In trapezoid ABCD, diagonal BD is perpendicular to the lateral side AB, angle ABD = angle BDC = 30 degrees

In trapezoid ABCD, diagonal BD is perpendicular to the lateral side AB, angle ABD = angle BDC = 30 degrees, find the length hell if the perimeter of the trapezoid is 60 cm.

Since, according to the condition ∠ADВ = ∠ВDC = 30, АВС = 180 – АDВ – ВDC = 180 – 30 – 30 = 120.

By condition BD is perpendicular to AB, so ∠ABD = 90, then ∠DBC = 120 – 90 = 30.

In the BCD triangle, the angles at the base of BD are equal, which means that the triangle is isosceles and BC = CD.

Let’s define the angle ВСD. ∠BCD = 180 – 30 – 30 = 120, then ∠BAD = 180 – 120 = 60.

The angles at the base of the AD trapezoid are 60, which means that the trapezoid is isosceles, AB + CD = BC.

Let’s draw two heights of the trapezoid from the vertices B and C.

Since the trapezoid is isosceles, the segments AH and MD, cut off by heights are equal, AH = MD.

In triangle ABE, leg AH is located opposite angle 30, therefore, AH = AB / 2, and in triangle CMD, similarly, MD = CD / 2.

Let AB = X cm, then AB = BC = CD = HM = X, and AH = MD = X / 2.

Then, knowing the perimeter of the trapezoid, we write:

P = AB + BC + CD + MD + NM + AH = 4 * AB + 2 * (AB / 2).

60 = 4 * X + 2 * X / 2 = 5 * X.

X = 60/5 = 12 cm.

Then the base AD = 12/2 + 12 + 12/2 = 24 cm.

Answer: BP = 24 cm.