91Ó°ÊÓ

The area of triangle having vertices at $\left(a,b\right)$, $\left(c,d\right)$ and $\left(e,f\right)$ is $\left|A\right|$, where $A=\frac{1}{2}\left|\begin{array}{ccc}a& b& 1\\ c& d& 1\\ e& f& 1\end{array}\right|$.

Clearly, the given polygon is divided into two triangular areas. Let first triangle have vertices $\left(4,5\right)$, $\left(-2,2\right)$and $\left(2,2\right)$ and second triangle have vertices $\left(4,5\right)$, $\left(2,2\right)$ and $\left(5,-2\right)$.

Here vertices of the triangle are $\left(a,b\right)=\left(4,5\right)$, $\left(c,d\right)=\left(-2,2\right)$and $\left(e,f\right)=\left(2,2\right)$. So, substitute 4 for a, 5 for b, -2 for c, 2 for d, 2 for e and 2 for f into the expression $A=\frac{1}{2}\left|\begin{array}{ccc}a& b& 1\\ c& d& 1\\ e& f& 1\end{array}\right|$.

$A=\frac{1}{2}\left|\begin{array}{ccc}4& 5& 1\\ −2& 2& 1\\ 2& 2& 1\end{array}\right|$

Find the determinant in this case by using expansion by minors.

$\begin{array}{c}A=\frac{1}{2}\left|\begin{array}{ccc}4& 5& 1\\ −2& 2& 1\\ 2& 2& 1\end{array}\right|\\ =\frac{1}{2}\left(4\left|\begin{array}{cc}2& 1\\ 2& 1\end{array}\right|−5\left|\begin{array}{cc}−2& 1\\ 2& 1\end{array}\right|+\left|\begin{array}{cc}−2& 2\\ 2& 2\end{array}\right|\right)\\ =\frac{1}{2}\left(4\left[2−2\right]−5\left[−2−2\right]+\left[2\left(−2\right)−2\left(2\right)\right]\right)\\ =\frac{1}{2}\left(4\left[0\right]−5\left[−4\right]+\left[−4−4\right]\right)\\ =\frac{1}{2}\left(0−20−8\right)\\ =\frac{1}{2}×\left(−28\right)\\ =−14\end{array}$

Since, area of triangle is an absolute value, therefore, area of first triangle is 14 square units.

Here vertices of the triangle are $\left(a,b\right)=\left(4,5\right)$, $\left(c,d\right)=\left(2,2\right)$and $\left(e,f\right)=\left(5,-2\right)$. So, substitute 4 for a, 5 for b, 2 for c, 2 for d, 5 for e and -2 for f into the expression$A=\frac{1}{2}\left|\begin{array}{ccc}a& b& 1\\ c& d& 1\\ e& f& 1\end{array}\right|$.

$A=\frac{1}{2}\left|\begin{array}{ccc}4& 5& 1\\ 2& 2& 1\\ 5& −2& 1\end{array}\right|$

Find the determinant in this case by using expansion by minors.

$\begin{array}{c}A=\frac{1}{2}\left|\begin{array}{ccc}4& 5& 1\\ 2& 2& 1\\ 5& −2& 1\end{array}\right|\\ =\frac{1}{2}\left(4\left|\begin{array}{cc}2& 1\\ −2& 1\end{array}\right|−5\left|\begin{array}{cc}2& 1\\ 5& 1\end{array}\right|+\left|\begin{array}{cc}2& 2\\ 5& −2\end{array}\right|\right)\\ =\frac{1}{2}\left(4\left[2−\left(−2\right)\right]−5\left[−2−5\right]+\left[2\left(−2\right)−2\left(5\right)\right]\right)\\ =\frac{1}{2}\left(4\left[4\right]−5\left[−7\right]+\left[−4−10\right]\right)\\ =\frac{1}{2}\left(16+35−14\right)\\ =\frac{1}{2}×\left(37\right)\\ =\text{18.5}\end{array}$

Since, area of triangle is an absolute value, therefore, area of second triangle is 18.5 square units.

Area of polygon is the sum of area of first triangle and area of second triangle. Since area of first triangle is 14 square units and area of second triangle is 18.5 square units, therefore, area of polygon is:

$\begin{array}{c}\text{Area}=14+18.5\\ =\text{32.5}\end{array}$

Therefore, area is 32.5 square units.