ABCD is a rectangle and BCE is an isosceles triangle in which BC = CE. What is the area of the given:

$\left(\frac{b}{2}+a\right) b$

We know that,

BC = CE = b

So, ABC is a right angle triangle and its area will be = \(\frac{1}{2}\) × b × b  = \(\frac{b^2}{2}\)

Now the area of quadrilateral ABCD = b × a

So, the area of the given figure = \(\frac{b^2}{2}\) + b × a = $\left(\frac{b}{2}+a\right) b$