To answer this question we will use the following diagram as a reference:
The semicircle's radius is 8in.
Notice that the area of the paperboard that remains is the area of the rectangular paper board minus the area of the semicircle.
The semicircle's area is:
[tex]A_s=\frac{\pi(8in)^2}{2}=\frac{64\pi}{2}in^2=32\pi in^2\approx100.48in^2.[/tex]
The rectangle's area is:
[tex]A_r=16in*25in=400in^2.[/tex]
Then the area of the paperboard that remains:
[tex]\begin{gathered} A=400in^2-100.48in^2 \\ =299.52in^2. \end{gathered}[/tex]
Answer:
[tex]299.52in^2.[/tex]
