In the triangle ABC, the medians BB₁ = 15 and CC₁ = 18, intersect at the point O, the angle BOS = 90 ° Find the area ABC?

Since BB1 and CC1 are the medians of the triangle ABC, point O divides them in a ratio of 2/1 starting from the vertices of the triangle.

Then ОВ = 10 cm, ОВ1 = 5 cm, ОВ = 12 cm, ОВ1 = 6 cm.

Triangles ОВ1С, ОВС, OC1В are rectangular, then, by the Pythagorean theorem:

BC ^ 2 = OC ^ 2 + OB ^ 2 = 144 + 100 = 244.

BC = 2 * √61 cm.

BC1 ^ 2 = OB ^ 2 + OC1 ^ 2 = 100 + 36 = 136.

BC1 = 2 * √34 cm.

Then AB = 2 * 2 * √34 = 4 * √34 cm.

CB1 ^ 2 = OC ^ 2 + OB1 ^ 2 = 144 + 25 = 169.

CB1 = 13 cm.

AC = 2 * CB1 = 2 * 13 = 26 cm.

Then Ravs = 26 + 4 * √34 + 2 * √61 = 68.94 cm.

Answer: The perimeter of the triangle is 68.94 cm.