Chapter 11.2, Problem 17PSB

To determine

To find the area of the shaded triangular region AFC.

Answer to Problem 17PSB

The area of the shaded triangular region AFC is $33$ .

Explanation of Solution

Given:

  $\begin{array}{l}AD=8,\\ BD=5,\\ BE=6,\\ EC=12.\end{array}$

Formula Used:

  $\begin{array}{l}Areaofrectangle=b\ast h\\ b=Base,\\ h=Height\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  $\begin{array}{l}Areaofrectangle=b\ast h\\ A(\square DBEF)=DB*BE=5*6=30\end{array}$

  $\begin{array}{l}DF=BE=6\\ Areaoftriangle=\frac{1}{2}\ast b\ast h\\ A(\Delta ADF)=\frac{1}{2}*AD*DF\\ =\frac{1}{2}*8*6\\ =24\end{array}$

  $\begin{array}{l}FE=BD=5\\ A(\Delta FEC)=\frac{1}{2}*EC*FE\\ =\frac{1}{2}*12*5\\ =30\end{array}$

  $\begin{array}{l}AB=AD+DB=8+5=13\\ BC=BE+EC=6+12\\ A(\Delta ABC)=\frac{1}{2}*BC*AB\\ =\frac{1}{2}*18*13\\ =117\end{array}$

  $\begin{array}{l}Areaofshadedregion=A(\Delta ABC)-A(\square DBEF)-A(\Delta FEC)-A(\Delta ADF)\\ =117-30-30-24\\ =117-84\\ =33\end{array}$