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Compute the shaded area shown on the picture.\FRAME{dtbpF}{2.9568in}{2.4267in%
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Solution: \ First we will compute the area of right triangle $ACD.$ \ \ $%
AC=15\unit{in}$ and $AD=17\unit{in}$.%
\begin{equation*}
A_{\text{ACD}}=\dfrac{15\unit{in}\cdot 17\unit{in}}{2}=\dfrac{255\unit{in}%
^{2}}{2}=127.5\unit{in}^{2}
\end{equation*}%
To get the area of the shaded region, we will subtract the areas of the
rectangle $ABFE$ and right triangles $DEF$ and $BCF$. \ 
\begin{eqnarray*}
A_{\text{ABFE}} &=&7\unit{in}\cdot 4\unit{in}=28\unit{in}^{2} \\
A_{\text{DEF}} &=&\dfrac{10\unit{in}\cdot 4\unit{in}}{2}=20\unit{in}^{2} \\
A_{\text{BCF}} &=&\dfrac{7\unit{in}\cdot 11\unit{in}}{2}=\dfrac{77\unit{in}%
^{2}}{2}=38.5\unit{in}^{2}
\end{eqnarray*}%
So the shaded area is when we subtract the white areas from the big right
triangle.%
\begin{eqnarray*}
A &=&A_{\text{ACD}}-\left( A_{\text{ABFE}}+A_{\text{DEF}}+A_{\text{BCF}%
}\right)  \\
&=&127.5\unit{in}^{2}-\left( 28\unit{in}^{2}+20\unit{in}^{2}+38.5\unit{in}%
^{2}\right) =127.5\unit{in}^{2}-86.5\unit{in}^{2}=41\unit{in}^{2}
\end{eqnarray*}

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