# Thread: Perplexing Mensuration Question - sphere in a cone

1. ## Perplexing Mensuration Question - sphere in a cone

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 $\displaystyle VD=VC$, $\displaystyle VO=VA$ and $\displaystyle OD=CA$.

Can anyone give me hints? Help is greatly appreciated!

2. Hello!

Here is a hint for you!

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

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

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.

3. Hello Gusbob!

So here's how it would be done:

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

$\displaystyle r + 64 + r^2 = 144$

$\displaystyle r + r^2 = 80$

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

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

Here is how I'll do it.

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

5. Originally Posted by PythagorasNeophyte
Hello Gusbob!

So here's how it would be done:

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

$\displaystyle r + 64 + r^2 = 144$

$\displaystyle r + r^2 = 80$

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:

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

Now square both sides:

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

Now solve for r.

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

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

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

$\displaystyle -80 +24r = 0$

$\displaystyle 24r = 80$

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

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

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

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

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

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

Can anyone help me?

7. 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:

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

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

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

...
The correct calculation:

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

8. 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:

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

CA = 5cm

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

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

$\displaystyle = 100\pi$

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

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

Am I on the right track?

9. 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:

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

CA = 5cm <<<<<<

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

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

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

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

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

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.