Expert: Josh - 6/3/2009

Question
the area of a regular octagon is 35 cm squared. What is the area of a similar octagon with sides sex times as large?

Answer
Amber,

I will refer you to the diagram at http://mathcentral.uregina.ca/QQ/database/QQ.09.01/laurie2.html

Given an octagon of length "b" inscribed in a square of size "w", the area of the octagon A is given by the difference between the area of the square and the area of the four triangular pieces. Again, refer to the figure.

So, we have A = w^2 - 2*b^2, where b=w/(2+sqrt(2)) from geometry.

Since w=b(2+sqrt(2)), we can write this as
A = 4 (1+sqrt(2)) b^2 after factorization,
in terms of "b", i.e., the length of any side of the octagon.

Since A is proportional to b^2, if b becomes 6b (six times longer), the new area will be 36 times larger.

Out of interest, you can set 4 (1+sqrt(2)) b^2 = 35 [cm^2] and solve for "b" to find the initial length of the octagon.