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.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
The figure above shows 2 circles. The larger circle has center A
[#permalink]
04 Aug 2021, 04:41
04 Aug 2021, 04:41
1
2
7
Bookmarks
00:00
Question Stats:
54% (03:27) correct
45% (02:50) wrong based on 33 sessions
Hide Show timer Statistics
Attachment:
Two circles.png [ 28.46 KiB | Viewed 3378 times ]
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. Also, points A, B, C, and E are collinear.
Quantity A |
Quantity B |
R |
4r |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
The figure above shows 2 circles. The larger circle has center A
[#permalink]
05 Aug 2021, 00:15
05 Aug 2021, 00:15
3
Explanation:
(Refer to the figure below)
Since, AM = ME = \(R\)
AME is an isosceles right triangle i.e. AE = \(\sqrt{2}R\)
Also, CD = DE = \(r\)
CDE is an isosceles right triangle i.e. CE = \(\sqrt{2}r\)
AE = \(R + r + \sqrt{2}r\)
\(\sqrt{2}R = R + r + \sqrt{2}r\)
\((\sqrt{2}-1)R = (\sqrt{2}+1)r\)
Col. A: \(R\)
Col. B: \(4r\)
Dividing both sides with \(r\);
Col. A: \(\frac{R}{r}\)
Col. B: 4
Col. A: \(\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}\)
Col. B: 4
Col. A: 5.83
Col. B: 4
Hence, option A
(Refer to the figure below)
Since, AM = ME = \(R\)
AME is an isosceles right triangle i.e. AE = \(\sqrt{2}R\)
Also, CD = DE = \(r\)
CDE is an isosceles right triangle i.e. CE = \(\sqrt{2}r\)
AE = \(R + r + \sqrt{2}r\)
\(\sqrt{2}R = R + r + \sqrt{2}r\)
\((\sqrt{2}-1)R = (\sqrt{2}+1)r\)
Col. A: \(R\)
Col. B: \(4r\)
Dividing both sides with \(r\);
Col. A: \(\frac{R}{r}\)
Col. B: 4
Col. A: \(\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}\)
Col. B: 4
Col. A: 5.83
Col. B: 4
Hence, option A
Attachments
Two circles.png [ 28.26 KiB | Viewed 3166 times ]
General Discussion
Re: The figure above shows 2 circles. The larger circle has center A
[#permalink]
04 Aug 2021, 05:13
04 Aug 2021, 05:13
hm, cannot figure this one out
