Question

points n and r both lie on circle o. line segment rq is tangent to the circle at point r. what is the perimeter of triangle ron? 10.0 units 15.0 units 18.7 units 23.7 units

Explanation:

Step1: Use the property of tangent

Since \(RQ\) is tangent to the circle at \(R\), \(OR\perp RQ\) (tangent - radius property). In right - triangle \(ORQ\), \(ON = OR=5\) (radii of the circle), \(NQ = RQ = 5\sqrt{3}\) (tangent - secant property: if two tangent segments are drawn from an external point to a circle, then the two segments are congruent). And \(OQ=ON + NQ=5 + 5\).

Step2: Find \(\angle Q\)

In right - triangle \(ORQ\), \(\sin\angle Q=\frac{OR}{OQ}\). Let \(OQ = x\), \(OR = 5\), \(RQ=5\sqrt{3}\). By the Pythagorean theorem \(OQ^{2}=OR^{2}+RQ^{2}\), but we can also use \(\tan\angle Q=\frac{OR}{RQ}\). \(\tan\angle Q=\frac{5}{5\sqrt{3}}=\frac{1}{\sqrt{3}}\), so \(\angle Q = 30^{\circ}\), then \(\angle ROQ=60^{\circ}\).

Step3: Prove \(\triangle RON\) is equilateral

Since \(ON = OR\) (radii) and \(\angle ROQ = 60^{\circ}\), \(\triangle RON\) is an equilateral triangle (a triangle with two equal sides and the included angle equal to \(60^{\circ}\) is equilateral). So \(RN=ON = OR = 5\).

Step4: Calculate the perimeter

The perimeter of \(\triangle RON\) is \(P=OR + ON+RN\). Substituting \(OR = ON=RN = 5\), we get \(P=5 + 5+5=15\).

Answer:

15.0 units