Exercise 33 Page 647

Let $r$ represent the radius of Octagon A and the apothem of Octagon B. We expect Octagon B to be bigger than the other one because the radius of Octagon B will be longer than its apothem.

We know that the central angle of an octagon is $45^{\circ }$ and that the apothem bisects this angle and the opposite side.

In the diagram, $x$ and $y$ represent half of the side lengths of Octagon A and Octagon B, respectively. Additionally, $a$ represents the apothem of Octagon A. Using the trigonometric ratios, we can rewrite these lengths in terms of $r.$

	Trigonometric Ratio	Rewrite
Octagon A	$\sin 22.5^{\circ }=\frac{x}{r}$	$x=r\sin 22.5^{\circ }$
	$\cos 22.5^{\circ }=\frac{a}{r}$	$a=r\cos 22.5^{\circ }$
Octagon B	$\tan 22.5^{\circ }=\frac{y}{r}$	$y=r\tan 22.5^{\circ }$

We will now find the perimeters of the octagons by multiplying the side lengths by the number of sides. Remember that $x$ and $y$ are half-lengths of the sides, so we need to multiply them by $2\cdot 8,$ or $16.$ Then we can use the formula for the area of a regular polygon.

	Perimeter, $p$	Apothem, $a$	Area$=\frac{1}{2}ap$
Octagon A	$16x=16r\sin 22.5^{\circ }$	$a=r\cos 22.5^{\circ }$	$A_A=\frac{1}{2}\cdot r\cos 22.5^{\circ }\cdot 16r\sin 22.5^{\circ }$ $\Downarrow$ $A_A=8r^2\cos 22.5^{\circ }\sin 22.5^{\circ }$
Octagon B	$16y=16r\tan 22.5^{\circ }$	$r$	$A_B=\frac{1}{2}\cdot r\cdot 16r\tan 22.5^{\circ }$ $\Downarrow$ $A_B=8r^2\tan 22.5^{\circ }$

The ratio of $A_B$ to $A_A$ tells us how many times bigger the area of Octagon B is. Therefore, the area $A_B$ of Octagon B is about $1.17$ times the area $A_A$ of Octagon A.