The diagonal AC of the trapezoid ABCD (AB || CD) divides it into two similar triangles.

The diagonal AC of the trapezoid ABCD (AB || CD) divides it into two similar triangles. Find SABCD if AB = 25cm, BC = 20cm, AC = 15cm.

use Heron’s formula
S of triangle = root of ((p * (p – a) * (p – b) * (p – c)), where p is a semiperimeter
p = (a + b + c) / 2
substituting the known values, we get
p = (25 + 20 + 15) / 2 = 60/2 = 30
S of triangle ABC = root of ((30 * (30 – 25) * (30 – 20) * (30 – 15))) = root of ((3 * 10 * 5 * 10 * 3 * 5) = 5 * 3 * 10 = 150
then in the likeness of triangles
k ^ 2 = S CDA / S ABC k = AD / BC = DC / AC = AC / AB = 15/25 = 3/5
9/25 = S CDA / 150 from here
S CDA = 150 * 9/25 = 6 * 9 = 54
S trapezoid = S ABC + S CDA = 150 + 54 = 204