A manufacturer drills a round hole of radius r through the center of a metal sphere of radius r. find the volume of the remainin
g metal “bead.”
1 answer:
The media attached below is the bottom/top view of drilled sphere. If you look it from sideways, you'll see that the drilled hole is in the form of cylindrical shape. So,
Volume of the drilled shape= volume of cylinder = πr²h, where h is 2r.
∴Volume of cylinder(V1) = πr²(2r) = 2πr³.
Volume of sphere(V2) = 4πr³/3.
Therefore volume of remaining metal bead = V1 - V2
= 2πr³ - 4πr³/3
= (6πr³ - 4πr³)/3
= 2πr³/3.
Hence volume of remaining bead is 2πr³/3
Volume of the drilled shape= volume of cylinder = πr²h, where h is 2r.
∴Volume of cylinder(V1) = πr²(2r) = 2πr³.
Volume of sphere(V2) = 4πr³/3.
Therefore volume of remaining metal bead = V1 - V2
= 2πr³ - 4πr³/3
= (6πr³ - 4πr³)/3
= 2πr³/3.
Hence volume of remaining bead is 2πr³/3
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A triangle MPN is right and <span>isosceles, because
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