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.

We can connect the line CA and AE are we get a 30-60-90 triangles in both the cases and then we can compute the lengths of the 2 sides reqd...

vowwwwthis takes me 10 menutes

Another way you might save a little time here is to

1. Draw radii OC and OE
2. Recognize that OCB and OEB are 30-30-120 triangles
3. Use the fact that the ratio of sides in such a triangle is 1:1:root(3) to get CB = EB = 5*root(3)

30-30-120 triangles obviously aren't the most common ones, but they aren't unheard of either. You can derive the ratio above by drawing a 30-30-120, splitting it into two 30-60-90 triangles by drawing a line from the middle of the longest side to the opposite corner, and then applying what you know about the side ratios in 30-60-90's.

Recognize that OCB and OEB are 30-30-120 triangles
Use the fact that the ratio of sides in such a triangle is 1:1:root(3) to get CB = EB = 5*root(3)

Great. Thanks for sharing.