The area was lost. Please find it.

Extend the seven slant edges to meet at the vertex $V$ of the non-truncated pyramid. Let $h$ be the distance from the midpoint of an $x+2$ side to that vertex. Then the height of each trapezium forming the sides is $\tfrac{2h}{x+2}$ , and so the lateral area is $L \; = \; 7 \times \tfrac12(x + x+2) \times \frac{2h}{x+2} \; = \; \frac{14(x+1)h}{x+2}$ If $d(x)$ is the distance from the midpoint of a heptagon of side $x$ to the centre, then $d(x) = \tfrac12x \cot\tfrac{\pi}{7}$ , and the vertical height of the vertex $V$ above the base is $H \; = \; \sqrt{h^2 - d(x+2)^2} \; = \; \frac{x+2}{14(x+1)}\sqrt{L^2 - 49(x+1)^2\cot^2\tfrac{\pi}{7}}$ so the vertical height of the frustrum of the pyramid is $K \; = \; \tfrac{2}{x+2}H \;= \; \frac{1}{7(x+1)}\sqrt{L^2 - 49(x+1)^2\cot^2\tfrac{\pi}{7}}$ and hence $(x+1)^2 \; = \; \frac{L^2}{49(K^2 + \cot^2\frac{\pi}{7})}$ The sum of the areas of the two heptagons is $7 \big[\tfrac12 x d(x) + \tfrac12(x+2)d(x+2)\big] \; = \; \tfrac74(x^2 + (x+2)^2)\cot\tfrac{\pi}{7} \; = \; \tfrac72((x+1)^2 + 1)\cot\tfrac{\pi}{7} \; = \; \frac{7}{2\tan\frac{\pi}{7}}\left[ \frac{L^2}{49(K^2 + \cot^2\frac{\pi}{7})} + 1\right]$ With $K = 9$ and $L = 63\sqrt{82}$ , we obtain $573.1076$ as the sum of the heptagonal areas, which makes the answer $\boxed{573}$ .