Two spheres of radii 17 and 7 are inscribed into a cone?

Question

The distance between their centers is 26.
What is area of cross-section of the cone
by the plane tangent to both spheres?

Answer 1

The radii and spacing of the spheres have been carefully selected to make this problem easy. The foci of the ellipse are at the points where the plane is tangent to the spheres, and the distance between them is 10. It turns out that we can add 7 to (1/2) of 10 to get the major axis, or 12. Then the area of the ellipse can be found with the expression:

π a²√(1-(c/a)²)

where a = 12, c = 5, or

12 π √119

Answer 2

that means the change in the radius of the cone is 10 in the distance 26. That means the cone opens at a slope of 10/26.

The cross section tangent to the sphere of radius 17. is going to have a radius of 17 + 17*(10/26) = 23.538 and an area of 1740.63.

The other cross section, will be [7 + 7(10/26)]^2 * pi = 295.124

I hope I read your question right. A picture would have helped a ton, hope this helps.

Answer 3

questions
1) if the two radii are 17 and 7, sum = 24. 26 > 24
. . . . this means the two sphere are not tangent to each other
2) what do you mean cross-section by the plane? . . tangent to both spheres?

Answer 4

If the sector is melted down right into a distinctive shape, in concept they might desire to have the comparable section. so first locate the part of the circle (4/3)*(3.14)*(7^3)=1436.03 next set this section to the hot formulation 1436.03=(a million/3)*(3.14)*(r^2)*(28) r^2= 40 9 Radius =7cm desire this enables have an incredible day!