An Oblique Cut through a Square Column

In my opinion the easiest way to tackle is determine the normal vector to the cut and then find the cosine that form that normal vector with axis. Once you get the cosine, we divide the area of the base by the cosine and we get the sought area.

Fortunately the data are so gentle that you even do not need a pencil, the unit vector normal to the plane is 4/9i -7/9j+4/9k so scalar product give us the value of the cosine 4/9 then the area will be 100/(4/9) equal to 225.

Consider the plane equation $x+\frac{7}{4}y=-z$ , so that the line of steepest descent through the origin $\left( 0,0 \right)$ projected onto the $xy$ plane is the line $y=\frac { 7 }{ 4 }x$ . Letting $x=1$ , we have $y=\frac { 7 }{ 4 }$ and $-z=1+\frac { 7 }{ 4 } y=\frac { 65 }{ 16 }$ so that we can compute the ratio of the area of the cross section with that of the square base

$\dfrac { \sqrt { 1+{ \left( \frac { 7 }{ 4 } \right) }^{ 2 }+{ \left( -\frac { 65 }{ 16 } \right) }^{ 2 } } }{ \sqrt { 1+{ \left( \frac { 7 }{ 4 } \right) }^{ 2 } } } =2.25$

Thus we immediately have the answer $225$

The constant $25$ is superfluous to this problem and can be any constant.

10x10x(1+(4/4)^2+(-7/4)^2)^0.5 =225