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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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21 May 2014, 22:58
Hey everyone, Hey Bunuel,
I totally understand your solution with the 306090 triangle. However, I tried to imagine this in my mind and stucked a little.
How is it possible to create a 306090 triangle in a rectangular? I mean every angle is 90 degree  Doesnt the diagonal seperate the angles into half? (45 degree)
Na, that can´t be true. The diagonal is not a "mirror axis" in the rectangular, right? So the angles are not 45.
Is there a common rule ? Or did you just saw all the Squared (3) and thought about a 306090 triangle?



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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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22 May 2014, 00:27
Attachment: File comment: inscribed rectangle
gmatrectangle.jpg [ 25.88 KiB  Viewed 8788 times ]
Let's check the extreme cases. Case 1: When rectangle is almost a line As shown in figure, if we reduce the height of rectangle to nearly zero, its perimeter will be 2r + 2r =4r Case 2: when rectangle is a square Perimeter will be maximum when it is square, in this case, perimeter will be 4sqrt(2)r All the values of perimeter between these extreme values are possible. 4r < perimeter < 4sqrt(2) r or, 4r <perimeter < 5.76r Now lets check the options: A. 2rsqr3 = 2* 1.73 *r = 3.46r B. 2r(sqr3 +1) = 5.46r C. 4rsqr2 = 5.76r D.4rsqr3 = 6.92r E.4r(sqr3+1) = 10.92r From the option , only option B is satisfying the criteria (4r <perimeter < 5.76r) Hence Answer is 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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22 May 2014, 04:27
gunna wrote: Hey everyone, Hey Bunuel,
I totally understand your solution with the 306090 triangle. However, I tried to imagine this in my mind and stucked a little.
How is it possible to create a 306090 triangle in a rectangular? I mean every angle is 90 degree  Doesnt the diagonal seperate the angles into half? (45 degree)
Na, that can´t be true. The diagonal is not a "mirror axis" in the rectangular, right? So the angles are not 45.
Is there a common rule ? Or did you just saw all the Squared (3) and thought about a 306090 triangle? In a rectangle for the diagonal to bisect the angle, the rectangle must be a square. In all other cases the diagonal does NOT bisect the angle.
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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 May 2014, 06:13
CHECK OUT THIS SMART APPROACH So we obviously know that the diameter of the circle will equal the diagonal of the inscribed triangle so let's call the side of the rectangle 'a' and 'b' where 'a' is different from 'b'. Now then we have that a^2 + b^2 = (2r)^2 We need to find 2 (a+b) (a+b)^2 = 4r^2 + 2ab Taking square root we have that a+b= 2r + sqrt (2ab) Now let's look at answer choices CDE are out because they have 4r instead of 2r. Between A and B since, a,b must be positive then sqrt (2ab) > sqrt (3) Therefore B stands Hope this helps Cheers! J GIMME SOME FREAKING KUDOS



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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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22 Jun 2014, 19:05
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.... Are you sure that a square has the largest perimeter possible ? I supposed that for any rectangle a square has the maximum available area while holding the minimum available perimeter. Correct me please if I'm wrong.



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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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22 Jun 2014, 23:35
VadimKlimenko wrote: 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.... Are you sure that a square has the largest perimeter possible ? I supposed that for any rectangle a square has the maximum available area while holding the minimum available perimeter. Correct me please if I'm wrong. Can you please elaborate what you're trying to say? Either by increasing or decreasing the dimension(s), we can make square a rectangle. Kindly refer below: For dimension "a", Area of Square \(= a^2\) & its perimeter = 4a If we increase just one side of the square to "a+1", it becomes a rectangle. Then Area of rectangle \(= a^2 + a\) & its perimeter = 4a+2 However, if you decrease one side to "a1", then also it becomes rectangle with Area of rectangle \(= a^2  a\) & its perimeter = 4a2
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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28 Jun 2014, 00:37
PareshGmat wrote: VadimKlimenko wrote: 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.... Are you sure that a square has the largest perimeter possible ? I supposed that for any rectangle a square has the maximum available area while holding the minimum available perimeter. Correct me please if I'm wrong. Can you please elaborate what you're trying to say? Either by increasing or decreasing the dimension(s), we can make square a rectangle. Kindly refer below: For dimension "a", Area of Square \(= a^2\) & its perimeter = 4a If we increase just one side of the square to "a+1", it becomes a rectangle. Then Area of rectangle \(= a^2 + a\) & its perimeter = 4a+2 However, if you decrease one side to "a1", then also it becomes rectangle with Area of rectangle \(= a^2  a\) & its perimeter = 4a2 I mean that if we are given a rectangular (a square is also a rectangular) with fixed perimeter X, then maximum area of this rectangular can be reached (am I using the appropriate word ?) only if it is a square with X/4 sides. And oppositely, for any given rectangular with fixed area S, the minim perimeter can be reached (?) if the rectangular is a square with sides SQR(S). But I have concerns why in the mentioned problem we have the largest (not least one) perimeter when the inscribed in the circle rectangular is square.



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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 Nov 2014, 00:16
Bunuel wrote: 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. I think what distracted me while solving this question was the statement, "if the rectangle is not square". I knew that the diagonal of a square will certainly be the diameter of the circle. But, now when I drawing inscribed rectangles in a circle, I notice that the diagonal of rectangle will certainly be the diameter of a circle. GMAC often says that in a PS question, every bit of data is useful. Why is the data "if the rectangle is not square" given in this question? Is it just a helper to say the triangles formed won't be a 454590 triangle and to use the answer choices and go ahead presuming the triangle to be a 306090 triangle?
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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 Nov 2014, 00:31
PareshGmat wrote: VadimKlimenko wrote: 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.... Are you sure that a square has the largest perimeter possible ? I supposed that for any rectangle a square has the maximum available area while holding the minimum available perimeter. Correct me please if I'm wrong. I mean that if we are given a rectangular (a square is also a rectangular) with fixed perimeter X, then maximum area of this rectangular can be reached (am I using the appropriate word ?) only if it is a square with X/4 sides. And oppositely, for any given rectangular with fixed area S, the minim perimeter can be reached (?) if the rectangular is a square with sides SQR(S). But I have concerns why in the mentioned problem we have the largest (not least one) perimeter when the inscribed in the circle rectangular is square. I mean that if we are given a rectangular (a square is also a rectangular) with fixed perimeter X, then maximum area of this rectangular can be reached (am I using the appropriate word ?) only if it is a square with X/4 sides.This is correct Say if dimension of square = x; its perimeter = 4x; its area = x^2 If dimensions of rectangle are (x+1) & (x1); again perimeter = 4x; its area = x^2  1 See the difference; perimeter remaining same; area would decrease.....
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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22 Jan 2015, 06:17
if square => perimeter = 4r root2 (pythogarus will do)
so our rectangle perimeter would be less than the above figure
=> we are left with A & B
Lets try to find the minimum perimeter....let the rectangle be so that its width is close to zero => perimeter = 2 * diameter =4r
=> our perimeter has to be >=4r
=> A is out.....2*root 3 is less than 4
=> ans B
Hope that helps..!



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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 2015, 09:59
I think there is a quick and elegant solution here that most people are missing.
It's important that it's not a square. Let's start there. The perimeter of a square that is inscribed in a circle is 4r√2. We know that when you inscribe a square in a circle, it has the maximum area and perimeter of any rectangle inscribed in that same circle. Therefore, the perimeter of this rectangle is less than 4r√2.
Next, let's see how we can minimize the perimeter: If the length of the rectangle is very close to the diameter of the circle and the width is meaningless, then the perimeter will roughly be equal to 4r. Therefore, the perimeter of this rectangle is greater than 4r.
So, we know that the perimeter is in between 4r and 4r√2.
C, D and E are above the upper bound.
So the answer is A or B.
A is clearly the smaller of the two. Let's see if it's above the lower bound...
Is 2r√3 > 4r? Is √3 > 2? Is ~1.7 > 2? No, it is not.
Therefore, A is out, and the answer is 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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16 Aug 2015, 02:55
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. As the third option doesn't include "root of 3", is there a way to prove that the internal triangle is 306090?



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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 Aug 2015, 05:25
alex1233 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? perimeter of the square inscribed in a circle with radius 'r' is 4*r*square root 2 perimeter of the smallest rectangle that can be inscribed within the circle with no or negligible area is 4r approx For any other rectangle within the circle, the perimeter falls within these limits. Hence, the correct option is 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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16 Mar 2016, 01:00
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 => 2r(1.732) = 3.464r B. 2r (sqr3 + 1) => 2r(1.732+1)=2r(2.732) = 5.464r C. 4r sqr2 => 4r(1.414) = 2r(2.828) =5.656r D. 4r sqr3 => 4r(1.732)= 2r (3.464)=6.928r E. 4r (sqr3 + 1) => 4r (2.732) = 10.928r The perimeter of the rectangle will be smaller then the perimeter of the circle i.e. 2(3.14)r =6.28r so D and E are out. The perimeter of the rectangle must be greater than 4r so A is out. Option C is for square, so C is out too. Left with option B only.
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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 Apr 2016, 05:16
We can prove that: \(2r \sqrt{8}\) > perimeter of the rectangle > \(2r \sqrt{4}\) How to prove? Simply by prove that: \(2 (L^2 + W^2) > (L+W)^2 > (L^2 +W^2)\) => \(\sqrt{2 (L^2 + W^2)} > \sqrt{(L+W)^2} > \sqrt{(L^2 +W^2)}\) => \(2\sqrt{2 (L^2 + W^2)} > 2\sqrt{(L+W)^2} > 2\sqrt{(L^2 +W^2)}\) or \(2r \sqrt{8}\) > perimeter of the rectangle > \(2r \sqrt{4}\) So B, \(\sqrt{8}>\sqrt{3} + 1 > 2\)
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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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15 Mar 2017, 10:27
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 used a similar line of reasoning as Anupreets's. Since the possible answer choices are increasing in value, I was thinking about the maximum possible perimeter for a rectangle with constraints (inscribed in circle). The maximum possible perimeter will be that of a rectangle which is also a square. The diagonal of a square inscribed in a circle will go through the center of the circle. We would have a 454590 triangle, and so we know the ratio of the sides of such a triangle is xxxSqrt2. x*Sqrt2 = 2r (given) > solve for x > x = sqrt2*r Therefore the perimeter of the inscribed square is 4*sqrt2*r Check the possible answer choices: answer c is the perimeter of the inscribed square and D and E are out as they are greater than the perimeter of the inscribed square. So we need to choose between A and B. I created a 37 questions set and this was the final question and I was running out of time so simply guessed. I guessed B incorrectly... and this could have been avoided by checking the minimum perimeter. It would have been just seconds longer... Create rectangle with smallest possible perimeter > very flat approx line with constraint length max 2r. So slightly bigger than 4r. Sqrt3*2*r is smaller than 4r so answer choice A is out. Choose answer B. I agree that Bunuel's solution is much faster, but it could have been wrong if the question stem DID NOT include "if the rectangle is NOT a square". The official answer from gmatprep Question pack is very lengthy and difficult.



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Re: A rectangle is inscribed in a circle of radius r. If the rectangle is [#permalink]
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10 Jan 2018, 09:57
Please see my approach to this question as per Sketch. 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.
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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 2018, 19:46
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. \(2r \sqrt 3\)
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 The perimeter of the rectangle is \(2r\sqrt{3}+2r=2r(\sqrt{3}+1)\). Answer: B. Is this always the case? That the right triangles within a rectangle are always 306090?



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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 2018, 20:06
Your question: Is this always the case? That the right triangles within a rectangle are always 306090? Answer NO See the figure Attachment:
IMG_20180518_083441.jpg [ 372.2 KiB  Viewed 524 times ]
thinkpad18 wrote: 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. \(2r \sqrt 3\)
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: The attachment Rectangle.png is no longer available The perimeter of the rectangle is \(2r\sqrt{3}+2r=2r(\sqrt{3}+1)\). Answer: B. Is this always the case? That the right triangles within a rectangle are always 306090?
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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 2018, 20:22
gmatbusters wrote: Your question: Is this always the case? That the right triangles within a rectangle are always 306090? Answer NO See the figure Attachment: IMG_20180518_083441.jpg Is this always the case? That the right triangles within a rectangle are always 306090? [/quote] Got it. Why is it 306090 in this case? (other than looking at the answers)?




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