Question

5a) what is the area of this circle? 5b) what is the perimeter of this shape? a=(2,1) b=(1,2) c=(-1,2) d=(-2,1) e=(-2,-1) f=(-1,-2) g=(1,-2) h=(2,-1)

Explanation:

Step1: Find radius of the circle

The diameter of the circle is given as 13n. The radius $r$ of a circle is half of its diameter. So, $r=\frac{13n}{2}$.

Step2: Use the area - formula for a circle

The area formula of a circle is $A = \pi r^{2}$. Substitute $r=\frac{13n}{2}$ into the formula: $A=\pi(\frac{13n}{2})^{2}=\pi\times\frac{169n^{2}}{4}=\frac{169\pi n^{2}}{4}$.

Step3: Find the lengths of the sides of the polygon using the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}$.
For side AB with $A=(2,1)$ and $B=(1,2)$:
$d_{AB}=\sqrt{(1 - 2)^{2}+(2 - 1)^{2}}=\sqrt{(-1)^{2}+1^{2}}=\sqrt{1 + 1}=\sqrt{2}$.
For side BC with $B=(1,2)$ and $C=(-1,2)$:
$d_{BC}=\sqrt{(-1 - 1)^{2}+(2 - 2)^{2}}=\sqrt{(-2)^{2}+0^{2}} = 2$.
For side CD with $C=(-1,2)$ and $D=(-2,1)$:
$d_{CD}=\sqrt{(-2+1)^{2}+(1 - 2)^{2}}=\sqrt{(-1)^{2}+(-1)^{2}}=\sqrt{2}$.
For side DE with $D=(-2,1)$ and $E=(-2,-1)$:
$d_{DE}=\sqrt{(-2 + 2)^{2}+(-1 - 1)^{2}}=\sqrt{0^{2}+(-2)^{2}}=2$.
For side EF with $E=(-2,-1)$ and $F=(-1,-2)$:
$d_{EF}=\sqrt{(-1 + 2)^{2}+(-2 + 1)^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$.
For side FG with $F=(-1,-2)$ and $G=(1,-2)$:
$d_{FG}=\sqrt{(1 + 1)^{2}+(-2 + 2)^{2}}=\sqrt{2^{2}+0^{2}}=2$.
For side GH with $G=(1,-2)$ and $H=(2,-1)$:
$d_{GH}=\sqrt{(2 - 1)^{2}+(-1 + 2)^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$.
For side HA with $H=(2,-1)$ and $A=(2,1)$:
$d_{HA}=\sqrt{(2 - 2)^{2}+(1 + 1)^{2}}=\sqrt{0^{2}+2^{2}}=2$.

Step4: Calculate the perimeter of the polygon

The perimeter $P$ of the polygon is the sum of the lengths of its sides.
$P=\sqrt{2}+2+\sqrt{2}+2+\sqrt{2}+2+\sqrt{2}+2=8 + 4\sqrt{2}$.

Answer:

The area of the circle is $\frac{169\pi n^{2}}{4}$, and the perimeter of the polygon is $8 + 4\sqrt{2}$.