A sphere has a radius of 10 cm. A right circular cone has a height of 10 cm and base diameter of 10 cm. The sphere and the cone stand on a horizontal surface. If a horizontal plane cuts both the sphere and the cone, the cross-sections will both be circles.

Find the height of the horizontal plane (from the bottom) (in cm) that gives circular cross-sections of the sphere and cone of equal area.

Consider first the vertical plane that includes the vertical axis of each shape. The intersection of this plane with these shapes is a circle and triangle. We next need to cut the circle and triangle with a horizontal line at a height that traverses the same interior width in each 2d shape.  The problem just got much simpler.  That width is a diameter, but let's rather focus on the radius.

For the triangle:

r(h) = 5 (1-h/10)  (at the base: h=0, the half width is 5, and at the top: h=10, it is 0)

For the circle:

r(h)=sqrt {100- (10-h)^2}  (since x^2 + y^2 =10^2 and r = x along this line).

Equating these gives:

h^2 - 20 h + 20 = 0

h = 10 (+/-) 4 sqrt (5) = 1.0557, 18.944 (choose the first root, since the second root gives a negative radius)

h = 1.0557 cm

radius = 4.472   (this is sqrt 20) and area = 62.83 cm^2

Edited on September 8, 2018, 4:22 am

Posted by Steven Lord on 2018-09-07 08:58:43