This one's a doozy! Here's an illustration of the described triangle PAB inscribed inside square PQRS.


- Point B divides side QR such that \(SB = w\) and \(RB = r\) and the sides of the square are r + w.

\(Area(△PQA) = \frac{1}{2}*w*(r+w) = \frac{1}{2}(w^2 + rw)\)

- Since sides PA and PB are equal (bc they're sides of an equilateral triangle), and sides QR and SR are both r+w (bc they're sides of a square), their 3rd sides must also be equal. SB = QA = w

This means BR = r, so ARB is a 45-45-90 triangle. The hypotenuse of a 45-45-90 triangle is = its side length *\( \sqrt{2} \), so \(AB = r * \sqrt{2} \).

\(Area(△ARB) = \frac{1}{2}r^2\)

- AP is the hypotenuse of PQA. Since we found one of the sides of the equilateral triangle = \(r*\sqrt{2} \), we can set up the Pythagorean formula:

\(AQ^2 + QP^2 = AP^2\)
\(w^2 + (r+w)^2 = {r*\sqrt{2}}^2 \)
\(w^2 + r^2+2rw+w^2 = 2r^2\)
\(2w^2 +2rw = r^2\)

Thus, we can substitute \(2w^2 +2rw\) for \(r^2\).

\(\frac {Area of ARB}{Area of PQA} \frac\)
\(= \frac {\frac{1}{2}r^2} {\frac{1}{2}(w^2 + rw)} \frac \)
\(=\frac {2w^2 +2rw}{w^2 + rw}\frac \)
\(=2\)