now, notice the picture below
is really just one rectangle and a circle, well, half a circle or a semi-circle
so.... if you get the area of the whole rectangle, 16*19
and subtract, the area of half the circle, you end up with the shaded area
since, what you'd be leftover is, the rectangle, with a semi-circular hole, that was subtracted
[tex]\bf \textit{area of a circle}\\\\
A=\pi r^2\qquad
\begin{cases}
r=radius=\frac{diameter}{2}\\
----------\\
diameter=16\\
r=\frac{16}{2}=8
\end{cases}\implies A=64\pi \\\\
-----------------------------\\\\
\textit{shaded area}\implies (16\cdot 19)\quad -\quad 64\pi [/tex]
now... let's say, you get an amount of hmmm [?]
ok, how much is that in percentage? well, 16*19 is 304
so, if we take 304 as the 100%, how much is [?] in percentage?
[tex]\bf \begin{array}{ccllll}
amount&\%\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
304&100\\
\boxed{?}&x
\end{array}\implies \cfrac{304}{\boxed{?}}=\cfrac{100}{x}[/tex]
solve for "x"
