Area of a cut column

Question

An infinite column with its axis along the $z$ axis, and with square cross section of side length $10$, is cut by the plane $4x - 7y + 4z = 25$. The cut is in the shape of a parallelogram. Find the area of the cut.

My approach: The area of the cut when projected parallel to the $z$ axis is the area of the square cross section, i.e. $100$. If $\theta$ is the angle between the $z$ axis and normal to the given plane, then the area of the cut (which is a parallelogram) is given by

$$ A = \dfrac{100}{\cos(\theta)}$$

Then normal vector to the plane is $(4, -7, 4)$, so

$$\cos(\theta)= \dfrac{4}{\sqrt{4^2 + (-7)^2 + 4^2 } } = \dfrac{4}{9}$$

Therefore, the required area is

$$A = \dfrac{100}{\left( \frac{4}{9} \right) } = 9 \times 25 = 225 $$

I would like to verify if this correct. Your comments, or alternative solutions are highly appreciated.

Answer 1

Alternative solution:
Take quadrilateral $ABCD$, where $A=(5,5,0), B=(-5,5,0), C=(-5,-5,0), D=(5,-5,0)$, as a base for the given infinite column. Now, the given plane will intersect an infinite column in these points:
$P_1=(5,5,10), P_2=(-5,5,20), P_3=(-5,-5,2.5)$ and $P_4=(5,-5,-7.5)$.

The area of the cut is obviously equal to the area of quadrilateral $P_1P_2P_3P_4$. It is easy to get $225$ for this area.