Diagonals of an isosceles trapezoid are perpendicular to each other. the length of its leg is 26 cm. an altitude from the vertex of an obtuse angle to the base divides the longer base into two parts, such that the shorter part is 10 cm long. what is the area of this trapezoid?

Here's how the answer is generated:

Let the smaller base be = a cm
Then bigger base = 20 + a cm

Sides of the bigger inner triangle formed by the diagonals = (20 + a) / sqr (2)

(from pythagorean theorem or using cos 45 rule)

Sides of the smaller inter triangle formed by the diagonals = a/sqr (2)
(from pythagorean theorem or using cos 45 rule)

height = 24 (pythagorean)

side triangle formed by leg and diagonals (pythagorean):

26^2 = (20 + a) ^ 2/2 + a^2/2
a=14

then
20 + a =34

A = 1/2 * 24 * (14 + 34) = 576 cm^2

The 3rd quadrant as 3x+g5ti=56i: 92d-543