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Question

Find the height of a cone with a radius of 12 in. and a volume of 408π408 \pi in 3^3.

Solution

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V=13Bh408π=13(π122)h408=13(144)h408=48h8.5=h\begin{align*} V&=\dfrac{1}{3}Bh\\ 408\pi&=\dfrac{1}{3}(\pi\cdot12^2)h\\ 408&=\dfrac{1}{3}(144)h\\ 408&=48h\\ 8.5&=h \end{align*}

Use the formula V=13BhV=\dfrac{1}{3}Bh where BB is the area of the base and hh is the height

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