# The bases of the trapezoid are 8 and 12, and one of the acute angles is 30 degrees.

The bases of the trapezoid are 8 and 12, and one of the acute angles is 30 degrees. The continuation of the sides intersect at an angle of 90 degrees. Find the height of the trapezoid.

According to the condition, the angle AOD = 90, then the triangle AOD is rectangular, with an angle ODA = 30. The leg AO lies opposite the angle 30, and therefore, its length is equal to half the length of the hypotenuse AD.

AO = AD / 2 = 12/2 = 6 cm.

Triangles AOD and BOC are similar in acute angle, the angle ODA = OCB = 30, as are the corresponding angles at the intersection of parallel straight lines DA and CB of the secant DO.

Then:

AD / BC = AO / BO.

12/8 = 6 / VO.

ВO = 8 * 6/12 = 4 cm.

In the triangle BOС, according to the Pythagorean theorem, the leg of the OС will be equal to:

OS ^ 2 = BC ^ 2 = ВO ^ 2 = 82 – 42 = 64 – 16 = 48.

OS = 4 * √3 cm.

AD / BC = DO / CO.

12/8 = DO / 4 * √3.

DO = 12 * 4 * √3 / 8 = 6 * √3 cm.

Then SD = DO – OС = 6 * √3 – 4 * √3 = 2 * √3 cm.

Let us determine the height of CH from the triangle DСН, which lies opposite the angle 30 and is equal to half the length of СD.

СН = СD / 2 = 2 * √3 / 2 = √3 cm.

Answer: The height of the trapezoid is √3 cm.

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