If youcut the cone in two halves, you will get an equilateral triangle its vertices are vertices of an hexagon and its height is 3R/2 and R is the radius of the sphere. Continue from there
A cone circumscribes a sphere and has its slant height equal to the diameter of its base.
Show that the volume of the cone is 9/4 the volume of the sphere.
Basically I need to show my professor that the volume of the cone is 9/4 the volume of the sphere.
Can someone help me with it or just give me like 3 solution steps so I can proceed. Thanks in advance.
My apology. the center of an inscribed circle is the intersection point of the bisectors. In an equilateral triangle a bisector is also a median. The intersection point of the bisectors is also the intersection point of the medians. Therefore, in your problem the radius of the inscribed circle is 1/3 the length of the median, which is also the height of the equilateral triangle. You need to correlate the side of the triangle with the radius of the circle via the length of the height.
I decipher 1/2drad3 as (1/2)*diameter*radical(3), radical here means sqrt
You got the radius of the sphere correctly assuming the height of the cone is one unit. but if you label the height of the cone h units, What the radius of the sphere would be.
Now you will need to relate h to the side of the equilateral triangle, which is the base of the triangle (why?)