Question

HELP I NEED THIS NOW!!!!

Answer 1

Answer:

So, The area of trapezoid = 4.413 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and m∠NMK = 65° , OT = R

So, m∠MNP = 180° - 65° = 115° ⇒ supplementary interior angles.

Join ON ⇒ ΔOSN is a right triangle at S.

And NP, NM are tangents to the circle

So, ON bisects the angle ∠MNP = ∠MNS

∠ONS = (∠MNS)/2 = 115/2 = 57.5°

So, tan (∠ONS) = OS/NS = R/(NS)

NS = R/tan (∠ONS)=R/(tan 57.5°) ≈ 0.637 R

∴ NP = 2 * NS = 1.274 R

Construct: a line parallel to ST from N to line MK

let the intersection point be Q  ⇒ NQ = 2R

So, Δ NQM is a right triangle at Q

tan (∠NMQ) = tan 65 = NQ/MQ

MQ = NQ/(tan 65)=2R/tan 65 ≈ 0.932 R

∴ MK = 2 MT = 2 (MQ + QT) ⇒ QT = NS

         = 2 (MQ + NS ) = 2( 0.932 R + 0.637 R) = 2 * 1.569 R

∴ MK = 3.139 R

Now, area of trapezoid = height × (sum of parallel sides/ 2)

Area = ST * (NP + MK)/2

        = 2R * (1.274 R + 3.139 R) / 2

        = 2R * 4.413 R /2

       = 4.413 R²

So, The area of trapezoid = 4.413 R²