 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A Cone and a Sphere (Posted on 2018-09-07) 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.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) attempted solution | Comment 1 of 6

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 Please log in:
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