In an isosceles trapezoid MHKP angle M = 90 degrees Angle K = 150 degrees, HK = 2 cm, diagonal MK

| September 17, 2021 | education

In an isosceles trapezoid MHKP angle M = 90 degrees Angle K = 150 degrees, HK = 2 cm, diagonal MK is perpendicular to the lateral side of KP. find the middle line of the trapezoid

The MK diagonal is perpendicular to the KР side, then the angle НKM = НKРR – MKР = 150 – 90 = 60.

Then, in a right-angled triangle MНK, the angle HMK = 90 – 60 = 30. The leg of the НK lies opposite the angle 30, then MK = 2 * НK = 2 * 2 = 4 cm.

In a right-angled triangle MKP, the angle PMK = НKM = 60 as the cross-lying angles at the intersection of parallel lines НK and MP secant MK.

Then Cos60 = MK / MP.

MР = MK / Cos60 = 4 / (1/2) = 8 cm.

Determine the length of the midline of the trapezoid. AB = (НK + MР) / 2 = (2 + 8) / 2 = 5 cm.

Answer: The length of the middle line is 5 cm.



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