The radius of a circle inscribed in a regular triangle is 7 find the side of this triangle.

The center of the inscribed circle is the intersection point of the bisectors (in a regular triangle, the bisectors, heights and medians are equal, and this is generally the same thing)
Bisectors are divided in a ratio of 2: 1 counting from the top
the radius drawn to the side at right angles is 1/3 of the entire bisector, so the entire bisector is 21
Let’s say we have a triangle ABE, AC is the bisector
We got a right-angled triangle ABC
In it we know the side and 3 angles (AC = 21, angle ACB = 90b, angle ABC = 60 degrees, because the triangle is correct, angle BAC = 30 degrees)
find the side AB by the sine theorem
AB = AC * sin C / sin B = 42 / √3
you can also find the side of a regular triangle by the formula
r = a * √3 / 6
express a and substitute 7 for r
a = 7 * 6 / √3 = 42 / √3
If we get rid of irrationality, then it will be like this:
42 / √3 * √3 / √3 = 42√3 / 3 = 14√3