Circular cones A and B are congruent. If the area of the base of cone B is 16π in2, then the volume of cone A is

if the area of the base of cone B is 16π, since both cones are congruent, then the area of the base of cone A is also 16π.

well, the base of cone A is a circle, and we know its diameter is "x", so its radius is half that, or x/2, and we also know that the area is 16π, thus

[tex]\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ -----\\ r=\frac{x}{2}\\ A=16\pi \end{cases}\implies 16\pi =\pi \left( \frac{x}{2} \right)^2\implies 16\pi =\pi \cdot \cfrac{x^2}{2^2} \\\\\\ \cfrac{16\pi }{\pi }=\cfrac{x^2}{4}\implies 16=\cfrac{x^2}{4}\implies 64=x^2\implies \sqrt{64}=x\implies \boxed{8=x}[/tex]

and since we know what "x" is, then the radius of it is (8)/2, or just 4, and its height is (8) + 2, or just 10, then

[tex]\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\ -----\\ r=4\\ h=10 \end{cases}\implies V=\cfrac{\pi (4)^2(10)}{3}\implies V=\cfrac{160\pi }{3}[/tex]