Right Square Pyramid with a special base - Reposted

Let each side of the square base be of length $a$ and the height of the pyramid be $h$ . Then $2πh=4a\implies \dfrac{h}{a}=\dfrac{2}{π}$ . Height of each lateral face is $\sqrt {h^2+\dfrac{a^2}{4}}=\dfrac{a}{2}\sqrt {1+\left (\dfrac{2h}{a}\right )^2}=\dfrac{a}{2}\sqrt {1+\dfrac{16}{π^2}}$ . So area of the lateral surface of the pyramid is $4\times \dfrac{1}{2}a\times \dfrac{a}{2}\sqrt {1+\dfrac{16}{π^2}}=a^2\sqrt {1+\dfrac{16}{π^2}}$ . Area of square base is $a^2$ . Therefore the required ratio is $\sqrt {1+\dfrac{16}{π^2}}\approx \boxed {1.618993}$ .

I provided two solutions for the square base.

Solution using coordinate geometry:

For $BD: y = a - x$ and for $AQ: y = \dfrac{1}{a}x \implies x = \dfrac{a^2}{a + 1} \implies$

$A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2}$ $\implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies$

$\boxed{a = \dfrac{1 + \sqrt{5}}{2} = \phi}$ .

Solution using similar triangles:

Since vertical angles are congruent and $AB \parallel CD \implies$ alternate interior angles are congruent $\implies \triangle{ABP} \sim \triangle{DPQ} \implies$

$\dfrac{a}{1} = \dfrac{x}{a - x} \implies x = \dfrac{a^2}{a + 1} \implies A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2}$

$\implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies \boxed{a = \dfrac{1 + \sqrt{5}}{2} = \phi}$

Let $\overline{OR} = r \implies$ $2\pi r = 4\phi \implies r = \dfrac{2\phi}{\pi}$

Let $T$ be the midpoint of $\overline{AD} \implies \overline{OT} = \dfrac{\phi}{2}$

In $\triangle{ROT}$ the slant height is $s = \sqrt{\dfrac{4\phi^2}{\pi^2} + \dfrac{\phi^2}{4}} =$ $\dfrac{\phi}{2\pi}\sqrt{16 + \pi^2} \implies$

$A_{s} = \dfrac{\phi^2}{\pi}\sqrt{16 + \pi^2}$ $\implies \dfrac{A_{s}}{A_{\square}} = \dfrac{\sqrt{16 + \pi^2}}{\pi} \approx \boxed{1.618993}$ .