1. **State the problem:** We have a solid made of a rectangular block with a square base 6 cm by 6 cm and height 10 cm, with a right pyramid on top having the same square base and apex V such that VA = VB = VC = VD = 5 cm. The solid is cut into two identical halves along the plane of symmetry VAFHCV. We need to find the surface area of one of the halves. 2. **Understand the solid:** - The rectangular block has base ABCD with side 6 cm and height 10 cm. - The pyramid sits on top of the block with apex V and base EFGH (same square 6 cm by 6 cm). - VA = VB = VC = VD = 5 cm means the pyramid edges from apex to base vertices are 5 cm. 3. **Calculate the total surface area of the solid before cutting:** - Surface area of the rectangular block (excluding the top face covered by the pyramid): - Base ABCD: area = $6 \times 6 = 36$ cm² - Four vertical faces: each $6 \times 10 = 60$ cm², total $4 \times 60 = 240$ cm² - Surface area of the pyramid: - Base EFGH is the same square 6 cm by 6 cm, area = 36 cm² (this is the top face of the block, so it is internal between block and pyramid, not exposed) - Four triangular faces of the pyramid. 4. **Calculate the slant height of the pyramid faces:** - The pyramid has apex V above the square base EFGH. - Since VA = 5 cm, and the base side is 6 cm, the center of the base is at the midpoint of EFGH. - The height of the pyramid $h_p$ can be found using Pythagoras: - Half diagonal of base square = $\frac{6\sqrt{2}}{2} = 3\sqrt{2}$ cm - Using $VA = 5$ cm as the edge from apex to vertex, the height $h_p$ satisfies: $$h_p = \sqrt{5^2 - (3\sqrt{2})^2} = \sqrt{25 - 18} = \sqrt{7} \approx 2.6458 \text{ cm}$$ 5. **Calculate the slant height of each triangular face:** - Each triangular face has base 6 cm. - The slant height $l$ is the height of the triangle from apex V to midpoint of a base edge. - Distance from center to midpoint of a side = half side = 3 cm. - Using Pythagoras: $$l = \sqrt{h_p^2 + 3^2} = \sqrt{7 + 9} = \sqrt{16} = 4 \text{ cm}$$ 6. **Area of one triangular face:** $$\frac{1}{2} \times 6 \times 4 = 12 \text{ cm}^2$$ - Total area of four triangular faces = $4 \times 12 = 48$ cm² 7. **Total surface area of the solid:** - Block sides + pyramid faces (excluding base of pyramid since it is on block): $$240 + 48 + 36 = 324 \text{ cm}^2$$ 8. **Cutting the solid into two halves along plane VAFHCV:** - The plane cuts the solid into two identical halves. - The cut exposes a new surface equal to the cross-section along the plane. 9. **Calculate the area of the cross-section (cut surface):** - The plane passes through points V, A, F, H, C, V. - This cross-section is a polygon formed by these points. - The cross-section includes: - Triangle VAF (pyramid side) - Rectangle AFHC (part of block side) - Triangle VHC (pyramid side) 10. **Calculate areas of these parts:** - Triangle VAF and VHC are right triangles with base 6 cm and height 5 cm (edges VA and VC). - Area of each triangle = $\frac{1}{2} \times 6 \times 5 = 15$ cm² - Rectangle AFHC has sides 6 cm (AF) and 10 cm (height of block). - Area = $6 \times 10 = 60$ cm² 11. **Total area of cross-section:** $$15 + 60 + 15 = 90 \text{ cm}^2$$ 12. **Surface area of one half after cutting:** - Half of the original surface area (excluding the cut) plus the cross-section area: - Original surface area without base = 324 cm² - Half of original surface area = $\frac{324}{2} = 162$ cm² - Add cross-section area = 90 cm² - Total surface area of one half = $162 + 90 = 252$ cm² **Final answer:** $$\boxed{252 \text{ cm}^2}$$