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 ?
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 $\displaystyle r$ you have $\displaystyle r^2+9=36$ thus, $\displaystyle r=\sqrt{27}=3\sqrt{3}$Originally Posted by totalnewbie
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 $\displaystyle 3\sqrt{3}$ (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, $\displaystyle 3\sqrt{3} 3\sqrt{3}=3x$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.
Sorry, I did you problem the other way around I put the cone into the sphere.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, $\displaystyle \frac{6\sqrt{3}}{6+3\sqrt{3}}=\frac{6\sqrt{3}}{6+3 \sqrt{3}}\frac{6-3\sqrt{3}}{6-3\sqrt{3}}=4\sqrt{3}-6$.
The diagram shows a cross section through the cone and cylinder.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 )
RonL