A pile of gravel at a building site is a cone whose base circumference is 88 $m$ and whose slant height is 50 $m$. Find its volume in $m^3$. Take π = 22/7.

From 2π$r=88,r=(88\times 7)/44=14m$. The vertical height = √$(50^2-14^2)=$ √(2500 − 196) = √2304 = 48 $m$. Volume = $(1/3)(22/7)(196)(48)=(1/3)(616)(48)=9856m^3$. (Leaving out the 1/3 gives $29568{\text{m}}^3$; using the slant height 50 $m$ in place of the vertical height 48 $m$ gives $10266.67{\text{m}}^3$; solving π$r=88$ instead of 2π$r=88$ gives $r=7m$ and $2464{\text{m}}^3.$)