The base AB of an isosceles triangle ABC = 12 cm, its medians AM and BK intersect at point O

The base AB of an isosceles triangle ABC = 12 cm, its medians AM and BK intersect at point O, the angle AOB = 120 *. Find these medians.

Consider the AOC triangle. Since the medians of an isosceles triangle are equal and at the point of intersection are divided in the ratio 2/1, then AO = CO, therefore the AOC triangle is isosceles, and its angles at the base will be equal: angle A = C = (180 – 120) / 2 = 300.

Then by the theorem of sines: AC / Sin 120 = AO / Sin 30.

12 / (√3 / 2) = AO / (1/2).

AO = 6 / (√3 / 2) = 12 / √3 = 4 * √3.

The medians of the triangle, at the point of intersection, are divided in a ratio of 2/1, then AO / OM = 2/1.

ОМ = AO / 2 = 2 * √3.

Then M = CK = 2 * √3 + 4 * √3 = 6 * √3.

Answer: The medians are 6 * √3 cm.