Chapter 12.3, Problem 46PPSE

Solution

To calculate: The area of the figure as shown below. What shapes can be area to be founded. Can the polygon be broken into these shapes and how many shapes will be needed to break the polygon into.

  

The area of polygon is 44 square units.

Given information:

The figure as follows.

  

Formula Used:

The area of triangle is calculated as $Area=\frac{1}{2}\left|\det A\right|$ where $\det \left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right]=a_1{b}_2{c}_3+b_1{c}_2{a}_3-\left(a_3{b}_2{c}_1+b_3{c}_2{a}_1+c_3{a}_2{b}_1\right)$ .

Calculation :

Draw the following diagram.

  

The area of the polygon can be divided into area of four triangles.

Split the polygon into four parts.

The area of triangle $AFE$ with coordinates $\left(2,5\right)$ , $\left(-4,3\right)$ and $\left(-4,1\right)$ .

  $\begin{array}{c}{\Delta }_1=\frac{1}{2}\left[\begin{array}{ccc}2& 5& 1\\ -4& 3& 1\\ -4& 1& 1\end{array}\right]\\ =\frac{1}{2}\left[2\left(3-1\right)-5\left(-4+4\right)+1\left(-4+12\right)\right]\\ =\frac{1}{2}\left[4-0+8\right]\\ =6\end{array}$

The area of triangle $AED$ with coordinates $\left(2,5\right)$ , $\left(-4,1\right)$ and $\left(1,-4\right)$ .

  $\begin{array}{c}{\Delta }_2=\frac{1}{2}\left[\begin{array}{ccc}2& 5& 1\\ -4& 1& 1\\ 1& -4& 1\end{array}\right]\\ =\frac{1}{2}\left[2\left(1+4\right)-5\left(-4-1\right)+1\left(16-1\right)\right]\\ =\frac{1}{2}\left[10+25+15\right]\\ =25\end{array}$

The area of triangle $ADC$ with coordinates $\left(2,5\right)$ , $\left(1,-4\right)$ and $\left(4,1\right)$ .

  $\begin{array}{c}{\Delta }_3=\frac{1}{2}\left[\begin{array}{ccc}2& 5& 1\\ 1& -4& 1\\ 4& 1& 1\end{array}\right]\\ =\frac{1}{2}\left[2\left(-4-1\right)-5\left(1-4\right)+1\left(1+16\right)\right]\\ =\frac{1}{2}\left[-10+15+17\right]\\ =11\end{array}$

The area of triangle $ABC$ with coordinates $\left(2,5\right)$ , $\left(4,3\right)$ and $\left(4,1\right)$ .

  $\begin{array}{c}{\Delta }_4=\frac{1}{2}\left[\begin{array}{ccc}2& 5& 1\\ 4& 3& 1\\ 4& 1& 1\end{array}\right]\\ =\frac{1}{2}\left[2\left(3-1\right)-5\left(4-4\right)+1\left(4-12\right)\right]\\ =\frac{1}{2}\left[4+0-8\right]\\ =2\end{array}$

The area of the polygon $ABCDEF$ is ${\Delta }_1+{\Delta }_2+{\Delta }_3+{\Delta }_4$ .

  $\begin{array}{c}{\Delta }_1+{\Delta }_2+{\Delta }_3+{\Delta }_4=6+25+11+2\\ =44\text{ sq units}\end{array}$