Sphere is into the cone.

Generator of the cone is 6 cm and height of the cone is 3 cm. How much is radius of the sphere ?

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- Jan 8th 2006, 09:12 AMtotalnewbieSphere into cone
Sphere is into the cone.

Generator of the cone is 6 cm and height of the cone is 3 cm. How much is radius of the sphere ? - Jan 8th 2006, 09:55 AMThePerfectHackerQuote:

Originally Posted by**totalnewbie**

- Jan 8th 2006, 10:06 AMtotalnewbie
I meant that: The slant height of an object (such as a cone, frustum, or pyramid) is the distance measured along a lateral face from the base to the apex.

- Jan 8th 2006, 10:42 AMThePerfectHackerQuote:

Originally Posted by**totalnewbie**

- Jan 8th 2006, 11:01 AMtotalnewbie
Oh my GOD!

I wanted to know radius of the sphere not cone!

Sorry, my mistake. Could you help ? - Jan 8th 2006, 01:02 PMThePerfectHackerQuote:

Originally Posted by**totalnewbie**

- Jan 8th 2006, 01:17 PMtotalnewbie
Well. I have a sphere and cone. I take the sphere and I put it in the cone. Then the sphere is inside the cone. And now I have to know the radius of the sphere which is inside the cone.

- Jan 8th 2006, 02:25 PMThePerfectHackerQuote:

Originally Posted by**totalnewbie**

Thus, $\displaystyle 27=3x$ thus, $\displaystyle x=9$. Thus the total length of diameter is $\displaystyle 9+3=12$ thus, the radius of sphere is 6. - Jan 9th 2006, 11:09 AMtotalnewbie
How is it possible that the radius of sphere is 6 when the radius of cone is sqrt(27) and the sphere is inside the cone. The radius of sphere must be smaller than the radius of cone.

- Jan 9th 2006, 02:19 PMThePerfectHackerQuote:

Originally Posted by**totalnewbie**

Now, I can finally do the problem properly.

There is an elegant theorem from geometry that the area of any triangle divided the its semiperimeter is the radius of the inscribed circle!

Thus, draw an isoseles triangle with lateral length 6 and height 3 thus, its base is $\displaystyle 6\sqrt{3}$. Thus, the area is$\displaystyle 9\sqrt{3}$ To find its perimeter use the pythagorean theorem and it was covered in this forum how to do that, thus it is (semiperimeter) $\displaystyle 6+3\sqrt{3}$. Thus, the radius is $\displaystyle r=\frac{6\sqrt{3}}{6+3\sqrt{3}}$.

But, $\displaystyle \frac{6\sqrt{3}}{6+3\sqrt{3}}=\frac{6\sqrt{3}}{6+3 \sqrt{3}}\frac{6-3\sqrt{3}}{6-3\sqrt{3}}=4\sqrt{3}-6$. - Jan 17th 2006, 05:13 AMtotalnewbie
This is not the right answer. The right one is 1,39 cm. Use the paper and do the drawing and you can see that radius of this sphere can't be 0,9 as you have said.

- Jan 17th 2006, 06:19 AMCaptainBlackQuote:

Originally Posted by**totalnewbie**

From this we see that Pythagoras's theorem tells us that:

$\displaystyle (3-r)^2=r^2+(3-r)^2/4$,

which has solutions:

$\displaystyle r \approx -19.3923\ \mbox{or}\ r \approx 1.3923$.

The first of these is non-physical so:

$\displaystyle r \approx 1.3923$.

(I must admit that I took a short cut to solving the equation, the thing is

obviousely a quadratic, but rather than risk error in rearranging it and

applying the quadratic formula I used QuickMath to do the algebra for me :D )

RonL - Jan 17th 2006, 03:02 PMThePerfectHacker
Curious, someone recheck my solution and tell me where I made a mistake? It is 3 posts back.

- Jan 17th 2006, 08:35 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

RonL