GeminiHeat wrote:
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. \((\frac{5}{3})\pi + 5\sqrt{3}\)
B. \((\frac{5}{3})\pi + 10\sqrt{√}3\)
C. \((\frac{10}{3})\pi + 5\sqrt{3}\)
D. \((\frac{10}{3})\pi + 10\sqrt{3}\)
E. \((\frac{10}{3})\pi + 20\sqrt{3}\)
Join CE, OC and OE
Since, CD is parallel to AB, \(x = 30\) (Alternate Interior angles are equal)
i.e. \(∠CBA = 60\)
Thus \(∠COE = 120\) (Central angle is twice the angle subtended between two chords)
Arc Length CE \(= \frac{120}{360}2π(5) = \frac{10π}{3}\)
Now, CBE is an equilateral triangle inscribed in a circle with O as the centroid of the triangle
i.e. \(OB = radius = \frac{2}{3}H\)
\(5 = \frac{2}{3}H\)
\(H = \frac{15}{2}\)
We know, height of an equilateral triangle is given by \(\frac{\sqrt{3}}{2}a\) where \(a\) is the side of equilateral triangle
Thus, \(\frac{\sqrt{3}}{2}a = \frac{15}{2}\)
\(a = \frac{15}{\sqrt{3}} = 5\sqrt{3}\)
Perimeter of Shaded region =\(\frac{10π}{3} + 5\sqrt{3} + 5\sqrt{3} = \frac{10π}{3} + 10\sqrt{3}\)
Hence, option D
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In the figure, circle O has center O, diameter AB.png [ 9.12 KiB | Viewed 3048 times ]