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? rn=qn rq=5sqroot3 on=5 10.0 units 15.0 units 18.7 units 23.7 units

Third option 18.7 units is the correct answer.

Step-by-step explanation:

We will draw a figure to understand the given parameters first.

Tangent RQ = 5√3

RN = QN and ON = 5

We have to find the perimeter of the triangle RON.

By the tangent theorem length of tangents are same when drawn from any external point.

Therefore QR = QN = 5√3

ON = 5 (given)

OR = ON = 5 (Radius of the circle)

Therefore perimeter of triangle RON = OR + ON + RN

P = 2×ON + RN = 2×5 + QN (RN = QN given)

P = 10 + 5√3 = 10 + 5×1.73 = 10 + 8.7 = 18.7 units

Therefore perimeter of ΔRON = 18.7 units

Option 3 is correct .i.e., Area of ΔRON = 18.7 units

Step-by-step explanation:

Given: Circle with centre O , RN = QN , RQ = units ,

          ON = 5 units , RQ is tangent.

To find: Perimeter of ΔRON.

Figure is attached.

Perimeter of Δ RON = OR + ON + RN    (1)

As, Q is an external point.

So, to satisfy the Condition RN = QN on graph. QN should also be a tangent from external point Q.

⇒ QN = QR (Tangent from same external point are equal in length)

⇒ QN = QR = RN

⇒ RN = 5 units

OR and ON are radius of same circle.

⇒ OR  = ON = 5 units

Subtstituting these value in eqn 1, we get

∴Perimeter of Δ RON = 5 + 5 + 5

                                    = 10 + 5

                                    = 10 + 5 × (1.73205)      (Value of    = 1.73025)  

                                    = 10 + 8.66025

                                    = 18.66025 units

                                    ≈ 18.7 units (Rounded off to nearest tenth)

Therefore, Option 3 is correct .i.e., Area of ΔRON = 18.7 units

0.6

39mph

step-by-step explanation: