Let $l_n$ be a geometric series of the side lengths of any equilateral two-dimensional figure and $A_n$ be a geometric series of the area of the said figure.
Prove that $q_A = {q_l}^2$, whereas $q_A = \frac{A_n}{A_{n-1}}$ and $q_l = \frac{l_n}{l_{n-1}}$

My attempt is the following: When assuming a simple square, $A_n = {l_n}^2$, therefore $$\frac{A_{n+1}}{A_n}=\frac{{l_{n+1}}^2}{{l_n}^2}$$ Which then results in $$\frac{A_n*q_A}{A_n}=\frac{({l_n*q_l})^2}{{l_n}^2}$$ Further simplified, that leaves me with $$q_A = {q_l}^2$$ So now, I have proven that the given relation is true for squares. However, I don't know how I can prove it for any two-dimensional figure.

Thank you for your help!

  • 1
    $\begingroup$ In general, given two similar polygons (which need not be equilateral) of areas $P$ and $Q$, if a side of length $p$ in the first polygon corresponds to a side of length $q$ in the second, then $$\frac{P}{Q} = \left(\frac{p}{q}\right)^2$$ The result you want follows. To see why the relation above holds, convince yourself that it works for triangles. Then, note that any polygon can be decomposed into triangles, and that similar polygons can be dissected in the same way to create pairs of similar triangles. $\endgroup$ – Blue Nov 2 '18 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.