In trapezium ABCD, the base of AD and BC are respectively equal to 15 and 5, the angle CDA = 60.

| September 2, 2021 | education

In trapezium ABCD, the base of AD and BC are respectively equal to 15 and 5, the angle CDA = 60. A straight line is drawn through the vertex B and the middle of the CD-point O until the intersection. with continuation of AD at point E. angle ABE = 90, angle CBE = 30. find the perimeter of the trapezoid.

Determine the value of the angle ВСD. The sum of the angles of the trapezoid at the lateral side is 180, then the angle ВСD = 180 – 60 = 120.

In the ВCO triangle, the ВOС angle = 180 – 120 – 30 = 30. Then the ВOС triangle is isosceles, and ВС = СО = 5 cm.

By condition, point O is the middle of the lateral side of CD, then DО = СО = 5 cm, and СD = 2 * СО = 2 * 5 = 10 cm.

By condition, AB is perpendicular to BE, then the angle ABC = ABE + OBC = 90 + 30 = 120.

Since the angle ABC = BCO = 120, the trapezoid ABCD is isosceles, which means AB = CD = 10 cm.

Then the perimeter of the triangle is:

P = AB + BC + CD + AD = 10 + 5 + 10 + 15 = 40 cm.

Answer: The perimeter of the trapezoid is 40 cm.



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