Question

1. in the figure at the right, sphere o with radius 15 mm is intersected by a plane 3 mm from the center. to the nearest tenth, find the radius of the cross section ⊙y.

Explanation:

Step1: Apply Pythagorean theorem

Let \(r\) be the radius of the cross - section. The radius of the sphere \(R = 15\) mm and the distance from the center of the sphere to the plane \(d=3\) mm. According to the Pythagorean theorem \(R^{2}=r^{2}+d^{2}\).
So, \(r^{2}=R^{2}-d^{2}\).

Step2: Substitute values

Substitute \(R = 15\) and \(d = 3\) into the formula: \(r^{2}=15^{2}-3^{2}\).

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Step3: Solve for \(r\)

Take the square root of both sides: \(r=\sqrt{216}\approx14.7\)

Answer:

\(14.7\) mm