Question

line segment on is perpendicular to line segment ml. what is the length of segment np? 2 units 4 units 1 unit 3 units

Explanation:

Step1: Use the Pythagorean theorem in right - triangle \(OPL\)

In right - triangle \(OPL\), \(OL = 5\) (radius of the circle), \(OP = 4\). According to the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = OL\), \(a = OP\), and \(b = PL\). So \(PL=\sqrt{OL^{2}-OP^{2}}\).
Substitute \(OL = 5\) and \(OP = 4\) into the formula: \(PL=\sqrt{5^{2}-4^{2}}=\sqrt{25 - 16}=\sqrt{9}=3\).

Step2: Use the property of perpendicular from the center to a chord

Since \(ON\perp ML\), by the property of a circle (a perpendicular from the center of a circle to a chord bisects the chord), \(MP = PL\). Also, in right - triangle \(OPN\) (because \(ON = OL = 5\) (radii of the same circle), \(OP = 4\)), and using the Pythagorean theorem in right - triangle \(OPN\) where \(ON\) is the radius (\(ON = 5\)) and \(OP = 4\). Let \(NP=x\), then \(x=\sqrt{ON^{2}-OP^{2}}\) (another way: since the perpendicular from the center to the chord bisects the chord \(ML\), and in right - triangle \(OPL\) we found \(PL = 3\), and also \(NP=PL\) (by congruent right - triangles \(OPL\) and \(OPN\) ( \(ON = OL\), \(OP\) is common, \(\angle OPL=\angle OPN = 90^{\circ}\)) )

Answer:

\(3\) units