Q41. - Algebra 2

Question

GEOMETRY Find the area of the polygon shown at the right.

Step-by-Step Solution

Answer

Area of the polygon is 32.5 square units.

Step 1 - Define area of triangle

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 $\left(5,-2\right)$. 

Step 2 - Calculate area of first triangle

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.

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

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

Step 3 - Calculate area of second triangle

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.

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

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

Step 4 - Calculate area of polygon

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.