The height of an equilateral triangle is 13√3. Find its perimeter.

Consider an equilateral triangle ABC, in which all sides are equal: AB = BC = AC.

Suppose that the height equal to 13√3 is drawn to the side of the AC, we will designate it as BO. Then the triangle ABO is rectangular, in which AB is the hypotenuse, BO and OA are the legs.

Let us denote the side AB by x.

Since in an equilateral triangle, the height is also the median of the side to which it is drawn, then AO = AC / 2 = AB / 2 = x / 2.

According to the Pythagorean theorem, in a right-angled triangle, the sum of the squares of the legs is equal to the square of the hypotenuse:

(13√3) ^ 2 + (x / 2) ^ 2 = x ^ 2;

507 + (x ^ 2) / 4 = x ^ 2;

2028 + x ^ 2 = 4 * x ^ 2;

2028 = 4x ^ 2 – x ^ 2;

2028 = 3x ^ 2;

x ^ 2 = 2028: 3;

x ^ 2 = 676;

x = 26.

We find that the side AB is equal to 26. This means that the sides BC and AC are equal to 26.

Let’s calculate the perimeter of a given triangle as the sum of the lengths of all its sides:

P = AB + BC + AC = 26 + 26 + 26 = 26 * 3 = 78.

Answer: 78.