We need to subtract the area of the semicircle that was cut out of the paperboard from its original area.
The original area of the paperboard was the area of a rectangle 32in long and 24 in wide:
[tex]32in\cdot24in=768in^{2}[/tex]
And the area of the semicircle, noticing that its radius r is 24in/2, is given by:
[tex]\frac{\pi r^2}{2}=\frac{3.14\cdot(12in)^{2}}{2}=\frac{3.14\cdot144in^2}{2}=226.08in^2[/tex]
Thus, the area of the paperboard that remains is:
[tex]768in^2-226.08in^2=(768-226.08)in^{2}=541.92in^{2}[/tex]
Therefore, the answer is 541.92 in².
