In an isosceles trapezoid ABCD, the height of the ВC forms an angle of 30 ° with the lateral side.

In an isosceles trapezoid ABCD, the height of the ВC forms an angle of 30 ° with the lateral side. A) Find the corners of the trapezoid; b) find the perimeter of the trapezoid if the smaller base and side are 5 cm; c) prove that the midpoints of the sides of the trapezoid are the vertices of the rhombus.

а.

Since the trapezoid is isosceles, its angles at the base are equal, the angle CDA = BAD = 30.

The sum of the angles at the side of the trapezoid is 180, then the angle ABC = BCD = 180 – 30 = 150.

Answer: The angles of the trapezoid are 30, 150, 150, 30.

b.

Let’s build the height of the HВ trapezoid АBСD. Triangle ABH is rectangular, then CosBAH = AH / AB.

AH = AB * Cos30 = 5 * √3 / 2 cm.

In an isosceles trapezoid, the height whitens the larger base into two segments, the length of the smaller one being equal to the half-difference of the lengths of the bases. AH = (AD – BC) / 2.

АD = 2 * АН + ВС = 5 * √3 + 5 = 5 * (1 + √3) see.

Then Ravsd = 5 + 5 + 5 + 5 * (1 + √3) = 15 + 5 * (1 + √3) = 5 * (4 + √3) cm.

Answer: The perimeter of the trapezoid is 5 * (4 + √3) cm.

с.

Let’s construct the diagonals AC and BD. In triangles ABC and ACD, the segments KM and OP are their midlines. Then the CM and OP are parallel to the AC, which means that the CM is parallel to the OP. Similarly, MP is parallel to OK. Then KMRO is a parallelogram.

KP is the middle line of the trapezoid, MO is its height, then KP is perpendicular to MO. In the parallelogram of the KMPO, the diagonals are perpendicular, hence the KMPO is a rhombus, as required.