 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.) solution | Comment 2 of 6 | Incorrect y coordinate in the below analysis results in considering a cone of incorrect radius. See subsequent comments:

The plane cut equal areas if it cuts the two solids in circles of the same radius. That allows us to make this a plane geometry problem by erecting a vertical plane containing the axis of the cone and the center of the circle. To simplify the equation of the resulting circle (on the surface of the sphere), let's put the origin at the center of the circle/sphere. The top and bottom of the sphere are  then each 10 cm away from this midline while the apex of the cone is on the midline.

The equation of the circle is x^2 + y^2 = 100 and in this instance x represents the radius of the circle the sphere cuts off on the horizontal cutting plane mentioned in the puzzle.

The cone (triangle cross section that is) is entirely below the x-axis except for the apex, which is on the x-axis. Half the distance between the two sides with the same value of y constitutes the radius of the cross-section circle of the cone, and that is exactly -y.

So we need to solve x^2 + y^2 = 100 where x = -y and x is positive:

2*y^2 = 100

y = -sqrt(50)

Since the bottom is 10 cm down from the origin, the cutting plane is 10 - sqrt(50) ~= 2.92893218813452 cm up from the floor.

Edited on September 8, 2018, 8:57 am
 Posted by Charlie on 2018-09-07 13:57:30 Please log in:
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