GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 07 Aug 2020, 10:07 ### GMAT 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

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

Track

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

Author Message
TAGS:

### Hide Tags

Intern  Joined: 18 Mar 2012
Posts: 43
GPA: 3.7
A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

12
1
115 00:00

Difficulty:   95% (hard)

Question Stats: 55% (02:28) correct 45% (02:42) wrong based on 824 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. $$4r (\sqrt 3 + 1)$$
Math Expert V
Joined: 02 Sep 2009
Posts: 65854
A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

66
1
39
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 132055 times ]
The perimeter of the rectangle is $$2r\sqrt{3}+2r=2r(\sqrt{3}+1)$$.

_________________
SVP  Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2447
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
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
3
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/670-to-710-a-long-journey-without-destination-still-happy-141642.html
##### General Discussion
Intern  Joined: 04 Mar 2012
Posts: 36
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

beautiful!! Thanks Bunnel.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

32
5
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

Intern  Joined: 04 Mar 2012
Posts: 36
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

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 V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

4
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

Intern  Joined: 05 Apr 2012
Posts: 38
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

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: 576
WE: Science (Education)
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

6
2
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: 209
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

1
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
Manager  Joined: 22 Apr 2013
Posts: 77
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

1
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°).

Bunuel your explanation is simple, clear, and didn't make my brain hurt. Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

3
1
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 124701 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

Manager  Joined: 04 Dec 2011
Posts: 55
Schools: Smith '16 (I)
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

2
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.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

1
1
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

Intern  Joined: 21 Mar 2013
Posts: 10
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 65854
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

mariofelix wrote:
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.
_________________
Manager  Joined: 09 Mar 2014
Posts: 53
Location: India
Concentration: General Management, Operations
GMAT 1: 760 Q50 V42
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

1
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°).

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.

Manager  B
Joined: 10 Mar 2013
Posts: 165
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
GPA: 3
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

1
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: 182
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

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 V
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: A rectangle is inscribed in a circle of radius r. If the rectangle is  [#permalink]

### Show Tags

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 Re: A rectangle is inscribed in a circle of radius r. If the rectangle is   [#permalink] 07 May 2014, 04:57

Go to page    1   2   3    Next  [ 46 posts ]

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