In an isosceles triangle ABC, angle B is equal to 120 degrees, points M and H are the midpoints of sides AB

| July 5, 2021 | education

In an isosceles triangle ABC, angle B is equal to 120 degrees, points M and H are the midpoints of sides AB and BC. accordingly, AC = 4√3. 1) find the area of the triangle ABC 2) find the distance between the midpoints of the segments AM and HC.

Triangle ABC is isosceles, AB = BC.

Let’s build the height, bisector and median BK, then AK = CK = AC / 2 = 4 * √3 / 2 = 2 * √3 cm. Angle ABK = 120/2 = 60.

In a right-angled triangle ABK, tg60 = AK / BK.

BK = AK / tg60 = 2 * √3 / √3 = 2 cm.

Determine the area of the triangle ABC.

Savs = AC * BK / 2 = 4 * √3 * 2/2 = 4 * √3 cm2.

The segment MH is the middle line of the triangle ABC then MH = AC / 2 = 4 * √3 / 2 = 2 * √3 cm.

The quadrilateral AMNC is an isosceles trapezoid, and the segment DE is its middle line, then DE = (MH + AC) / 2 = (2 * √3 + 4 * √3) / 2 = 3 * √3 cm.

Answer: The area is 4 * √3 cm2, the distance is 3 * √3 cm.



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