Solution:

Let the tangent lengths from $A, B, C, D, E$ to the incircle be $a, b, c, d, e.$ Then $a + b = 5,$ $b + c = 6,$ $c + d = 6,$ $d + e = 6,$ and $e + a = 7.$ The middle equations give $d = b$ and $e = c,$ so $c + a = 7;$ with $a + b = 5$ and $b + c = 6$ this yields $a = 3,$ $b = d = 2,$ $c = e = 4.$ If $r$ is the inradius, the interior angle at a vertex with tangent length $t$ satisfies $\tan\frac{\theta}{2} = \frac{r}{t},$ and the half-angles sum to half of $540^\circ:$ $$\arctan\frac{r}{3} + 2\arctan\frac{r}{2} + 2\arctan\frac{r}{4} = 270^\circ.$$

Let $\beta = \arctan\frac{r}{2}$ and $\gamma = \arctan\frac{r}{4}.$ Then $\arctan\frac{r}{3} = 270^\circ - 2(\beta + \gamma),$ and since $\tan(270^\circ - \theta) = \cot\theta,$ we get $\frac{r}{3} = \cot 2(\beta + \gamma).$ With $T = \tan(\beta + \gamma) = \frac{r/2 + r/4}{1 - r^2/8} = \frac{6r}{8 - r^2},$ the identity $\frac{r}{3} = \frac{1 - T^2}{2T}$ becomes $2rT = 3(1 - T^2),$ and substituting $T$ and clearing denominators gives $$5r^4 - 84r^2 + 64 = 0,$$ so $r^2 = 16$ or $r^2 = \frac{4}{5}.$ For $r^2 = \frac{4}{5}$ every half-angle is well under $54^\circ,$ so the half-angle sum falls far short of $270^\circ;$ this root is extraneous. Hence $r = 4.$

The semiperimeter is $s = \frac{5 + 6 + 6 + 6 + 7}{2} = 15,$ so the area is $rs = 4 \cdot 15 = 60.$