In triangle ABC AA1 and BB1-medians AA1 = 12 BB1 = 15 Medians intersect at point O angle AOB = 120 Sabc-?

univerkov | March 28, 2021 | education

Since AA1 and BB1 are medians, at point O they are divided into segments in a ratio of 2/1 starting from vertices A and B.

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

ОА = 2 * АА1 / 3 = 2 * 12/3 = 8 cm, ОА1 = АА1 – ОА = 12 – 8 = 4 cm.

Determine the area of the triangle AOB.

Saov = ОА * ОВ * Sin120 / 2 = 8 * 10 * √3 / 4 = 20 * √3 cm2.

Determine the area of the triangle AOB1.

Saov1 = ОА * ОВ1 * Sin60 = 8 * 5 * √3 / 4 = 10 * √3 cm2.

Then Savv1 = Saov + Saov1 = 20 * √3 + 10 * √3 = 30 * √3 cm2.

The median BB1 divides the ABC triangle into two equal triangles, then Savs = 2 * Savv1 = 2 * 30 * √3 = 60 * √3 cm2.

Answer: The area of the triangle ABC is 60 * √3 cm2.

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