In trapezoid ABCD (AD and BC – bases) angle A = 90g. angle С = 135gr. AB = 2 Find the centerline

| February 9, 2021 | education

In trapezoid ABCD (AD and BC – bases) angle A = 90g. angle С = 135gr. AB = 2 Find the centerline of the trapezoid if its diagonal is perpendicular to the side.

Let’s draw the height of the CH. Quadrangle ABСН is a rectangle, then CH = AB = 2 cm.

In a right-angled triangle СDН, the angle is DСН = ВСD – ВСН = 135 – 90 = 450.

Then the angle СDН= 90 – 45 = 450, which means that the triangle СDН is isosceles, DН = CH = 2 cm.

Consider a triangle AСD, in which angle C = 900, angle D = 450, then angle A = 180 – 90 – 45 = 450, which means that the triangle is isosceles, AC = СD.

In an isosceles triangle of AСD, the height of CH is also the median of the triangle, then AH = DН = 2 cm, and then BC = AH = 2 cm.

AD = AН + DН = 2 + 2 = 4 cm.

Determine the length of the midline of the trapezoid.

KM = (BC + AD) / 2 = (2 + 4) / 2 = 3 cm.

Answer: The middle line of the trapezoid is 3 cm.



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