length makes an angle 55° with the base, so the height $h$ satisfies: $$\cos 55^\circ = \frac{h}{\text{sloping edge length}} \implies h = \text{sloping edge length} \times \cos 55^\circ$$ We don't know the sloping edge length yet, so first find it. 4. **Find the sloping edge length:** The base square has side 20 cm, so half the diagonal is: $$\frac{20\sqrt{2}}{2} = 10\sqrt{2} \approx 14.142 \text{ cm}$$ The apex V is right above the center of the square. The sloping edge VA stretches from vertex A to apex V. From apex V down to center O of square, vertical height is $h$, and from center O to vertex A on the base is 14.142 cm. The sloping edge length $l$ is the hypotenuse of right triangle VOA: $$l = \sqrt{h^2 + (10\sqrt{2})^2} = \sqrt{h^2 + 200}$$ 5. **Using the angle 55° between sloping edge and base plane:** $$\cos 55^\circ = \frac{h}{l} = \frac{h}{\sqrt{h^2 + 200}}$$ Square both sides: $$\cos^2 55^\circ = \frac{h^2}{h^2 + 200}$$ Rearrange: $$h^2 + 200 = \frac{h^2}{\cos^2 55^\circ}$$ $$h^2 \left( \frac{1}{\cos^2 55^\circ} - 1 \right) = 200$$ Calculate: $$\cos 55^\circ \approx 0.5736, \quad \cos^2 55^\circ \approx 0.329$$ $$\frac{1}{0.329} - 1 = 3.04 - 1 = 2.04$$ So, $$h^2 \times 2.04 = 200 \implies h^2 = \frac{200}{2.04} \approx 98.04$$ $$h \approx 9.90 \text{ cm}$$ 6. **Calculate slant height (triangle on each face):** The slant height is the height of each triangular face along the midpoint of a side to apex. In a square pyramid, slant height $s$ relates to height $h$ and half the side length: $$s = \sqrt{h^2 + 10^2} = \sqrt{9.90^2 + 100} = \sqrt{98.01 + 100} = \sqrt{198.01} \approx 14.07 \text{ cm}$$ 7. **Calculate surface area:** - Base area: $20 \times 20 = 400$ cm² - Area of each triangular face: $\frac{1}{2} \times \text{base side} \times s = \frac{1}{2} \times 20 \times 14.07 = 140.7$ cm² - There are 4 triangular faces. Total area of triangles: $$4 \times 140.7 = 562.8 \text{ cm}^2$$ Total surface area: $$400 + 562.8 = 962.8 \text{ cm}^2$$ 8. **Round to 2 significant figures:** $$963 \approx 960 \text{ cm}^2$$ **Final answer:** The surface area of the pyramid is approximately **960 cm²**.