AboutJosh Expertise When I work through problems, I emphasize principles and key ideas which I believe are worth noting. I will try to answer questions in the following areas, but not at the advanced level. Algebra. Sequences & Series. Trigonometry. Functions & Graphs. Coordinate Geometry. Quadratic Polynomials. Exponentials & Logarithms. Basic Calculus. Probability, Permutations and Combinations. Mathematical Induction. Complex numbers. Physics problems.
Experience
Experience: I have worked as a teaching assistant in college. My hope is that more people will share knowledge without boundary, give help without seeking recognition or monetary rewards.
Supplementary Website: See a selection of past questions in my maths repository under "Question Archive"
Education Credentials: Bachelor degree in Engineering Science."Everyone struggles with something."
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.
