Answer: The required area is 84 square feet.
Step-by-step explanation: We are given to find the area of the two-dimensional cross section that is parallel to face ABC.
Since ABC is a right-angled triangle at ∠B = 90°, so its parallel face will also be a right-angled triangle.
And the area of the parallel face must also be equal to the area of ΔABC.
Applying Pythagoras theorem, we have from ΔABC that
[tex]AC^2=AB^2+BC^2\\\\\Rightarrow BC^2=AC^2-AB^2\\\\\Rightarrow BC^2=25^2-7^2\\\\\Rightarrow 625-49\\\\\Rightarrow BC^2=576\\\\\Rightarrow BC=24.[/tex]
So, area of triangle ABC is
[tex]A=\dfrac{1}{2}\times base\times altitude=\dfrac{1}{2}\timesB\times BC=\dfrac{1}{2}\times 7\times 24=84~\textup{sq ft.}[/tex]
Thus, the area of the parallel face is 84 sq. ft.
