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Retired Moderator
Joined: 16 Apr 2020
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Location: India
WE:Education (Education)
The figure above shows 2 circles. The larger circle has center A
[#permalink]
Updated on: 04 Sep 2026, 09:29
Updated on: 04 Sep 2026, 09:29
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Question Stats:
58% (03:15) correct
41% (02:17) wrong based on 36 sessions
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Two circles.png [ 28.46 KiB | Viewed 6688 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
Originally posted by KarunMendiratta on 04 Aug 2021, 04:41.
Last edited by Carcass on 04 Sep 2026, 09:29, edited 1 time in total.
Last edited by Carcass on 04 Sep 2026, 09:29, edited 1 time in total.
UPDATED
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: 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 6426 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
