Exercise 54 Page 66

We will draw a polygon $ABCDEFG$ with the following vertices. Let's plot these points on the coordinate plane and connect them to draw the polygon. Now, we will find the lengths of the polygon by using the Distance Formula. In the formula, $XY$ is the distance between the points $X$ and $Y,$ $(x_1,y_1)$ is the coordinates of $X,$ and $(x_2,y_2)$ is the coordinates of $Y.$ Now, let's find the length of each side of the polygon $ABCDEFG$ by applying the formula. We will substitute the points $A(1,1)$ and $B(10,1),$ to find the length $AB$ to start. Thus the length $AB$ is $9$ units. We can find the other lengths in the same way.

Side	Coordinates	$\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$	Length
$AB$	$A(1,1)$ and $B(10,1)$	$\sqrt{(1-10{)}^2+(1-1{)}^2}$	$AB=9$
$BC$	$B(10,1)$ and $C(10,8)$	$\sqrt{(10-10{)}^2+(1-8){)}^2}$	$BC=7$
$CD$	$C(10,8)$ and $D(7,5)$	$\sqrt{(10-7{)}^2+(8-5{)}^2}$	$CD=3\sqrt{2}$
$DE$	$D(7,5)$ and $E(4,5)$	$\sqrt{(7-4{)}^2+(5-5{)}^2}$	$DE=3$
$EF$	$E(4,5)$ and $F(4,8)$	$\sqrt{(4-4{)}^2+(5-8{)}^2}$	$EF=3$
$FG$	$F(4,8)$ and $G(1,8)$	$\sqrt{(4-1{)}^2+(8-8{)}^2}$	$FG=3$
$GA$	$G(1,8)$ and $A(1,1)$	$\sqrt{(1-1{)}^2+(8-1{)}^2}$	$GA=7$

Finally, we can find the perimeter of the polygon $ABCDEFG$ by adding all seven sides of it together. Thus, the perimeter is $32+3\sqrt{2}$ units. Next, we will find its area by separating it into small rectangles and a triangle as the following. We will find the area of each rectangle $A_R$ and the area of the triangle $A_T$ by using the formulas. In the formula, $b$ is the base and $h$ is the height.

Area	Base	Height	$A=bh$ or $A=\frac{bh}{2}$	Result
$A_1$	$b=3$	$h=7$	$A_1=3\cdot 7$	$21$
$A_2$	$b=3$	$h=4$	$A_2=3\cdot 4$	$12$
$A_3$	$b=3$	$h=4$	$A_3=3\cdot 4$	$12$
$A_4$	$b=3$	$h=3$	$A_4=\frac{3\cdot 3}{2}$	$4.5$

Finally, we can find the total area $A$ by adding the smaller areas together. Thus, the area of the polygon is $49.5$ units square.