Find the side of an (equilateral) triangle inscribed in a circle whose radius is 4√3 / 2.

Let’s build the height BH, which is also the median and bisector of the triangle, then AH = CH = AC / 2.

The center of the circumscribed circle near a regular triangle is the point of intersection of the medians, which are divided at point O in the ratio 2/1.

Then OH = BО / 2 = R / 2 = (4 * √ (3/2)) / 2 = 2 * √ (3/2) see.

BН = BО + ОН = 4 * √ (3/2) + 2 * √ (3/2) = 6 * √ (3/2) cm.

Let the length of the side of the triangle be X cm, then AB = X cm, AH = X / 2 cm.

By the Pythagorean theorem, BH ^ 2 = AB ^ 2 – AH ^ 2.

54 = X ^ 2 – X ^ 2/4 = 3 * X ^ 2/4.

X ^ 2 = 72

X = AB = 6 * √2 cm.

Answer: The side of the triangle is 6 * √2 cm.