8. (15 points) In a quadrilateral composed of 2013 small equilateral triangles with side lengths of 1, the minimum perimeter is


【Solution】Solution: According to the problem, equilateral triangles form two types of quadrilaterals: parallelograms and trapezoids. Parallelograms require an even number of triangles, but 2013 is odd, so only trapezoids, specifically isosceles trapezoids, can be formed.

Let the length of the lower base of the trapezoid be $a$ and the length of the upper base be $b$. Then the length of the legs is $(a-b)$, so the perimeter is $(a+b)+2$ $(a-b)$.

Since $a^{2}-b^{2}=(a+b)(a-b)=2013=3 \times 11 \times 61$, the product is fixed, and the difference is smaller when the sum is smaller, so: $(a+b) \times 2$ $(a-b)=2 \times 3 \times 11 \times 61=61 \times 66$,

When $a+b=61$ and $2(a-b)=66$, the difference is smaller, and the sum is smaller, resulting in the minimum perimeter: $66+61=127$. (No need to solve for $a=47, b=14$). Therefore, the answer is 127.