Triangle ABC is equilateral with sides 10 cm long. Triangle DEF is formed by drawing lines parallel to, and 1 cm away from, the sides of tri

My solution assumes that DEF is bigger than ABC, though you don't actually specify this in the question.

First, look at the diagram:

With reference to this we have:

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WOW Alan,  that looks very impressive!    

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),  $$L=10-2\sqrt{3}.$$

I like your solution, zacismyname......I would do this exactly the same way......

Hi Zacismyname,

Welcome to the forum.    :D

I think I may nickname you Zac   LOL