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

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

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16 May 2012, 06:44
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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. $$4r (\sqrt 3 + 1)$$
[Reveal] Spoiler: OA
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A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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16 May 2012, 07:57
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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 88835 times ]
The perimeter of the rectangle is $$2r\sqrt{3}+2r=2r(\sqrt{3}+1)$$.

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

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16 May 2012, 08:30
beautiful!! Thanks Bunnel.
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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16 May 2012, 08:32
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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)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 04 Mar 2012 Posts: 50 Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink] ### Show Tags 17 May 2012, 07:51 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: 7941 Location: Pune, India Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink] ### Show Tags 17 May 2012, 08:06 1 This post received KUDOS Expert's post 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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17 May 2012, 14:39
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

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

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18 Oct 2012, 00:41
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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....
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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18 Oct 2012, 02:28
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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 $$2r\sqrt{3}<4r$$.
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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01 May 2013, 11:24
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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
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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01 May 2013, 21:13
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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.
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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02 May 2013, 08:04
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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 82158 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)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 04 Dec 2011 Posts: 80 Schools: Smith '16 (I) Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink] ### Show Tags 23 Sep 2013, 00:37 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: 7941 Location: Pune, India Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink] ### Show Tags 23 Sep 2013, 02:32 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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26 Jan 2014, 14:35
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)
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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27 Jan 2014, 00:23
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.
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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09 Apr 2014, 23:46
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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.

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

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02 May 2014, 18:54
This is a classic example of using the answer choices to help solve the problem. THX Bunuel and Karishma!
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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07 May 2014, 01:17
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 ?
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]

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

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