Volume of a cone

**shadow_2145** · April 20th 2008, 01:48 PM

A cone without base is made from a quarter-circle. The base of the cone is a circle of radius 3 cm. What is the volume of the cone? Explain your reasoning.

Any help would be appreciated, thanks!

**Jhevon** · April 20th 2008, 01:55 PM

Originally Posted by shadow_2145

A cone without base is made from a quarter-circle. The base of the cone is a circle of radius 3 cm. What is the volume of the cone? Explain your reasoning.

Any help would be appreciated, thanks!

the circumference of the cone will be 3. figure out the radius from this, and then you can find the volume

EDIT: ah, i misread. "circumference" for "radius". don't know what i was thinking...

**Moo** · April 20th 2008, 02:02 PM

Hello,

A cone is actually similar to a portion of circle

So if we report to this sketch, the base of the cone is the left circle, whose radius is 3 (and whose perimeter is $2 \pi 3=6 \pi$ . The right part will be a quarter of circle, let's say of radius P, and which corresponds to the height of the cone !

The perimeter of the complete circle corresponding to the quarter circle is $2 \pi P$
Hence the perimeter of the quarter circle is $\frac{2 \pi P}{4}=\frac{\pi P}{2}$

This perimeter corresponds to the perimeter of the right circle, so we have :

$\frac{\pi P}{2}=6 \pi \Longleftrightarrow P=12$

The volume of a cone is given by $\frac{\text{Area of the base} \times \text{height}}{3}$

So here, the area of the base, which is the area of a circle, radius 3, will be $\pi 3^2=9 \pi$

-------> the volume of the cone is $\frac{12*9 \pi}{3}=36 \pi$

**galactus** · April 20th 2008, 02:08 PM

I think I am interpreting this right. From the diagram, we know the resulting radius is going to be 3.

Therefore, the curved portion of the cut out is the circumference of the cone base. $2{\pi}(3)=6{\pi}$ .

We can use this and $s=h{\theta}$

$s=6{\pi}, \;\ {\theta}=\frac{\pi}{2}$

Now, we can solve for h and find the height of the cone and then its volume.

$6{\pi}=h(\frac{\pi}{2}), \;\ h=12$

$V=\frac{\pi}{3}(3)^{2}(12)=36{\pi}$

**Soroban** · April 20th 2008, 06:58 PM

Hello, shadow_2145!

A cone without base is made from a quarter-circle.
The base of the cone is a circle of radius 3 cm.
What is the volume of the cone?

The radius of the base is 3 cm; the circumference is $6\pi$ cm.
This is the length of arc of the quarter-circle.

Code:

      * *
      |     *  6π
      |       *
    r |        *
      |
      |         *
      + - - - - *
           r

Then: . $r\left(\frac{\pi}{2}\right) \:=\:6\pi\quad\Rightarrow\quad r \,=\,12\text{ cm}$

The side view of the cone:

Code:

            *
           /|\
          / | \
         /  |  \ 12
        /  h|   \
       /    |    \
      * - - + - - *
               3

We have: . $h^2 + 3^2 \:=\:12^2\quad<br /> \Rightarrow\quad h \,=\,3\sqrt{15}$

Therefore: . $V \;=\;\frac{\pi}{3}r^2h \;=\;\frac{\pi}{3}(3^2)(3\sqrt{15}) \;=\;9\pi\sqrt{15}\text{ cm}^3$

**Moo** · April 20th 2008, 10:56 PM

Hello Soroban,

Are you sure the side view of a cone is a triangle ?

**galactus** · April 21st 2008, 04:39 AM

Soroban is correct. We both used the slant height as the vertical height. I can't believe I done that.

**Moo** · April 21st 2008, 04:57 AM

Oh I see

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