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Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]
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 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 16484 times ]

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

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

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

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]
17 May 2012, 08:06

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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

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

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]
08 Sep 2012, 14:12

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.

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

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]
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 [#permalink]
29 Mar 2013, 10:09

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]
01 May 2013, 11:24

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

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

Answer: B.

Bunuel your explanation is simple, clear, and didn't make my brain hurt.

Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]
02 May 2013, 08:04

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Expert's post

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 10245 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)

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

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Re: A rectangle is inscribed in a circle of radius r. If the [#permalink]
23 Sep 2013, 02:32

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Expert's post

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.

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

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

Expert's post

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.