In an isosceles triangle ABC AB = AC = 13 cm, BC = 10 cm. Find the distance from the point

In an isosceles triangle ABC AB = AC = 13 cm, BC = 10 cm. Find the distance from the point of intersection of the medians of the triangle to the vertex A

AB = AC = 13; AM is the median in the ABC triangle, drawn to the BC side, BM = MC, but since AB = BC, this is the median AM and the height to the BC side.

Consider a triangle (rectangular) ABM: find AM ^ 2 = AB ^ 2 – BM ^ 2 = 13 ^ 2 – (10/2) ^ 2 = 169 – 5 ^ 2 = 169 – 25 = 144, whence the height (the median is ) AM = √144 = 12 cm.

Distance OA = (2/3) * AM = (2/3) * 12 cm = 8 cm.