1. ## Puzzled no.2 lol

A sphere of ice cream 6cm in diameter melts and flows into a cone whose depth is equal to three times its radius. If the melted ice cream fills the cone completely and assuming that there is no change in the colume of the ice cream. Find the depth of the cone giving your answer correct to one decimal place.

2. The volume of the sphere of ice cream is $\displaystyle \frac{4}{3}{\pi}r^{3}=\frac{4}{3}{\pi}(3)^{3}=36{\ pi}$

The volume of the cone is: $\displaystyle \frac{1}{3}{\pi}r^{2}h$

Remember, you're told the height is 3 times the radius.

So, you have the equation:

$\displaystyle \frac{1}{3}{\pi}(\frac{h}{3})^{2}h=36{\pi}$

Solve for h.

3. Hello, Damon!

You're expected to know some volume formulas:

. . Volume of a sphere of radius $\displaystyle r\!:\;\;V \:=\:\frac{4}{3}\pi r^3$
. . Volume of a cone of radius $\displaystyle r$ and height $\displaystyle h\!:\;\;V \:=\:\frac{1}{3}\pi r^2h$

A sphere of ice cream 6 cm in diameter melts and flows into a cone
whose depth is equal to three times its radius.
If the melted ice cream fills the cone completely,
find the depth of the cone to one decimal place.

The radius of the sphere of ice cream is $\displaystyle 3$ cm.
The volume of the ice cream is: .$\displaystyle V \:=\:\frac{4}{3}\pi (3^3) \:=\:36\pi$ cm³.

The cone has a height which is three times its radius: .$\displaystyle h = 3r$
. . Hence, its volume is: .$\displaystyle V \:=\:\frac{1}{3}\pi r^2h \;=\;\frac{1}{3}\pi r^2(3r) \:=\:\pi r^3$

Since the ice cream completely fills the cone,
. . we have: .$\displaystyle \pi r^3\:=\:36\pi\quad\Rightarrow\quad r \:=\:\sqrt[3]{36}$

Therefore, the height is: .$\displaystyle h \:=\:3r\:=\:3\sqrt[3]{36}\:\approx\:9.9$ cm.