A Norman window has the shape of a rectangle surmounted by a semicircle Problem 
A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is $24$ ft, express the area, $A$, of the window as a function of the width, $x$, of the window.

I came up with the following but I feel like it's off.
$$A(x)=24x-4x^2+\frac{\pi\left(\frac x2\right)^2}2$$
Can anyone verify?
 A: Given the width $x$, the perimeter of the semicircle is $\pi \frac x2$ and the bottom is $x$.  The vertical side of the rectangle is then $\frac 12(24-x(1+\frac \pi 2))$  The area of the semicircle is $\frac \pi 8 x^2$ so the total area is $\frac \pi 8 x^2+\frac 12(24-x(1+\frac \pi 2))x$  It looks like you missed the fact that there are two vertical sides to the rectangle and didn't subtract the perimeter of the semicircle from $24$ to get the perimeter of the rectangle.  That is a guess as you didn't show how you found your answer.
A: If the width is $x$, so that the radius is $x/2$, then the semi-circumference is $\pi x/2$, so the semi-circle plus the bottom of the window have a total length of $x+\pi x/2$.  That means the amount remaining for the two vertical segments is $24-(x+\pi x/2)$, so each of those has length $12-(x+\pi x/2)/2$.  The area of the rectangular part is therefore $x(12-(x+\pi x/2)/2)$, and the area of the semi-circular part is $\pi(x/2)^2/2$.  So the total area is
$$
x\left( 12 - \frac{2+\pi}4 x \right) + \frac{\pi x^2}8 = 12x - \frac{4+\pi}8 x^2.
$$
