Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 18 Mar 2012
Posts: 47
GPA: 3.7

A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
16 May 2012, 07:44
Question Stats:
56% (01:34) correct 44% (02:07) wrong based on 793 sessions
HideShow timer Statistics
A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle? A. \(2r \sqrt 3\) B. \(2r (\sqrt 3 + 1)\) C. \(4r \sqrt 2\) D. \(4r \sqrt 3\) E. \(4r (\sqrt 3 + 1)\)
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 48110

A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
16 May 2012, 08:57
alexpavlos wrote: A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?
A. \(2r \sqrt 3\)
B. \(2r (\sqrt 3 + 1)\)
C. \(4r \sqrt 2\)
D. \(4r \sqrt 3\)
E. \(4r (\sqrt 3 + 1)\) A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle. Now, since each option has \(\sqrt{3}\) in it, then let's check a right triangle where the angles are 30°, 60°, and 90° (standard triangle for the GMAT), because its sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). So, in this case we would have: Attachment:
Rectangle.png [ 15.55 KiB  Viewed 102895 times ]
The perimeter of the rectangle is \(2r\sqrt{3}+2r=2r(\sqrt{3}+1)\). Answer: B.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2653
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
18 Oct 2012, 01:41
Largest perimeter is possible when rectangle is Square. Since option C is when it is square and the question itself says rectangle is not a square, C as well as D and E are out as D and E are greater than C. Our answer must be less than C. So only A and B left. Now do reverse engineering. B looks more plausible than A as the perimeter of rectangle must of the form a+b....
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html




Intern
Joined: 04 Mar 2012
Posts: 47

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
16 May 2012, 09:30
beautiful!! Thanks Bunnel.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8204
Location: Pune, India

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
16 May 2012, 09:32
alexpavlos wrote: A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?
A. 2r sqr3 B. 2r (sqr3 + 1) C. 4r sqr2 D. 4r sqr3 E. 4r (sqr3 + 1)
Any smart, quick way of solving this one other than brute force? Since the radius of the circle is r, the diameter must be 2r. Now imagine the rectangle. The diameter must be the hypotenuse of the right angled triangle of the rectangle. Say, if its sides are a and b, \((2r)^2 = a^2 + b^2 = 4r^2\) So when you square a and b and sum them, you should get 4r^2 The given options are the perimeter of the rectangle i.e. 2(a+b). So I ignore 2 of the options and try to split the leftover into a and b. The obvious first choices are options (B) and (E) since we can see that we can split the sum into 3 and 1. \(a + b = r\sqrt{3} + r\) Now check: \((\sqrt{3}r)^2 + r^2 = 4r^2\) That is what we wanted. Hence, the answer is (B)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 04 Mar 2012
Posts: 47

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
17 May 2012, 08:51
Just wondering, why is it mentioned in question that "if rectangle is not square"  will the answer be different had the rectangle inscribed in circle be a square ?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8204
Location: Pune, India

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
17 May 2012, 09:06
gmihir wrote: Just wondering, why is it mentioned in question that "if rectangle is not square"  will the answer be different had the rectangle inscribed in circle be a square ? They had to mention it because otherwise, the answer could have been (C) too. \((\sqrt{2}r)^2 + (\sqrt{2}r)^2 = 4r^2\) Hence \(2(a+b) = 2*(\sqrt{2}r + \sqrt{2}r) = 4\sqrt{2}r\)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 05 Apr 2012
Posts: 43

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
17 May 2012, 15:39
gmihir wrote: Just wondering, why is it mentioned in question that "if rectangle is not square"  will the answer be different had the rectangle inscribed in circle be a square ? hello gmihir my understanding is that They had to mention that it was not a square because in a square diagonals are angle bisectors making 45° ANGLES in this case the 30/60 /90 special right can not be applied in a rectangle diagonals are not angle bisectors hope this helps best regards keiraria



Director
Joined: 22 Mar 2011
Posts: 604
WE: Science (Education)

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
18 Oct 2012, 03:28
gurpreetsingh wrote: Largest perimeter is possible when rectangle is Square. Since option C is when it is square and the question itself says rectangle is not a square, C as well as D and E are out as D and E are greater than C.
Our answer must be less than C. So only A and B left. Now do reverse engineering. B looks more plausible than A as the perimeter of rectangle must of the form a+b.... The perimeter of the rectangle must be greater than \(4r\). Imagine a "flattened/smashed" rectangle, such that one of the sides is approaching the size of a diameter (\(2r\)), and the other side is approaching 0. When the two sides collapse into the diameter, the perimeter would be \(4r\). Or, if the two sides of the rectangle are \(a\) and \(b\), \(a+b\) must be greater than the diagonal of the rectangle, which is \(2r\). Therefore, you can immediately eliminate A, since \(2r\sqrt{3}<4r\).
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Manager
Joined: 12 Dec 2012
Posts: 221
Concentration: Leadership, Marketing
GMAT 1: 540 Q36 V28 GMAT 2: 550 Q39 V27 GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
01 May 2013, 12:24
VeritasPrepKarishma wrote: alexpavlos wrote: A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?
A. 2r sqr3 B. 2r (sqr3 + 1) C. 4r sqr2 D. 4r sqr3 E. 4r (sqr3 + 1)
Any smart, quick way of solving this one other than brute force? Since the radius of the circle is r, the diameter must be 2r. Now imagine the rectangle. The diameter must be the hypotenuse of the right angled triangle of the rectangle. Say, if its sides are a and b, \((2r)^2 = a^2 + b^2 = 4r^2\) So when you square a and b and sum them, you should get 4r^2 The given options are the perimeter of the rectangle i.e. 2(a+b). So I ignore 2 of the options and try to split the leftover into a and b. The obvious first choices are options (B) and (E) since we can see that we can split the sum into 3 and 1. \(a + b = r\sqrt{3} + r\) Now check: \((\sqrt{3}r)^2 + r^2 = 4r^2\) That is what we wanted. Hence, the answer is (B) Thanks Karishma , but could you please elaborate more ? I still do not understand
_________________
My RC Recipe http://gmatclub.com/forum/thercrecipe149577.html
My Problem Takeaway Template http://gmatclub.com/forum/thesimplestproblemtakeawaytemplate150646.html



Manager
Joined: 22 Apr 2013
Posts: 83

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
01 May 2013, 22:13
Bunuel wrote: Now, since each option has \(\sqrt{3}\) in it, then let's check a right triangle where the angles are 30°, 60°, and 90° (standard triangle for the GMAT), because its sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
Answer: B.
Bunuel your explanation is simple, clear, and didn't make my brain hurt.
_________________
I do not beg for kudos.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8204
Location: Pune, India

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
02 May 2013, 09:04
TheNona wrote: VeritasPrepKarishma wrote: alexpavlos wrote: A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?
A. 2r sqr3 B. 2r (sqr3 + 1) C. 4r sqr2 D. 4r sqr3 E. 4r (sqr3 + 1)
Any smart, quick way of solving this one other than brute force? Since the radius of the circle is r, the diameter must be 2r. Now imagine the rectangle. The diameter must be the hypotenuse of the right angled triangle of the rectangle. Say, if its sides are a and b, \((2r)^2 = a^2 + b^2 = 4r^2\) So when you square a and b and sum them, you should get 4r^2 The given options are the perimeter of the rectangle i.e. 2(a+b). So I ignore 2 of the options and try to split the leftover into a and b. The obvious first choices are options (B) and (E) since we can see that we can split the sum into 3 and 1. \(a + b = r\sqrt{3} + r\) Now check: \((\sqrt{3}r)^2 + r^2 = 4r^2\) That is what we wanted. Hence, the answer is (B) Thanks Karishma , but could you please elaborate more ? I still do not understand The method makes a guess based on the format of the expected answer. Make a circle and inscribe a rectangle in it. The diagonal of the rectangle will be the diameter of the circle. Attachment:
Ques3.jpg [ 4.88 KiB  Viewed 96175 times ]
We need to find the perimeter of the rectangle i.e. 2(a + b) We know that \(a^2 + b^2 = (2r)^2 = 4r^2\) So what can a and b be? \(3r^2 + r^2 = 4r^2\) (So \(a = \sqrt{3}r, b = r\)) or \(2r^2 + 2r^2 = 4r^2\) (So \(a = \sqrt{2}r, b = \sqrt{2}r\)) etc Now, looking at the options, we see that 2(a + b) can be \(2(\sqrt{3}r + r)\) i.e. option (B)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 04 Dec 2011
Posts: 67

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
23 Sep 2013, 01:37
This Question came up on my prep and I got stuck as I didn't look at answer choices. I just wrote L^2+B^2 = (2R)^2 and I had no clue how to solve it in 2 mins. I understand that we need to work on shortcuts and tricks, but just for academic purposes can someone show me how to manipulate this equation to answer choice algebraically? the only next step I could think was (L+B)^2 = 4R^2  2LB and had no clue how to go further.
_________________
Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!
1 Kudos = 1 thanks Nikhil



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8204
Location: Pune, India

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
23 Sep 2013, 03:32
nikhil007 wrote: This Question came up on my prep and I got stuck as I didn't look at answer choices. I just wrote L^2+B^2 = (2R)^2 and I had no clue how to solve it in 2 mins. I understand that we need to work on shortcuts and tricks, but just for academic purposes can someone show me how to manipulate this equation to answer choice algebraically? the only next step I could think was (L+B)^2 = 4R^2  2LB and had no clue how to go further. Note that the question says "which of the following could be the perimeter of the rectangle" The rectangle can be made in many ways and the perimeter would be different in these cases. The answer option gives us one such perimeter value. Hence there is no single algebraic method of obtaining the "correct answer". You have to look at options and say which one CAN be the perimeter value.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 21 Mar 2013
Posts: 11

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
26 Jan 2014, 15:35
if we assume r=2,5 => 2r= hypothenuse triangle = 5 ... we can assume the other 2 sides to be 3 and 4 so 5^2 = 4^2 + 3^2 ... then the perimeter would be 2 * ( 3 + 4 ) = 14.
if we pick option C we get ... 4r sqr (2) = 4 * 2,5 * 1,4 = 14.
Am i missing something ?
[quote="Bunuel"][quote="fameatop"][quote="Bunuel"][quote="alexpavlos"]A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?
A. 2r sqr3 B. 2r (sqr3 + 1) C. 4r sqr2 D. 4r sqr3 E. 4r (sqr3 + 1)



Math Expert
Joined: 02 Sep 2009
Posts: 48110

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
27 Jan 2014, 01:23



Manager
Joined: 09 Mar 2014
Posts: 53
Location: India
Concentration: General Management, Operations
GPA: 3.2
WE: Engineering (Energy and Utilities)

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
10 Apr 2014, 00:46
Bunuel wrote: alexpavlos wrote: A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?
A. 2r sqr3 B. 2r (sqr3 + 1) C. 4r sqr2 D. 4r sqr3 E. 4r (sqr3 + 1)
Any smart, quick way of solving this one other than brute force? A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle. Now, since each option has \(\sqrt{3}\) in it, then let's check a right triangle where the angles are 30°, 60°, and 90° (standard triangle for the GMAT), because its sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). Answer: B. Hi bunuel I might have an alternate solution for this one. Correct me if I am wrong. If the rectangle inscribed in a circle of radius r is a square, then its perimeter would be 4r\(\sqrt{2}\). However, it is given that the rectangle is not square. Therefore, the answer cannot be C. Also, for a circle of given radius, rectangle with maximum perimeter that can be incribed is a square. Therefore, the Perimeter of rectangle will be less than 4r\(\sqrt{2}\). So, D and E are eliminated. For A, If P = 2r\(\sqrt{3}\). Therefore (L+B) = r\(\sqrt{3}\). But (L+B) should be greater than 2r because, L, B, 2r form a triangle. Therefore A is also eliminated. Clearly, the answer is B.



Manager
Joined: 10 Mar 2013
Posts: 234
GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
02 May 2014, 19:54
This is a classic example of using the answer choices to help solve the problem. THX Bunuel and Karishma!



Manager
Joined: 28 Apr 2014
Posts: 241

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
07 May 2014, 02:17
VeritasPrepKarishma wrote: nikhil007 wrote: This Question came up on my prep and I got stuck as I didn't look at answer choices. I just wrote L^2+B^2 = (2R)^2 and I had no clue how to solve it in 2 mins. I understand that we need to work on shortcuts and tricks, but just for academic purposes can someone show me how to manipulate this equation to answer choice algebraically? the only next step I could think was (L+B)^2 = 4R^2  2LB and had no clue how to go further. Note that the question says "which of the following could be the perimeter of the rectangle" The rectangle can be made in many ways and the perimeter would be different in these cases. The answer option gives us one such perimeter value. Hence there is no single algebraic method of obtaining the "correct answer". You have to look at options and say which one CAN be the perimeter value. Ok Krishna so as I see it , this question can only and only be solved by the process of elimination. Right ?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8204
Location: Pune, India

Re: A rectangle is inscribed in a circle of radius r. If the rectangle is
[#permalink]
Show Tags
07 May 2014, 05:57
himanshujovi wrote: Ok Krishna so as I see it , this question can only and only be solved by the process of elimination. Right ? Yes, since you have multiple possible answers, you have to look for one which is possible. Note that you don't really need to eliminate every choice till you arrive at the correct answer. You need to intelligently guess which options can give you the correct answer.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Re: A rectangle is inscribed in a circle of radius r. If the rectangle is &nbs
[#permalink]
07 May 2014, 05:57



Go to page
1 2 3
Next
[ 42 posts ]



