Answer
(a) $A=(\frac{x}{4})^2+(\frac{10-x}{4})^2$
(b) $x=5cm$
Work Step by Step
(a) The side length of the first square is $\frac{x}{4}$ and its area is $(\frac{x}{4})^2$.
The side length of the second square is $\frac{10-x}{4}$ and its area is $(\frac{10-x}{4})^2$.
The total area would be $A=(\frac{x}{4})^2+(\frac{10-x}{4})^2$
(b) As shown in the graph, the minimum of the area happens when $x=5cm$ with $A=3.125cm^2$