= Solution
Write $\theta_i$ for the internal <angle> at $v_i$, using the numbering in part (b), and use the <hyperbolic plane> with <Gaussian curvature> $-1$. The necessary and sufficient conditions are \b[equal <lengths> for each pair of equally labelled <polygon sides>, and total <angle> $2\pi$ at each vertex <equivalence class>]. Explicitly, the two <angle> conditions are
$$
\boxed{\theta_0+\theta_1+\theta_3+\theta_4+\theta_6+\theta_8=2\pi,\qquad
\theta_2+\theta_5+\theta_7+\theta_9=2\pi.}
$$
Necessity follows because the identification must preserve <arc length> along each seam, and a smooth interior point of a <Riemannian surface> has total <angle> $2\pi$. For sufficiency, equally long <hyperbolic geodesic> segments can be identified by <isometries>. Across their interiors, two half-disc <neighbourhoods> join to a hyperbolic disc. At an identified vertex, the incident sectors form a cyclic disc with total <angle> $2\pi$, so no <cone singularity of a surface metric> remains. These <local isometries> therefore extend the <Riemannian metric> smoothly everywhere. This is the <smooth hyperbolic polygon gluing criterion>.
The ten internal <angles> sum to $4\pi$. The <area of a hyperbolic geodesic polygon> consequently gives
$$
\boxed{\operatorname{Area}(X)=(10-2)\pi-4\pi=4\pi.}
$$
Equivalently, the <Gauss-Bonnet theorem> gives $-\operatorname{Area}(X)=2\pi\chi(X)=-4\pi$.
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