GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
21 May 2021, 08:24
21 May 2021, 08:24
1
Expert Reply
8
Bookmarks
00:00
Question Stats:
59% (03:11) correct
40% (03:17) wrong based on 27 sessions
Hide Show timer Statistics
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. \((\frac{10}{3})\pi + 20\sqrt{3}\)
Show: ::
Attachment:
Perimeter.PNG [ 3.44 KiB | Viewed 4236 times ]
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
21 May 2021, 22:31
21 May 2021, 22:31
3
1
Bookmarks
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
Attachments
In the figure, circle O has center O, diameter AB.png [ 9.12 KiB | Viewed 3185 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 3118 times ]
