We have :

\(\dfrac{1}{3}+\dfrac{1}{31}+\dfrac{1}{35}+\dfrac{1}{37}+\dfrac{1}{47}+\dfrac{1}{53}+\dfrac{1}{61}\)

\(=\dfrac{1}{3}+\left(\dfrac{1}{31}+\dfrac{1}{35}+\dfrac{1}{37}\right)+\left(\dfrac{1}{47}+\dfrac{1}{53}+\dfrac{1}{61}\right)\)

\(< \dfrac{1}{3}+\left(\dfrac{1}{30}+\dfrac{1}{30}+\dfrac{1}{30}\right)+\left(\dfrac{1}{45}+\dfrac{1}{45}+\dfrac{1}{45}\right)\)

\(=\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{1}{2}\)

=> \(\dfrac{1}{3}+\dfrac{1}{31}+\dfrac{1}{35}+\dfrac{1}{37}+\dfrac{1}{47}+\dfrac{1}{53}+\dfrac{1}{61}< \dfrac{1}{2}\)

There are three ways to choose hundred of digits 

There are three dozen digit selectors

There are two ways to select the unit - number 

It is possible to write three digit whole number 

3 x 3 x 2 = 18 ( number )

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