Because \(BC=5cm\) => \(AD=5cm\)

The total area of triangle AMD and triangle BCN is:

\(\left(\dfrac{2.5}{2}\right).2=10\left(cm^2\right)\)

Then the area of trapezoid MNCD is:

\(60-10=50\left(cm^2\right)\)

So the answer is: \(50cm^2\)

The rectangular or trapezoidal length is large 

60 : 5 = 12 ( cm )

That trapezoidal baby is :

12 - 2 - 2 = 8 ( cm )

Trapezoidal area MNCD is 

\(\dfrac{\left(8+12\right)\times5}{2}=50\)( cm² )

This is the 1st level solution 

Because ABCD is a rectangle so AD = BC = 5cm

The total area of the two triangles AMD and BNC are:

\(\dfrac{5\cdot2}{2}+\dfrac{5\cdot2}{2}=5+5=10\left(cm^2\right)\)

The area of the MNCD quadrilateral is:

\(60-10=50\left(cm^2\right)\)

Answer: 50cm²