Lesson A sphere placed in an inverted cone

A sphere placed in an inverted cone

Problem 1

Let's consider vertical axial section of our configuration.


In vertical section, we have inverted isosceles triangle ABC with the base BC at the top 
and vertex A.  The base BC is 5 + 5 = 10 cm long.  The height (or the altitude) of the triangle from A to BC is 12 cm.
We also have a circle, inscribed into the triangle.
This circle is the section of the sphere, placed inside the cone.



Let start calculating the length of the lateral sides AB and AC.
They are the hypotenuses of right angled triangles, so

    AB = AC =  =  =  = 13 cm.



Obviously, the area of triangle ABC is

     =  = 60 square units.    (1)    (half the product the base and the altitude)



We can calculate the area of triangle ABC in other way as half the product of its perimeter 
and the radius of this inscribed circle

     = .    (2)   



Left sides of equations (1) and (2) are equal ( both represent  ).

Therefore, we can write this equation

    60 = ,  or  120 = P*r     (3)

after reducing the factor 1/2 in both sides.



The perimeter is easy to calculate:  P = AB + AC + BC = 13 + 13 + 10 = 36 cm.



So, equation (3) takes the form

    120 = 36*r,

which gives  

    r =  =  = 3  cm.


Thus we found the radius of the circle and of the sphere.


ANSWER. The radius of the sphere is 3  cm.