Answer:
<u>So, The area of trapezoid = 4.413 R²</u>
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.
<u>Join </u>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
<u>∴ NP = 2 * NS = 1.274 R</u>
<u>Construct: a line parallel to ST from N to line MK</u>
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
<u>∴ MK = 3.139 R</u>
<u></u>
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
