# Find area of cross section of cylinder by the plane $x$

I am working on my scholarship exam practice (assume high school/pre-university math background) and I think I got half way through but I am not sure how I could continue.

Let $$r$$ be a positive constant. Consider the cylinder $$x^2 +y^2 \leq r^2$$, and let $$C$$ be the part of the cylinder that satisfies $$0\leq z\leq y$$. Consider the cross section of $$C$$ by the plane $$x=t$$ ($$-r\leq t\leq r$$), and express its area in terms of $$r, t$$.

So below is what I have got and stuck there, not sure if I am on the right track. Could you please show or hint on how I can get to the area of this cross section? The key answer $$\frac {1}{2} (r^2-t^2)$$ is also provided. Apologies my drawing may be amateur.

The cross section of C by the plane $$x = t$$ is a triangle. When $$t = 0$$ $$(x = 0)$$ , this is an isosceles right triangle with sides r, r and $$r \sqrt 2$$: cross section of cylinder by the plane x = 0. When $$t \neq 0$$, this is an isosceles right triangle with cathetus $$\sqrt {r ^ 2-t ^ 2}$$: cross section of cylinder by the plane x = t. So, area of the cross section is $$\frac{1}{2}(r^2 - t^2)$$.