A cone without base is made from a quartercircle. 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!
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A cone without base is made from a quartercircle. 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!
Hello,
A cone is actually similar to a portion of circle
Attachment 5961
So if we report to this sketch, the base of the cone is the left circle, whose radius is 3 (and whose perimeter is $\displaystyle 2 \pi 3=6 \pi$. 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 $\displaystyle 2 \pi P$
Hence the perimeter of the quarter circle is $\displaystyle \frac{2 \pi P}{4}=\frac{\pi P}{2}$
This perimeter corresponds to the perimeter of the right circle, so we have :
$\displaystyle \frac{\pi P}{2}=6 \pi \Longleftrightarrow P=12$
The volume of a cone is given by $\displaystyle \frac{\text{Area of the base} \times \text{height}}{3}$
So here, the area of the base, which is the area of a circle, radius 3, will be $\displaystyle \pi 3^2=9 \pi$
> the volume of the cone is $\displaystyle \frac{12*9 \pi}{3}=36 \pi$
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. $\displaystyle 2{\pi}(3)=6{\pi}$.
We can use this and $\displaystyle s=h{\theta}$
$\displaystyle s=6{\pi}, \;\ {\theta}=\frac{\pi}{2}$
Now, we can solve for h and find the height of the cone and then its volume.
$\displaystyle 6{\pi}=h(\frac{\pi}{2}), \;\ h=12$
$\displaystyle V=\frac{\pi}{3}(3)^{2}(12)=36{\pi}$
Hello, shadow_2145!
Quote:
A cone without base is made from a quartercircle.
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 $\displaystyle 6\pi $ cm.
This is the length of arc of the quartercircle.
Then: .$\displaystyle r\left(\frac{\pi}{2}\right) \:=\:6\pi\quad\Rightarrow\quad r \,=\,12\text{ cm}$Code:* *
 * 6π
 *
r  *

 *
+     *
r
The side view of the cone:We have: .$\displaystyle h^2 + 3^2 \:=\:12^2\quadCode:*
/\
/  \
/  \ 12
/ h \
/  \
*   +   *
3
\Rightarrow\quad h \,=\,3\sqrt{15}$
Therefore: .$\displaystyle V \;=\;\frac{\pi}{3}r^2h \;=\;\frac{\pi}{3}(3^2)(3\sqrt{15}) \;=\;9\pi\sqrt{15}\text{ cm}^3$
Hello Soroban,
Are you sure the side view of a cone is a triangle ?
Soroban is correct. We both used the slant height as the vertical height. I can't believe I done that.
Oh I see :D