Question

1. the following shapes are a circle and some regular polygons that have their centers marked. all units are given in feet. which of these shapes has the greatest area? f. circle with radius 1 j. octagon with central - angle 67.5° and radius 1 g. square with apothem 1 k. octagon with apothem 1 h. rhombus - like shape with central - angle 45° and radius 1

Explanation:

Step1: Calculate area of circle

The area formula for a circle is $A = \pi r^{2}$. Given $r = 1$, then $A_{F}=\pi\times1^{2}=\pi\approx 3.14$.

Step2: Calculate area of square (G)

The distance from the center of the square to a side is 1. The side - length $s$ of the square is 2. The area formula for a square is $A=s^{2}$, so $A_{G}=2^{2}=4$.

Step3: Calculate area of rhombus (H)

The distance from the center of the rhombus to a side is 1. If the central angle is $90^{\circ}$ (since the angles at the center are $45^{\circ}$ each), and the apothem $a = 1$. The side - length $s$ of the rhombus can be found using right - triangle relationships. The area of a rhombus $A=\frac{1}{2}d_{1}d_{2}$. Since the apothem is 1 and the central angle is $90^{\circ}$, the diagonals $d_{1}=d_{2}=2$, so $A_{H}=\frac{1}{2}\times2\times2 = 2$.

Step4: Calculate area of octagon (J)

The area formula for a regular polygon is $A=\frac{1}{2}aP$, where $a$ is the apothem and $P$ is the perimeter. The apothem $a = 1$. The central angle of a regular octagon is $\theta=\frac{360^{\circ}}{8}=45^{\circ}$. If the apothem is 1, the side - length $s = 2\tan(22.5^{\circ})$. The perimeter $P = 8s=8\times2\tan(22.5^{\circ})$. Using the half - angle formula $\tan(22.5^{\circ})=\sqrt{2}-1$, $P = 16(\sqrt{2}-1)$. Then $A_{J}=\frac{1}{2}\times1\times16(\sqrt{2}-1)=8(\sqrt{2}-1)\approx8(1.414 - 1)=8\times0.414 = 3.312$.

Step5: Calculate area of octagon (K)

The apothem $a = 1$. The central angle of a regular octagon is $\frac{360^{\circ}}{8}=45^{\circ}$. The perimeter $P$ of a regular octagon with apothem $a = 1$: The side - length $s = 2\tan(22.5^{\circ})$, and $P=8s$. The area $A_{K}=\frac{1}{2}aP=\frac{1}{2}\times1\times8\times2\tan(22.5^{\circ})=8(\sqrt{2}-1)\approx3.312$.

Answer:

G. Square with area 4