In an isosceles trapezoid, the acute angle is 60 degrees, the bisector of this angle divides the smaller

| September 10, 2021 | education

In an isosceles trapezoid, the acute angle is 60 degrees, the bisector of this angle divides the smaller base equal to 16 cm in half. You need to find the middle line of the trapezoid.

Since AK is the bisector of the angle BAD, it cuts off on a smaller base a segment equal to the lateral side. BK = AK. According to the condition, point K is the middle of the base BC, then BK = CK = BC / 2 = 16/2 = 8 cm, then AB = CD = 8 cm.

Let’s draw the Height BH. In a right-angled triangle ABH, we determine the length of the leg AH.

Cos60 = AH / AB. AH = AB * Cos60 = 8 * 1/2 = 4 cm.

Since the trapezoid is isosceles, the length AH is equal to the half-difference of the lengths of the bases.

AH = (AD – BC) / 2.

AD = 2 * AH + BC = 2 * 4 + 16 = 24 cm.

Determine the length of the midline. КР = (ВС + АD) / 2 = (16 + 24) / 2 = 20 cm.

Answer: The length of the middle line is 20 cm.



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