this looks more like a maths olympiad type of problem.

this problem can be solved using differentiation to find the maximum area. however, since the problem has been put under E Maths, and E Maths students do not learn differentiation, here is a possible way to solve using non-calculus method.

let length of shaded rectangle be x.
by considering similar triangles, the width of the shaded rectangle can be found to be ...
(3/5)*(4-(4/5)*x)
= 2.4-0.48x
therefore area of shaded rectangle
= x times (2.4-0.48x)
= 2.4x-0.48x^2

since the expression is quadratic, we can do completing the square to find that the maximum area occurs when x=2.5 .

for maximum area of shaded rectangle,
length = 2.5 cm
width = 1.2 cm
area = 3 sq cm