# 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.