# The diagonal AC of the trapezoid ABCD, angle B = 90 degrees, divides it into two right-angled isosceles triangle

**The diagonal AC of the trapezoid ABCD, angle B = 90 degrees, divides it into two right-angled isosceles triangles with angle ACD = 90 degrees. Find the midline of a trapezoid if the area of the triangle is АСD = 144 cm2**

Consider a right-angled triangle АСD, in which АС = СD, and its area, by condition, is equal to 144 cm2.

Sasd = AC * CD / 2 = AC2 / 2 = 144.

AC = √ 2 * 144 = 12 * √2 cm.

AC = CD = 12 * √2 cm.

By the Pythagorean theorem, we define the hypotenuse AD, which is the great base of the trapezoid.

AD ^ 2 = AC ^ 2 + CD ^ 2 = 2 * AC ^ 2 = 2 * 144 * 2 = 576.

AD = 24 cm.

Consider a right-angled triangle ABC, in which AB = BC, AC = 12 * √2.

Let AB = BC = X cm.

Then, by the Pythagorean theorem, AC ^ 2 = AB ^ 2 + BC ^ 2 = 2 * X ^ 2 = (12 * √2) 2.

288 = 2 * X ^ 2.

X = √144 = 12 cm.

AB = BC = 12 cm.

Then, the middle line of the trapezoid is: KM = (BC + AD) / 2 = (12 + 24) / 2 = 18 cm.

Answer: The middle line of the trapezoid is 18 cm.