Here's an alternative method of solution.
Assuming that the triangle DEF lies 'outside' the triangle ABC, the area of DEF is equal to the area of ABC plus the area of three trapezia.
So, calling the length of side of DEF L,
$$\frac{1}{2}L^{2}\sin 60 = \frac{1}{2}10^{2}\sin 60+\frac{3}{2}(10+L),$$
Multiplying throughout by 4,
$$L^{2}\sqrt{3}=100\sqrt{3}+6(10+L),$$
$$\sqrt{3}(L^{2}-100)=\sqrt{3}(L+10)(L-10)=6(10+L),$$
$$\sqrt{3}(L-10)=6,$$
$$L=2\sqrt{3}+10,$$
so the length of the perimeter is $$6\sqrt{3}+30.$$
Assuming that DEF lies 'within' ABC leads to, (following the same procedure), $$L=10-2\sqrt{3}.$$
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