Pentacular

By the law of cosines, $\overline{BF}^2=50^2+50^2-2\cdot50^2\cos\left(\frac{2\pi}{5}\right)$ .

That is, $\overline{BF}\approx58.8$ , so the answer is $\boxed{58}$ .

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Area of triangle ABC = (1/2) BC x AB x sin(108 degree)

Area of triangle BCF = (1/2) BC x BF x sin(54 degree)

Observe that both triangle, when duplicated, form the rhombus ABCF.

Therefore, their areas are equal.

AB sin(108 degree) = BF sin(54 degree)

50 sin(108 degree) = BF sin(54 degree)

BF = 50 sin(108 degree) / sin(54 degree) = 100 cos(54 degree)

The result follows from putting it in a calculator.

The last step can be ignored if the double-angle formula is unknown.