The base AB of an isosceles triangle ABC is equal to 12cm; its median AM and BK intersect at point
| June 2, 2021 | education
The base AB of an isosceles triangle ABC is equal to 12cm; its median AM and BK intersect at point O and the angle AOB = 120 degrees. Find the medians
The medians drawn from the angles of the base of an isosceles triangle are equal, CK = AM, then AO = CO.
From the triangle AOC, by the cosine theorem, AC ^ 2 = OA ^ 2 + CO ^ 2 – 2 * AO * CO * Cos120.
144 = 2 * AO ^ 2 – 2 * AO ^ 2 * (-1 / 2).
144 = 2 * AO ^ 2 * (1 + 0.5).
AO ^ 2 = 72 / 1.5 = 48.
AO = 6 * √3 cm.
Then OM = 6 * √3 / 2 = 3 * √3 cm.
AM = CK = 6 * √3 + 3 * √3 = 9 * √3 cm.
Answer: The length of the medians AM and CK are equal to 9 * √3 cm.
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