In a right-angled triangle ABC, BC = 9 cm, the medians of the triangle intersect at point O

In a right-angled triangle ABC, BC = 9 cm, the medians of the triangle intersect at point O, OB = 10 cm. Find the area of triangle ABC.

By the property of the medians of the triangle, at the point O, the point of their intersection, they are divided in a ratio of 2/1 starting from the vertex, then BO / MO = 2/1.

MO = BО / 2 = 10/2 = 5 cm, then BM = MO + BО = 5 + 10 = 15 cm.

The BCM triangle is rectangular, then by the Pythagorean theorem, CM ^ 2 = BM ^ 2 – BC ^ 2 = 225 – 81 – 144.

CM = 12 cm.

Since BM is the median, then AM = CM = 12 cm, and then AC = 2 * CM = 2 * 12 = 24 cm.

Determine the area of ​​the triangle ABC.

Savs = AC * BC / 2 = 24 * 9/2 = 108 cm2.

Answer: The area of ​​the triangle is 108 cm2.

In a right-angled triangle ABC AB = 6 cm, AC = 10 cm. Points F and T are the midpoints of sides AB and BC, respectively. Calculate the area of ​​the triangle BFT.