Question

a regular octagon has a radius of 6 ft and a side length of 4.6 ft. what is the approximate area of the octagon? 71 ft² 101 ft² 110 ft² 202 ft²

Explanation:

Step1: Recall area formula for regular polygon

$A=\frac{1}{2}aP$, where $a$ is the apothem and $P$ is the perimeter. For a regular octagon with radius $r = 6$ ft and side - length $s=4.6$ ft, the apothem $a$ can be found using the right - triangle formed by the radius and half of the side - length. But we can also use the formula $A = 2(1 +\sqrt{2})s^{2}$ directly. Here we'll use $A=\frac{1}{2}aP$. First, find the perimeter $P$.
$P=ns$, where $n = 8$ (number of sides of an octagon) and $s = 4.6$ ft. So $P=8\times4.6=36.8$ ft.
The apothem $a$ can be found using the right - triangle formed by the radius and half of the side - length. Let the radius be $r$, half of the side - length be $\frac{s}{2}$. Using the Pythagorean theorem, $a=\sqrt{r^{2}-(\frac{s}{2})^{2}}$. Substituting $r = 6$ ft and $s = 4.6$ ft, we have $\frac{s}{2}=\frac{4.6}{2}=2.3$ ft. Then $a=\sqrt{6^{2}-2.3^{2}}=\sqrt{36 - 5.29}=\sqrt{30.71}\approx5.54$ ft.

Step2: Calculate the area

$A=\frac{1}{2}aP$. Substitute $a\approx5.54$ ft and $P = 36.8$ ft into the formula.
$A=\frac{1}{2}\times5.54\times36.8=5.54\times18.4 = 101.936\approx101$ ft².

Answer:

101 ft²