In the figure, circle \(O\) has center \(O,\) diameter \(AB\) and a radius of \(5.\) Line \(CD\) is parallel to the diam

[M7MBA]

In the figure, 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. \(\dfrac53\pi+5\sqrt3\)

B. \(\dfrac53\pi+10\sqrt3\)

C. \(\dfrac{10}3\pi+5\sqrt3\)

D. \(\dfrac{10}3\pi+10\sqrt3\)

E. \(\dfrac{10}3\pi+20\sqrt3\)

Answer: D

Source: Manhattan GMAT

[Scott@TargetTestPrep]

Perimeter.PNG

In the figure, 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. \(\dfrac53\pi+5\sqrt3\)

B. \(\dfrac53\pi+10\sqrt3\)

C. \(\dfrac{10}3\pi+5\sqrt3\)

D. \(\dfrac{10}3\pi+10\sqrt3\)

E. \(\dfrac{10}3\pi+20\sqrt3\)

Answer: D

Solution:

Since CD is parallel to AB, x = 30 degrees. The perimeter of the shaded region consists of equal chords BC and BE, as well as arc CAE. Let’s calculate the length of chord BC by drawing AC. We see that triangle ABC is a 30-60-90 triangle with hypotenuse AB = 10 (notice AB is the diameter of the circle with radius of 5). Therefore, AC, the shorter leg, is ½ x 10 = 5, and BC, the longer leg, is 5 x √3 = 5√3. Since BE = BC, BE = 5√3 also. Finally, for the length of arc CAE, notice that its inscribed angle is angle CBE = 2x = 60 degrees. The measure of arc CAE, in degrees, is twice its inscribed angle, so arc CAE = 120 degrees and its length is ⅓ of the circumference of the circle:

⅓ x 10π = 10π/3

Therefore, the perimeter of the shaded region is:

BC + BE + arc CAE = 5√3 + 5√3 + 10π/3 = 10√3 + 10π/3

Answer: D