Given: klmn is a trapezoid m∠k = 90°, m∠n = 45° lk = lm = 10 find: kn, area of klmn

Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and isosceles, because
m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to sq. units.

The graph isn’t that clear