Find the radius of a circle inscribed in an equilateral triangle if its side is 3√2.

Since the triangle ABC is equilateral, the center of the inscribed circle is the intersection point of the bisectors, medians and heights of the triangle.

An equilateral triangle has all internal angles of 60.

Then the area of the triangle ABC is equal to: Saavs = AB * AC * Sin60 / 2 = (3 * √2) 2 * (√3 / 4) =

9 * √3 / 2 cm2.

Also Savs = AC * BH / 2.

BН = 2 * Saс / АС = 2 * (9 * √3 / 2) / 3 * √2 = 6 * √3 / √2 = 3 * √6 / 2 cm.

Point O divides the median BH in the ratio 2/1, starting from the top, then the radius OH = BH / 3 = 3 * √6 / (2 * 3) = √6 / 2 cm.

Answer: The radius of the inscribed circle is √6 / 2 cm.