perimeter of shaded region

Is the answer = D ?

Can you please check if the options are correct !

nidhis.1408 wrote:In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

A (5/3)� + 5√3
B (5/3)� + 10√3
C (10/3)� + 5√3
D (10/3)�+ 10√3
E (10/3)� + 20√3


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I've amended the answer choices to reflect how they appear in the actual question.

Perimeter of the shaded region = arc CAE + CB + EB.

Since AB || CD, ∠DCB and ∠CBA are congruent.
Thus, x=30, and ∠CBE = 60.

An inscribed angle is formed by two chords.
Thus, ∠CBE is an inscribed angle.
The degree measurement of an inscribed angle = 1/2 the degree measurement of the intercepted arc.
∠CBE intercepts arc CAE.
The, arc CAE = 120 degrees.
Since the entire circle = 360 degrees, the length of arc CAE = 120/360 = 1/3 of the entire circumference.
C = 2�r = 10�.
Thus, the length of arc CAE = (1/3)(10�) = (10/3)�.

The correct answer must include (10/3)�.
Eliminate A and B.

Since ∠CBA = ∠ABE, the shaded sector above the diameter is congruent to the shaded sector below the diameter.
Thus, CB = EB.
Since OC=5 and AB=10, the length of CB must be between 5 and 10.
Thus, CB and EB are each between 5 and 10, implying that CB + EB is between 10 and 20.

Of the remaining answer choices, only D works:
(10/3)� + 10√3 ≈ (10/3)� + 10(1.7) ≈ (10/3)� + 17.

The correct answer is D.

length of arc CAE + CB + BE
But CB = BE
=> length of arc CAE + 2CB

1) to determine length of arc CAE
x = 30 degrees => angle CBE = 60 degrees
So COE = 120 degrees
So arc CAE = 1/3(2*PI*5) = 10PI/3

2) to determine CB.
CBA is right angled triangle with right angle at C.
apply sine rule
AB/SinC = CB/SinA => 10/1 = CB/Sin60
=> CB = 5root(3)

so ans should be (10/3)�+ 10√3

user123321