# Two similar cones have base radii of 7.5 cm and 9 cm. The larger cone has a volume of `198pi cm^3` . Find the volume of the smaller cone.I know the answer to this because my math teacher explained...

Two similar cones have base radii of 7.5 cm and 9 cm. The larger cone has a volume of `198pi cm^3` . Find the volume of the smaller cone.

I know the answer to this because my math teacher explained it to me but I didn' ask and don't get what to find. Don't you just simply find the volume by using the ratio factor then simplify it then cube it?

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The larger cone has a radius of 9 cm and a volume of 198pi cm^3. With these, let's determine the height of the cone using the formula of volume.

`V= 1/3 pir^2h`

Substitute the values of V and r.

`198pi = 1/3 pi (9)^2h`

`198pi=27pi h`

To solve for h, divide both sides by `27pi` .

`(198pi)/(27pi)=(27pih)/(27pi)`

`22/3 = h`

So the height of the larger cone is 22/3 cm.

Since the two cones are similar, let's use ratio and proportion to solve for the height of the smaller cone.

`h_S/r_S=h_L/r_L`

Substitute `r_s=7.5 cm` , `r_L=9 cm` and `h_L=22/3 cm` .

`h_s/7.5 = (22/3)/9`

`h_s/7.5=22/27`

Then, multiply both sides by 7.5 to isolate `h_s` .

`7.5*h_s/7.5 = (22/27)*7.5`

`h_s=6.11`

So the height of the smaller cone is 6.11 cm.

Then substitute the radius and height of the smaller cone to the formula of volume.

`V = 1/3pi r^2 h= 1/3pi(7.5)^2(6.11) = 114.56pi`

**Hence, the volume of the smaller cone is `114.56pi cm^3` .**

If the ratio of radii is 7.5:9, then the ratio of the volumes is `7.4^3:9^3=125:216`

(The ratio of corresponding lengths in similar figures is constant and called the scale factor. The ratio of corresponding areas is the square of the scale factor, while the ratio of common volumes is the cube of the scale factor.)

Then letting x be the required volume:

`125/216=x/(198pi) ==> x=(1375pi)/12=114.58bar(3)pi`

**The required volume is `114.58bar(3)pi"cm"^3` **

(The difference of this answer to mjripalda's is due to rounding error in determining the height.)