Question

a regular heptagon (7 sides) has an area of 540 cm². a second regular heptagon has an area of 60 cm². the perimeter of the larger heptagon is p cm. what is the perimeter of the smaller heptagon?
a $\frac{p}{81}$
b $\frac{p}{18}$
c $\frac{p}{9}$
d $\frac{p}{6}$
e $\frac{p}{3}$

Explanation:

Step1: Recall the ratio of areas and perimeters of similar polygons

For two similar polygons, if the ratio of their areas is $A_1:A_2$, the ratio of their corresponding - side lengths (and perimeters since perimeter is a linear - measure) is $\sqrt{A_1}:\sqrt{A_2}$. Let the area of the larger heptagon be $A_1 = 540\ cm^2$ and the area of the smaller heptagon be $A_2=60\ cm^2$. Let the perimeter of the larger heptagon be $p$ and the perimeter of the smaller heptagon be $q$.

Step2: Calculate the ratio of the areas

The ratio of the areas is $\frac{A_1}{A_2}=\frac{540}{60} = 9$.

Step3: Calculate the ratio of the perimeters

Since the ratio of the perimeters of two similar polygons is $\sqrt{\frac{A_1}{A_2}}$, and $\sqrt{\frac{540}{60}}=\sqrt{9}=3$. So, $\frac{p}{q}=3$, which implies $q = \frac{p}{3}$.

Answer:

E. $\frac{p}{3}$