The cross-sectional area is indeed $A_c = \pi d^2$ but the explanation on this page is not precise in this regard. You basically consider a single particle of diameter $d$ and its collisions with other particle of the same size.
Simple considerations in 2D
The chance for it to collide with another particle is proportional to the center of that other particle being in a circle with radius $2 r = d$ around the center of the particle under consideration.
Just think of it that way: What is the neighbourhood of a particle that another particle with the same radius would have to be in for a collision? Obviously they would collide if the two radii would touch or the distance would be smaller. Thus $\pi (2r)^2 = \pi d^2$ is the area the center of the second sphere must lie in for a collision between the two particles.
More general considerations in 3D
More generally in three-dimensional space one has to consider the area perpendicular to the relative velocity of the two particles. And integrate over this cross-sectional area
$$ d A_c = r \, dr \, d \phi $$
depending on the position of the two particles. Introducing an angle $\psi$ in between the line connecting the two centers and the relative velocity can be calculated according to
$$ r = d \, \sin \psi \hspace{2cm} dr = d \, \cos \psi d \psi$$
Now integrating over all potential angles in direction of the relative velocity $0 \leq \phi \leq 2 \pi$ and $0 \leq \psi \leq \frac{\pi}{2}$ considering the identity $\sin \psi \, \cos \psi = \frac{\sin ( 2 \psi)}{2}$ we yield
$$ A_c = \int\limits_{\phi = 0}^{2 \pi} d \phi \int\limits_{\psi = 0}^{\frac{\pi}{2}} \frac{d^2 \, \sin ( 2 \psi)}{2} d \psi = \pi d^2 $$
