Question

Two regular pentagons have perimeters of 8cm and 16cm. What is the ratio of their areas?

Answer 1

Since they are "regular", all the sides are the same length

Answer 2

lets call these sides x

Answer 3

that means \(p = 5x\)
and each side is: \({p \over 5} = x\)

Answer 4

the area is just the addition of triangles
My drawing is terrible but all of them should be the same

Answer 5

Using my drawing a little trig and by adding up the area of the 5 triangles we get that

\[A \approx 1.72 * x^2 \]

Answer 6

so A/A is \[1.72 X^2 \over 1.72 x^2\]

Answer 7

the 1.72's cancel and you end up with \[X^2 \over x^2\]

Answer 8

Lets say X = 8 and x = 16

Answer 9

just plug them in and simplify and you're good

Answer 10

If you tell me what you get I will verify your answer

Answer 11

1:2