# How does the volume of a cone change if its radius and height are both tripled?

Aug 19, 2016

The volume of the cone with changed dimensions is $27$ times the volume of the original cone.

#### Explanation:

Volume of a Cone $= \pi {r}^{2} \times \frac{h}{3}$ where $r = r a \mathrm{di} u s \mathmr{and} h = h e i g h t$
Let us say New $R = 3 r$ and New $H = 3 h$

So volume of the cone with changed dimensions

$= \pi {R}^{2} \times \frac{H}{3}$

Plugging the values $R = 3 r$ and $H = 3 h$

we get
volume of the cone with changed dimensions

$= \pi {\left(3 r\right)}^{2} \times \frac{3 h}{3}$

$= 27 \pi {r}^{2} \frac{h}{3}$

Hence the volume of the cone with changed dimensions is $27$ times the volume of the original cone.

Aug 19, 2016

Becomes 27 times greater than the original volume

#### Explanation:

Since volume of a cone $V = \frac{\pi \cdot h \cdot {r}^{2}}{3}$

In the original situation this is the formula.

Now we have 3r (for new radius at the bottom) and 3h as new height.

If you insert these into above equation, you will get

$V = 27 \cdot \frac{\pi \cdot h \cdot {r}^{2}}{3}$