Right Cone

[vampirewitchreine]

15. The volume of a right cone is \(\displaystyle 144 \pi\) and the area of its base is \(\displaystyle 36\pi\). Find the radius, height and slant height of the cone.



What I currently have is the radius.

Since the area of a circle is: \(\displaystyle 2\pi r^2\)
I did:
\(\displaystyle \frac {36 \pi}{2 \pi}= \frac {2 \pi r^2}{2 \pi}\)
\(\displaystyle \sqrt {18} = \sqrt {r^2}\)
\(\displaystyle 4.24 ~\approx r\)


What I've started to work out as I've been typing:

And the area of a cone is: \(\displaystyle \frac {1}{3} Bh \Rightarrow \frac{1}{3} \pi r^2 h\)
So I have
\(\displaystyle 114 \pi = \frac{1}{3} \pi r^2 h\)
\(\displaystyle 114 \pi = \frac{1}{3} \pi (4.24)^2 h\)
\(\displaystyle 114 \pi = \frac{1}{3} \pi (17.98) h\)
(And now I'm at a stand still.... could someone please help me work from here? If you could help me find the height, I can see if I can find the slant on my own.)

 

[Deleted member 4993]

15. The volume of a right cone is \(\displaystyle 144 \pi\) and the area of its base is \(\displaystyle 36\pi\). Find the radius, height and slant height of the cone.



What I currently have is the radius.

Since the area of a circle is: \(\displaystyle 2\pi r^2\) <<<< No area of a circle = π * r²

Now fix rest of the problem
I did:
\(\displaystyle \frac {36 \pi}{2 \pi}= \frac {2 \pi r^2}{2 \pi}\)
\(\displaystyle \sqrt {18} = \sqrt {r^2}\)
\(\displaystyle 4.24 ~\approx r\)


What I've started to work out as I've been typing:

And the area of a cone is: \(\displaystyle \frac {1}{3} Bh \Rightarrow \frac{1}{3} \pi r^2 h\) <<< B is given to you B = 36 * π

144 * π = 1/3 * (36 * π) * h

h = (144 * π)/[1/3 * (36 * π)] = 12

Why make life harder than it needs to be!!!

So I have
\(\displaystyle 114 \pi = \frac{1}{3} \pi r^2 h\)
\(\displaystyle 114 \pi = \frac{1}{3} \pi (4.24)^2 h\)
\(\displaystyle 114 \pi = \frac{1}{3} \pi (17.98) h\)
(And now I'm at a stand still.... could someone please help me work from here? If you could help me find the height, I can see if I can find the slant on my own.)

.

 

[soroban]

Hello, vampirewitchreine!

15. The volume of a right cone is \(\displaystyle 144 \pi\) and the area of its base is \(\displaystyle 36\pi\).
Find the radius, height and slant height of the cone.

Since the area of a circle is: \(\displaystyle 2\pi r^2\) . No!

We know that: .\(\displaystyle V \:=\:\frac{1}{3}Bh \)

So we have: .\(\displaystyle 144\pi \:=\:\frac{1}{3}(36\pi)h \quad\Rightarrow\quad \boxed{h \,=\,12}\)

The area of the circular base is: /\(\displaystyle B \,=\,\pi r^2\)
. . Hence: .\(\displaystyle 36\pi \,=\,\pi r^2 \quad\Rightarrow\quad \boxed{r\,=\,6}\)

 

[vampirewitchreine]

Okay, so I messed up the formula and mashed up Circumference with the Area.

My corrected equations did end up with the exact same results as soroban (thank you)

\(\displaystyle \frac {36 \pi}{\pi}= \frac {\pi r^2}{\pi}\)
\(\displaystyle \sqrt {36} = \sqrt {r^2}\)
\(\displaystyle r = 6\)


\(\displaystyle 114 \pi = \frac{1}{3} \pi r^2 h\)
\(\displaystyle 114 \pi = \frac{1}{3} (36 \pi)h\)
\(\displaystyle 114 \pi = (12 \pi) h\)
\(\displaystyle \frac {144 \pi}{12 \pi} = \frac {12 \pi h}{12 \pi}\)
\(\displaystyle 12 = h \)



So my slant height comes from Pythagorean Theorem:
\(\displaystyle s^2 = 6^2 + 12^2\)
\(\displaystyle s^2 = 36 + 144\)
\(\displaystyle \sqrt {s^2} = \sqrt {180}\)
\(\displaystyle s ~\approx 13.42\)



And Dennis..... isn't that suppose to be You're? Not Yer? Thought text talk was a no-no in the forum





On another note. Lets say that I have the Area of a square pyramid and the height. Would I divide out the height first then multiply both sides by 3 and get what the area of the base is? (And square root that answer to get what one length is?)

 

[Mrspi]

Okay, so I messed up the formula and mashed up Circumference with the Area.

My corrected equations did end up with the exact same results as soroban (thank you)

\(\displaystyle \frac {36 \pi}{\pi}= \frac {\pi r^2}{\pi}\)
\(\displaystyle \sqrt {36} = \sqrt {r^2}\)
\(\displaystyle r = 6\)


\(\displaystyle 114 \pi = \frac{1}{3} \pi r^2 h\)
\(\displaystyle 114 \pi = \frac{1}{3} (36 \pi)h\)
\(\displaystyle 114 \pi = (12 \pi) h\)
\(\displaystyle \frac {144 \pi}{12 \pi} = \frac {12 \pi h}{12 \pi}\)
\(\displaystyle 12 = h \)



So my slant height comes from Pythagorean Theorem:
\(\displaystyle s^2 = 6^2 + 12^2\)
\(\displaystyle s^2 = 36 + 144\)
\(\displaystyle \sqrt {s^2} = \sqrt {180}\)
\(\displaystyle s ~\approx 13.42\)



And Dennis..... isn't that suppose to be You're? Not Yer? Thought text talk was a no-no in the forum





On another note. Lets say that I have the Area of a square pyramid and the height. Would I divide out the height first then multiply both sides by 3 and get what the area of the base is? (And square root that answer to get what one length is?)

You have the AREA of a square pyramid....

What would be the basic formula you would start with?

From what I see, you seem to be confusing AREA with VOLUME.

Please clarify...

 

[vampirewitchreine]

You have the AREA of a square pyramid....

What would be the basic formula you would start with?

Area for a square pyramid is \(\displaystyle \frac {1}{2}ps+B\)

From what I see, you seem to be confusing AREA with VOLUME.

Please clarify...

>.< I seem to be doing this a lot lately...... So now I need to find the slant as well as the area of the base.....


The exact problem (from the book) is:
A pyramid (assuming that it's a square pyramid due to that's what this section is on) has a height of 15 ft and a volume (my bad, it was volume..... I posted wrong.) of 450 ft³. What is the area of the base of the pyramid?

Correct Hazel: just wanted to c if u'd notice :idea:

So most students just glance over it and don't think anything of it?

Oh, yeah.... and see.. you'd