A right circular cone has a volume of 71,148 π cubic centimeters and the area of

Choice D is correct. The volume, $V$, of a right circular cone is given by the formula $V=\frac{1}{3}\pi r^{2}h$, where $\pi r^2$ is the area of the circular base of the cone and $h$ is the height. It’s given that this right circular cone has a volume of $\mathrm{71,148}\pi$ cubic centimeters and the area of its base is $\mathrm{5,929}\pi$ square centimeters. Substituting $\mathrm{71,148}\pi$ for $V$ and $\mathrm{5,929}\pi$ for $\pi r^2$ in the formula $V=\frac{1}{3}\pi r^{2}h$ yields $\mathrm{71,148}\pi =\left(\frac{1}{3}\right)\left(\mathrm{5,929}\pi \right)\left(h\right)$. Dividing each side of this equation by $\mathrm{5,929}\pi$ yields $12=\frac{h}{3}$. Multiplying each side of this equation by $3$ yields $36=h$. Let $s$ represent the slant height, in centimeters, of this cone. A right triangle is formed by the radius, $r$, height, $h$, and slant height, $s$, of this cone, where $r$ and $h$ are the legs of the triangle and $s$ is the hypotenuse. Using the Pythagorean theorem, the equation $r^2+h^2=s^2$ represents this relationship. Because $\mathrm{5,929}\pi$ is the area of the base and the area of the base is $\pi r^2$, it follows that $\mathrm{5,929}\pi =\pi r^2$. Dividing both sides of this equation by $\pi$ yields $\mathrm{5,929}=r^2$. Substituting $\mathrm{5,929}$ for $r^2$ and $36$ for $h$ in the equation $r^2+h^2=s^2$ yields $\mathrm{5,929}+{36}^2=s^2$, which is equivalent to $\mathrm{5,929}+\mathrm{1,296}=s^2$, or $\mathrm{7,225}=s^2$. Taking the positive square root of both sides of this equation yields $85=s$. Therefore, the slant height of the cone is $85$ centimeters.

Choice A is incorrect. This is one-third of the height, in centimeters, not the slant height, in centimeters, of this cone.

Choice B is incorrect. This is the height, in centimeters, not the slant height, in centimeters, of this cone.

Choice C is incorrect. This is the radius, in centimeters, of the base, not the slant height, in centimeters, of this cone.