points n and r both lie on circle o. line segment rq is tangent to the circle at point r. what is the…

Explanation:

Step1: Recall tangent - radius property

Since RQ is tangent to the circle at point R, $\angle{ORQ}=90^{\circ}$. Let the radius of the circle be $r = ON=OR = 5$.

Step2: Use the Pythagorean theorem in right - triangle ORQ

In right - triangle ORQ, $OR = 5$ and $RQ=5\sqrt{3}$. By the Pythagorean theorem $OQ=\sqrt{OR^{2}+RQ^{2}}=\sqrt{25 + 75}=\sqrt{100}=10$. And $ON = 5$, so $NQ=OQ - ON=10 - 5 = 5$.

Step3: Prove $\triangle{ORN}$ is equilateral

Since $ON = OR = 5$ and $\angle{ORQ}=90^{\circ}$, $\cos\angle{ROQ}=\frac{OR}{OQ}=\frac{5}{10}=\frac{1}{2}$, so $\angle{ROQ}=60^{\circ}$. Because $ON = OR$, $\triangle{ORN}$ is an isosceles triangle. And $\angle{ROQ}=60^{\circ}$, so $\triangle{ORN}$ is an equilateral triangle, then $RN = OR=ON = 5$.

Step4: Calculate the perimeter of $\triangle{RON}$

The perimeter of $\triangle{RON}$ is $P=OR + RN+ON$. Substituting $OR = RN=ON = 5$ into the formula, we get $P=5 + 5+5=15$.

Answer:

15.0 units