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.