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

this is not the answer this is just a question. is this middle school math?

step-by-step explanation:

c.   x=2.4cm

step-by-step explanation:

similar triangles are triangles which have the same shape but not the same size. this means their angle measures are equal but their side lengths are not instead they are proportional. to solve, we will set up a proportion and solve.

a proportion is two equal ratios set equal to each other. we form the ratios by finding corresponding sides (sides which match to each other on both triangles by their position). out ratios will be one side of the small triangle over the corresponding side on the big triangle as shown below:

.

to solve the proportion, we'll cross multiply by multiplying numerator with denominator across the equal sign.