# Sphere into cone

• Jan 8th 2006, 10:12 AM
totalnewbie
Sphere 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, 10:55 AM
ThePerfectHacker
Quote:

Originally Posted by totalnewbie
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 cone ?

I want to help but do not know what a generator of a cone is?
• Jan 8th 2006, 11:06 AM
totalnewbie
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, 11:42 AM
ThePerfectHacker
Quote:

Originally Posted by totalnewbie
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.

I am assuming you are using a right-cone (with a right angle) but then this is application of Pythagorean Theorem, try to imagine that because I cannot draw it. The lateral lengh is 6 that is your hypotenuse and the height is 3. Thus call the radius $r$ you have $r^2+9=36$ thus, $r=\sqrt{27}=3\sqrt{3}$
• Jan 8th 2006, 12:01 PM
totalnewbie
Oh my GOD!
I wanted to know radius of the sphere not cone!
Sorry, my mistake. Could you help ?
• Jan 8th 2006, 02:02 PM
ThePerfectHacker
Quote:

Originally Posted by totalnewbie
Oh my GOD!
I wanted to know radius of the sphere not cone!
Sorry, my mistake. Could you help ?

Yes, but you need to rephrase the problem I do not understand
• Jan 8th 2006, 02:17 PM
totalnewbie
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, 03:25 PM
ThePerfectHacker
Quote:

Originally Posted by totalnewbie
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.

There is a theorem that when two chord interest in the circle the product of their lengths are equal. Now you have a cone in a sphere imagine that to be a triangle with lateral length 6 height 3 inscribed a circle, thus the radius of this triangle is $3\sqrt{3}$ (this was covered before). From the vertex of the triangle draw the diameter of the circle, thus the diameter and the base side of the triangle intersect. But by the theorem I just told you multiply their segments, thus 1/2base of traingle*1/2base of triangle=Height of triangle*Remainder of diameter. Thus, $3\sqrt{3} 3\sqrt{3}=3x$
Thus, $27=3x$ thus, $x=9$. Thus the total length of diameter is $9+3=12$ thus, the radius of sphere is 6.
• Jan 9th 2006, 12:09 PM
totalnewbie
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, 03:19 PM
ThePerfectHacker
Quote:

Originally Posted by totalnewbie
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.

Sorry, I did you problem the other way around :( I put the cone into the sphere.

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

But, $\frac{6\sqrt{3}}{6+3\sqrt{3}}=\frac{6\sqrt{3}}{6+3 \sqrt{3}}\frac{6-3\sqrt{3}}{6-3\sqrt{3}}=4\sqrt{3}-6$.
• Jan 17th 2006, 06:13 AM
totalnewbie
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, 07:19 AM
CaptainBlack
Quote:

Originally Posted by totalnewbie
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 ?

The diagram shows a cross section through the cone and cylinder.

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

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

which has solutions:

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

The first of these is non-physical so:

$r \approx 1.3923$.

(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, 04:02 PM
ThePerfectHacker
Curious, someone recheck my solution and tell me where I made a mistake? It is 3 posts back.
• Jan 17th 2006, 09:35 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker

[snip]

Thus, draw an isoseles triangle with lateral length 6 and height 3 thus, its base is $6\sqrt{3}$.

[snip]