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In the figure, circle O has center O, diameter AB and a radius of 5.
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21 May 2021, 08:24
21 May 2021, 08:24
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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. \((\frac{10}{3})\pi + 20\sqrt{3}\)
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.
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21 May 2021, 22:31
21 May 2021, 22:31
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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 3408 times ]
Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
22 May 2021, 10:08
22 May 2021, 10:08
3
Solution:
Construction: Join AC & AE.
As we know AB is the diameter, Angle ACB & Angle AEB will be 90 degrees(Inscribed angle theorem)
Angle CBA= 30 degrees (Alternate Angle theorem)
Thus Triangle BCA is a 30-60-90 triangle
And using the 30-60-90 theorem, we can find, length of CB=5\(\sqrt{3}\)
As, Triangle BEA is similar to Triangle BCA (One side and two angles are equal)
Thus, EB=5\(\sqrt{3}\)
Therefore, I know that the two side of the highlighted region: 5\(\sqrt{3}\)+5\(\sqrt{3}\)=10\(\sqrt{3}\)
And we see 10\(\sqrt{3}\) only in option D
IMO D
Hope This helps!
Construction: Join AC & AE.
As we know AB is the diameter, Angle ACB & Angle AEB will be 90 degrees(Inscribed angle theorem)
Angle CBA= 30 degrees (Alternate Angle theorem)
Thus Triangle BCA is a 30-60-90 triangle
And using the 30-60-90 theorem, we can find, length of CB=5\(\sqrt{3}\)
As, Triangle BEA is similar to Triangle BCA (One side and two angles are equal)
Thus, EB=5\(\sqrt{3}\)
Therefore, I know that the two side of the highlighted region: 5\(\sqrt{3}\)+5\(\sqrt{3}\)=10\(\sqrt{3}\)
And we see 10\(\sqrt{3}\) only in option D
IMO D
Hope This helps!
Attachments
Capture.JPG [ 19.31 KiB | Viewed 3309 times ]
