The locations of M and N are arbitrary, so we can choose them to make the problem easy to solve.
Let M = B. Then ∆CDM = ∆CDB = 18 cm².
Let N = D. Then ∆ABN = ∆ABD = 23 cm².
Diagonal BD divides the trapezoid into triangles ABD and CBD. The area of the trapezoid is the sum of the areas of these triangles:
... 23 cm² + 18 cm² = 41 cm² . . . . the area of the trapezoid.
