SOLUTION: A right circular cylinder with altitude equal to the diameter of its base is inscribed in a sphere. (a) compare the lateral area of the cylinder with the area of the sphere. (b) Co

Algebra ->  Volume -> SOLUTION: A right circular cylinder with altitude equal to the diameter of its base is inscribed in a sphere. (a) compare the lateral area of the cylinder with the area of the sphere. (b) Co      Log On


   

Question 1181732: A right circular cylinder with altitude equal to the diameter of its base is inscribed in a sphere. (a) compare the lateral area of the cylinder with the area of the sphere. (b) Compare the volume of the sphere with that of the cylinder.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


In cross section, the diagram is a square (the cylinder) inscribed in a circle (the sphere).

Draw the right triangle with vertices at (a) the center of the square/circle, (b) the center of the top of the cylinder, and (c) the point of contact of the square and the circle.

That triangle is an isosceles right triangle, so the hypotenuse (the radius of the sphere) is sqrt(2) times the radius of the cylinder.

So let r be the radius of the cylinder; then r*sqrt(2)is the radius of the sphere.

The height of the cylinder is 2r.

(a) Surface areas....

Cylinder:
Sphere: 4%28pi%29%28r%2Asqrt%282%29%29%5E2+=+8%28pi%29r%5E2

The ratio of the surface areas is 6:8 = 3:4

(b) Volumes....

Cylinder: %28pi%29%28r%5E2%29%28h%29+=+%28pi%29%28r%5E2%29%282r%29+=+2%28pi%29r%5E3
Sphere: %284%2F3%29%28pi%29%28r%2Asqrt%282%29%29%5E3+=+%28%288%2F3%29sqrt%282%29%29%28pi%29r%5E3

The ratio of the volumes is 2:(8/3)sqrt(2) or 1:(4/3)sqrt(2)