# In triangle ABC, the medians BB1 and CC1 intersect at point O and are equal to 15 cm and 18 cm, respectively

**In triangle ABC, the medians BB1 and CC1 intersect at point O and are equal to 15 cm and 18 cm, respectively. Find the perimeter of triangle ABC if the angle is BOC = 90 degrees.**

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

Then OC = 2 * CC1 / 3 = 2 * 18/3 = 12 cm, OC1 = CC1 / 3 = 6 cm.

OB = 2 * BB1 / 3 = 2 * 15/3 = 10 cm, OB1 = BB1 / 3 = 5 cm.

Since CC1 is perpendicular to BB1, then in a right-angled triangle BOC according to the Pythagorean theorem, BC ^ 2 = OC ^ 2 + OB ^ 2 = 144 + 100 = 244.

BC = √136 = 2 * √61 cm.

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

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

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

BC1 = 2 * √34 cm, then AB = 4 * √34 cm.

Then the perimeter of the triangle ABC is:

Ravs = 2 * √61 + 4 * √34 + 13 cm.

Answer: The perimeter is 2 * √61 + 4 * √34 + 13 cm.