The Sierpinski carpet is a $1\times 1$ square with the centre $9^{th}$ removed, then the centre $9^{th}$ of the eight remaining disjoint squares removed, and so on $n$ times, with $n\to\infty$, as described here: Wolfram
If you take a section through a Sierpinski carpet half-way up you have the Cantor set; an infinite closed perfect set of zero measure. In fact the square fractal's area is given by $\lim_{n\to\infty}(\frac{8}{9})^n$ which appears to limit to $0$, implying the measure across at any height is zero.
If instead of repeatedly removing the central $9^{th}$, we first remove the central $9^{th}$, then remove the centre $25^{th}$ of the $8$ remaining smaller squares. Then remove the centre $49^{th}$ of the remaining $96$ smaller still squares, and so on, then the area remaining is given by half of the Wallis Product: $$\prod_{n=1}^{\infty}\frac{(2n+1)^2-1}{(2n+1)^2}=\frac{\pi}{4}$$
Edd Pegg describes this here
What is the Lebesgue measure $\lambda(y)$ of a horizontal section through this set as a function of $y$, the height from the baseline at which the section is taken?
For the original Sierpinski carpet the measure of any section looks to be closely related to the 2-adic metric of $y$, provided we don't take $n$ to infinity.
For the pi-carpet (let's call it) it looks to be something like half of the length of a chord across a circle, taken at the $(2n+1)$-adic metric distance from the circle's widest point. The reason I say this is that every width of a circle is still represented, they are just ordered by some metric similar to the $2$-adic but growing to $2n+1$ at each base step instead of $2n$.