The problem states that r = 5.
Thus, circumference = 2�r = 10�.
Perimeter of shaded region = arc CAE + CB + EB.
Arc CAE:
Since AB || CD, x=30.
Thus, ∠CBE = 60.
∠CBE is an inscribed angle that intercepts arc CAE.
When an inscribed angle intercepts an arc, the degree measurement of the intercepted arc is twice that of the inscribed angle.
Thus, arc CAE = 2*60 = 120 degrees.
Since 120/360 = 1/3, arc CAE is 1/3 of the circumference.
Thus, arc CAE = (1/3) * 10� = (10/3)�.
CB + EB:
The length of CB is more than the radius of 5 but less than the diameter of 10.
Same for EB.
CB + EB ≈ 8+8 ≈ 16.
Thus:
Perimeter ≈ (10/3)� + 16.
Looking at the answer choices:
Only answer choice D works:
(10/3)� + 10√3 ≈ (10/3)� + 10*(1.7) ≈ (10/3)� + 17.
The correct answer is D.
