Perplexing Mensuration Question - sphere in a cone

**PythagorasNeophyte** · September 26th 2010, 12:33 AM

Hello everyone.

This maths question is somewhat beyond my ken. One part of it talks about similarity while the other talks about mensuration. Still, I could not find an interrelationship between the clues given.

The image is at Mensuration Question | Flickr

What I know intuitively is that there is a Pythagoras' Theorem involved in finding the radius of the sphere. However, VD, which is 8cm, is solely a part of the hypotenuse. If I have the value of the hypotenuse, then I can use the Pythagoras' Theorem to find the radius of the cone.

Also, since triangle VDO is similar to triangle VCA, then it would be $VD=VC$ , $VO=VA$ and $OD=CA$ .

Can anyone give me hints? Help is greatly appreciated!

**Gusbob** · September 26th 2010, 12:53 AM

Hello!

Here is a hint for you!

Let the radius of the sphere be r.
Since VD = 8cm, by the pythagoras theorem, VO is $\sqrt{64+r^2}$ . Since OC is the radius, it has also length r.

Now VC = VO + OC = 12 cm. Substituting, you'll get $r + \sqrt{64 + r^2} = 12$

Move the r over, square both sides, and solve accordingly.

I did this quickly, so you may want to check the working for the equations.

**PythagorasNeophyte** · September 26th 2010, 01:10 AM

Hello Gusbob!

Thank you for your reply.

So here's how it would be done:

$r + \sqrt{64 + r^2} = 12$

$r + 64 + r^2 = 144$

$r + r^2 = 80$

Oh no... it seems that I have to solve two unknowns. The answer may be wrong.

**Gusbob** · September 26th 2010, 01:15 AM

You made a mistake squaring: you need to square $r + \sqrt{64+r^2}$ as a whole, or move the r over with the 12.

Here is how I'll do it.

$r + \sqrt{64 + r^2} = 12$
$\sqrt{64+r^2} = 12 -r$
$64 + r^2 = 144 - 24r + r^2$

**earboth** · September 26th 2010, 01:17 AM

Originally Posted by PythagorasNeophyte

Hello Gusbob!

Thank you for your reply.

So here's how it would be done:

$r + \sqrt{64 + r^2} = 12$

$r + 64 + r^2 = 144$

$r + r^2 = 80$

Oh no... it seems that I have to solve two unknowns. The answer may be wrong.

Sorry ... but what you've done is nearly a crime.

As Gusbob had told you, you have to move the linear r to the RHS first:

$\sqrt{64+r^2}=12-r$

Now square both sides:

$64+r^2=144-24r+r^2$

Now solve for r.

**PythagorasNeophyte** · September 27th 2010, 02:16 AM

Oh, I see. It seems that I had flopped my algebra.

$64 + r^2 = 144 - 24r + r^2$

$64-144 + r^2 - r^2 +24r = 0$

$-80 +24r = 0$

$24r = 80$

$r = 3\left(\frac{1}{3}\right)$

Now that I had found the radius of the sphere, I must find the volume of it:

$\dfrac{4}{3}\pi r^3$

$= \dfrac{4}{3}\pi (\dfrac{10}{3)}^3$

$= \dfrac{4}{3}\pi \dfrac{100}{9}$

$= \dfrac{400}{27}\pi$

Can anyone help me?

**earboth** · September 27th 2010, 04:26 AM

Originally Posted by PythagorasNeophyte

Oh, I see. It seems that I had flopped my algebra.

....
Now that I had found the radius of the sphere, I must find the volume of it:

$\dfrac{4}{3}\pi r^3$

$= \dfrac{4}{3}\pi \left(\dfrac{10}{3}\right)^3$

$= \dfrac{4}{3}\pi \dfrac{100}{9}$ <--- You have squared the value of the radius

...

The correct calculation:

$= \dfrac{4}{3}\pi \cdot \dfrac{1000}{27} = \dfrac{4000}{81} \pi$

**PythagorasNeophyte** · September 29th 2010, 12:20 AM

I am very sorry, earboth.

So now I would have to find the radius of the cone.

I discern that triangle VDO is similar to triangle VCA.

So:

$\dfrac{CA}{3\left(\frac{1}{3}\right)}$ $= \dfrac{12}{8]$ (corr. sides of similar triangles)

CA = 5cm

Volume of the cone is $\dfrac{1}{3}\pi r^2h$

= $\dfrac{!}{3}\pi5^2(!2)$

$= 100\pi$

So the volume of cone that is not occupied by the sphere is:

$100\pi - \dfrac{4000}{81} \pi$

Am I on the right track?

**earboth** · September 29th 2010, 01:46 AM

Originally Posted by PythagorasNeophyte

I am very sorry, earboth. <--- why?

So now I would have to find the radius of the cone.

I discern that triangle VDO is similar to triangle VCA.

So:

$\dfrac{CA}{3\left(\frac{1}{3}\right)}$ $= \dfrac{12}{8}$ (corr. sides of similar triangles)

CA = 5cm <<<<<<

Volume of the cone is $\dfrac{1}{3}\pi r^2h$

= $\dfrac{1}{3}\pi5^2(12)$

$= 100\pi$ <<<<<<

So the volume of cone that is not occupied by the sphere is:

$100\pi - \dfrac{4000}{81} \pi$

Am I on the right track?

Perfect!

If you simplify your result you'll see that the sphere nearly takes half of the volume of the cone.

(Btw: I've corrected your LaTeX)

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