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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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°).

So, in this case we would have:

Attachment:

Rectangle.png [ 15.55 KiB | Viewed 72313 times ]

The perimeter of the rectangle is \(2r\sqrt{3}+2r=2r(\sqrt{3}+1)\).

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)
_________________

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]

Show Tags

18 Oct 2012, 01:41

12

This post received KUDOS

1

This post was BOOKMARKED

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....
_________________

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]

Show Tags

18 Oct 2012, 03:28

3

This post received KUDOS

1

This post was BOOKMARKED

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.

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. \((\sqrt{2}r)^2 + (\sqrt{2}r)^2 = 4r^2\)

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]

Show Tags

01 May 2013, 22:13

1

This post received KUDOS

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.
_________________

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 65782 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)
_________________

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.
_________________

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]

Show Tags

10 Apr 2014, 00:46

1

This post received KUDOS

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.

Re: A rectangle is inscribed in a circle of radius r. If the [#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 ?

Re: A rectangle is inscribed in a circle of radius r. If the [#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

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]

Show Tags

08 Sep 2012, 15:12

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°).

So, in this case we would have:

Attachment:

Rectangle.png

The perimeter of the rectangle is \(2r\sqrt{3}+2r=2r(\sqrt{3}+1)\).

Answer: B.

Hi Bunuel, Using your method i am getting B as the answer but when i used Pythagoras, the answer i am getting is C. Can you kindly clarify my doubt. Waiting for your response.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]

Show Tags

29 Mar 2013, 11:09

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....

I don't know what to say. Just simple logic - no calculations. Absolute GMAT style. Tusi chha gaye badshao

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]

Show Tags

01 May 2013, 12:24

1

This post was BOOKMARKED

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
_________________

Re: A rectangle is inscribed in a circle of radius r. If the [#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!

Re: A rectangle is inscribed in a circle of radius r. If the [#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)

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 ?

\(\sqrt{2}\) only approximately is 1.4 but we are not asked about approximate perimeter. Check what would be the exact perimeter in this case.
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...