Given ABCD -trapezoid, AB = CD AC perpendicular to CD, BC = 7 cm AD = 25 cm, AC = 20 cm find: CD, area ABC.

In the right-angled triangle ACD, according to the Pythagorean theorem, we find the leg CD.
CD ^ 2 = AD ^ 2-AC ^ 2;
Substitute the numerical values of the lengths of the hypotenuse and leg.
CD ^ 2 = 25 ^ 2-20 ^ 2 = 625-400 = 225;
CD = √225 = 15 (cm);
We find the area of the triangle ABC according to Heron’s formula:
S = √p (p-AB) (p-BC) (p-AC); p is a semi-perimeter.
p = (AB + BC + AC) / 2;
p = (15 + 7 + 20) / 2 = 42/2 = 21 (cm).
S = √21 (21-15) (21-7) (21-20) = √21 * 6 * 14 * 1 = √1764 = 42 (cm ^ 2).