# Mensuration

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### Re: Mensuration

I am certain that whatever textbook has this problem has formulas for the volume of a cylinder of radius r and height h, $V= \pi r^2h$, and for a cone of radius r and height h, $\frac{1}{3}\pi r^2h$. In this problem, r= x and h= m.

As for surface area, the base is a circle of radius x so has area $\pi x^2$. The vertical side can be "unrolled" to make a rectangle with height m and length the circumference of that circle, $2\pi x$. The surface area of the cone is a little harder. As the last part suggests, you could cut from the circumference to the center then flatten it to get a part of a circle with radius $\sqrt{x^2+ m^2}$. The circumference of such a circle would be $2\pi\sqrt{x^2+ m^2}$ but the cone has circumference $2\pi x$ so it is only $\frac{x}{\sqrt{x^2+m^2}}$ of that circle so has area $\frac{x}{\sqrt{x^2+ m^2}}(\pi (x^2+ m^2)= \pi x\sqrt{x^2+ m^2}$.

HallsofIvy

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