Question 1181732
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In cross section, the diagram is a square (the cylinder) inscribed in a circle (the sphere).<br>
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.<br>
That triangle is an isosceles right triangle, so the hypotenuse (the radius of the sphere) is sqrt(2) times the radius of the cylinder.<br>
So let r be the radius of the cylinder; then r*sqrt(2)is the radius of the sphere.<br>
The height of the cylinder is 2r.<br>
(a) Surface areas....<br>
Cylinder: {{{2(pi)r^2+2(pi)(r)(h) = 2(pi)r^2+2(pi)(r)(2r) = 6(pi)r^2}}}
Sphere: {{{4(pi)(r*sqrt(2))^2 = 8(pi)r^2}}}<br>
The ratio of the surface areas is 6:8 = 3:4<br>
(b) Volumes....<br>
Cylinder: {{{(pi)(r^2)(h) = (pi)(r^2)(2r) = 2(pi)r^3}}}
Sphere: {{{(4/3)(pi)(r*sqrt(2))^3 = ((8/3)sqrt(2))(pi)r^3}}}<br>
The ratio of the volumes is 2:(8/3)sqrt(2) or 1:(4/3)sqrt(2)<br>