A sphere of radius $15$ $15$ inches rest on a circular hole of diameter $10$ $10$ inches cut into a board. How far does the sphere extend below the surface of the board?

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Answer 1 · 2015-10-12T20:26:45

$15 - 10 \sqrt{2} \approx 0.858$ $15 - 10sqrt(2) ~~ 0.858$ inches

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In this diagram:

$a = \sqrt{15^{2} - 5^{2}} = \sqrt{225 - 25} = \sqrt{200} = 10 \sqrt{2}$ $a = sqrt(15^2 - 5^2) = sqrt(225-25) = sqrt(200) = 10sqrt(2)$

So the sphere extends $15 - a = 15 - 10 \sqrt{2}$ $15 - a = 15 - 10sqrt(2)$ inches below the upper surface of the board.

A sphere of radius 1515 inches rest on a circular hole of diameter 1010 inches cut into a board. How far does the sphere extend below the surface of the board?