In the triangle ABC, AA1 and BB1 are medians, AA1 = 12 cm, BB1 = 15 cm. The medians intersect at point O

| April 5, 2021 | education

In the triangle ABC, AA1 and BB1 are medians, AA1 = 12 cm, BB1 = 15 cm. The medians intersect at point O, and the angle AOB = 120 degrees. Find the area of the triangle.

Consider the part ABA1B1, consisting of 4 ABO triangles; AA1O; A1B1O; AOB1:

C1 = AO * OB * sin 120 °; C2 = OB * A1O * sin (180 ° – 120 °) = OB * A1O * sin 60 °; C3 = A1O * OB1 * sin 120 °; C4 = OB1 * OA * sin (180 ° – 120 °) = OB1 * OA * sin 60 °.

AO = 2/3 * 12 = 8 cm; ОВ = 2/3 * 15 cm = 10 cm; ОА1 = 1/3 * 12 = 4 cm; ОВ1 = 1 / * 15 = 5 cm.The areas are equal:

C1 = 8 * 10 * sin 120 ° / 2 = 80 * √3 / 2 * 2 = 20 * √3;

C2 = 10 * 4 * √3 / 2 * 2 = 10 * √3; C3 = 4 * 5 * √3 / 2 * 2 = 5 * √3; C4 = 5 * 8 * √3 / 2 * 2 = 10 * √3; C0 = C1 + C2 + C3 + C4 = (20 + 10 + 5 + 10) = 45 (cm2).

C (ABC) = 4/3 * C0 = 60 cm2.



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