In triangle ABC BC = A AC = B angle C = alpha, medians AA1 and BB1 intersect at point O

In triangle ABC BC = A AC = B angle C = alpha, medians AA1 and BB1 intersect at point O, find the area of triangle AOB1.

The area of a triangle is equal to half the product of the sides of the triangle by the sine of the angle between them.

Sас = (АС * ВС * SinАСВ) / 2 = (a * b * Sin α) / 2.

Let’s draw the median CC1.

By the property of the medians of the triangle, they divide the ABC triangle into six equal triangles. Then Saov1 = Saavs / 6 = ((a * b * Sin α) / 2) / 6 = (a * b * Sin α) / 12 cm2.

Answer: The area of the triangle AOB1 is (a * b * Sin α) / 12 cm2.