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For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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For a circle with center point P, cord XY is the perpendicular bisector of radius AP (A is a point on the edge of the circle). What is the length of cord XY? (1) The circumference of circle P is twice the area of circle P. (2) The length of Arc XAY = \(\frac{2\pi}{3}\). Attachment:
Untitled.png [ 4.33 KiB  Viewed 12502 times ]
How come the answer is D? I have drawn these pictures as they were not provided with the questions. Even though with my guess work I have selected A which is incorrect. Can someone please let me know how to solve this? Also, I understand this will include a concept of 306090 degree triangle  any idea which angles to assign 30 and 60 degrees?
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Originally posted by enigma123 on 08 Feb 2012, 18:44.
Last edited by Bunuel on 28 Nov 2017, 20:24, edited 2 times in total.
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For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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For a circle with center point P, cord XY is the perpendicular bisector of radius AP (A is a point on the edge of the circle). What is the length of cord XY?From the diagram and the stem: AZ=ZP=r/2. In a right triangle ZPX ratio of ZP to XP is 1:2, hence ZPX is a 306090 right triangle where the sides are in ratio: \(1:\sqrt{3}:2\). The longest leg is ZX which corresponds with \(\sqrt{3}\) and is opposite to 60 degrees angle. Thus <XPY=60+60=120 (1) The circumference of circle P is twice the area of circle P > \(2\pi{r}=2*\pi{r^2}\) > \(r=1\) > \(XZ=\frac{\sqrt{3}}{2}\) > \(XY=2*XZ=\sqrt{3}\). Sufficient. (2) The length of Arc XAY = \(\frac{2\pi}{3}\) > \(\frac{2\pi}{3}=\frac{120}{360}*2\pi{r}\) > \(r=1\), the same as above. Sufficient. Answer: D. Attachment:
Chord.PNG [ 24.43 KiB  Viewed 16281 times ]
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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11 Feb 2012, 12:29
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Sorry Bunuel  in your explanation, how come longest leg be ZX? I think it should be XP because that's opposite to 90 degree angle. Also, do you mind telling me how did you find out which side will correspond to 60 degree and 30 degree angle?
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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11 Feb 2012, 12:45
enigma123 wrote: Sorry Bunuel  in your explanation, how come longest leg be ZX? I think it should be XP because that's opposite to 90 degree angle. Also, do you mind telling me how did you find out which side will correspond to 60 degree and 30 degree angle? XP is hypotenuse, which obviously is the longest side but the longest leg is ZX (so the second longest side). In a right triangle where the angles are 30°, 60°, and 90° the 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°). Since the ratio of the leg ZP to the hypotenuse XP is 1:2, then ZP (the shortest side) corresponds to 1 and thus is the opposite of the smallest angle 30°, which means that another leg ZX corresponds to \(\sqrt{3}\). Hope it's clear.
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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11 Feb 2012, 13:52
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Sorry Bunuel  still struggling. How did you get XZ = sqrt3/2? Apologies for been a pain.
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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17 Jul 2013, 09:16
Bunuel wrote: Attachment: Chord.PNG For a circle with center point P, cord XY is the perpendicular bisector of radius AP (A is a point on the edge of the circle). What is the length of cord XY?From the diagram and the stem: AZ=ZP=r/2. In a right triangle ZPX ratio of ZP to XP is 1:2, hence ZPX is a 306090 right triangle where the sides are in ratio: \(1:\sqrt{3}:2\). The longest leg is ZX which corresponds with \(\sqrt{3}\) and is opposite to 60 degrees angle. Thus <XPY=60+60=120 (1) The circumference of circle P is twice the area of circle P > \(2\pi{r}=2*\pi{r^2}\) > \(r=1\) > \(XZ=\frac{\sqrt{3}}{2}\) > \(XY=2*XZ=\sqrt{3}\). Sufficient. (2) The length of Arc XAY = 2pi/3 > \(\frac{2\pi}{3}=\frac{120}{360}*2\pi{r}\) > \(r=1\), the same as above. Sufficient. Answer: D. Hi Bunuel, How did you assume ZPX is a 306090 right triangle just from the ratio of ZP to XP (1:2). How can we assume in any triangle if the two sides are in the ratio 1:2, it will be a 306090 triangle? I thought we have to know we have to know that the triangle is 306090 triangle beforehand to calculated the third side based on the ratio of two given sides.



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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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17 Jul 2013, 10:31
keenys wrote: Bunuel wrote: For a circle with center point P, cord XY is the perpendicular bisector of radius AP (A is a point on the edge of the circle). What is the length of cord XY?From the diagram and the stem: AZ=ZP=r/2. In a right triangle ZPX ratio of ZP to XP is 1:2, hence ZPX is a 306090 right triangle where the sides are in ratio: \(1:\sqrt{3}:2\). The longest leg is ZX which corresponds with \(\sqrt{3}\) and is opposite to 60 degrees angle. Thus <XPY=60+60=120 (1) The circumference of circle P is twice the area of circle P > \(2\pi{r}=2*\pi{r^2}\) > \(r=1\) > \(XZ=\frac{\sqrt{3}}{2}\) > \(XY=2*XZ=\sqrt{3}\). Sufficient. (2) The length of Arc XAY = 2pi/3 > \(\frac{2\pi}{3}=\frac{120}{360}*2\pi{r}\) > \(r=1\), the same as above. Sufficient. Answer: D. Hi Bunuel, How did you assume ZPX is a 306090 right triangle just from the ratio of ZP to XP (1:2). How can we assume in any triangle if the two sides are in the ratio 1:2, it will be a 306090 triangle? I thought we have to know we have to know that the triangle is 306090 triangle beforehand to calculated the third side based on the ratio of two given sides. Notice that since XY is the perpendicular to AP, then ZPX is a right triangle with right angle at Z. So, we have that side:hypotenuse=1:2, which means that we have 306090 triangle, where the ratio of the sides is \(1:\sqrt{3}:2\).
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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17 Jul 2013, 10:50
Thanks for you reply Bunuel.
However, I still did not understand. Here we have angle XZP=90, XP=r and ZP=r/2.
We do not know that the other angles are 60 and 30 respectively. How can we use the ratio of two sides not three to conclude that it is a 306090 triangle?
Should we know beforehand that it is a 306090 triangle to use two sides to calculate the third one?



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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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17 Jul 2013, 10:57
Bunuel wrote: keenys wrote: Thanks for you reply Bunuel.
However, I still did not understand. Here we have angle XZP=90, XP=r and ZP=r/2.
We do not know that the other angles are 60 and 30 respectively. How can we use the ratio of two sides not three to conclude that it is a 306090 triangle?
Should we know beforehand that it is a 306090 triangle to use two sides to calculate the third one? When we know two sides in a right triangle the third one is fixed. We have side:hypotenuse=1x:2x > third side = \(\sqrt{(2x)^2x^2}=\sqrt{3}*x\), so the sides are in the ratio: \(1:\sqrt{3}:2\) > 306090 triangle. Does this make sense? Thanks Bunuel. Now it makes complete sense. I missed the last part in calculating the third side using Pythagoras.



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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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03 Sep 2013, 00:27
Quote: We have side:hypotenuse=1x:2x > third side = \(\sqrt{(2x)^2x^2}=\sqrt{3}*x\) I wonder if this is just today...that I am looking at this perfectly clear explanation and still do not get it. I did a couple of minutes later. So first of  thanks for detailed explanations Bunuel and others. I just wanted to add that \((2x)^2x^2 = 3x^2\) for those who look at the formula with a predetermined mind so focused on that formula and as a result forget to calculate this basic stuff, perhaps wondering where that \(\sqrt{3}*x\) came from. It is possible it is just me, but it often appears to me that it is not. This is one of those..."duuuhhh"s
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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03 Sep 2013, 05:19
Bunuel wrote: keenys wrote: Thanks for you reply Bunuel.
However, I still did not understand. Here we have angle XZP=90, XP=r and ZP=r/2.
We do not know that the other angles are 60 and 30 respectively. How can we use the ratio of two sides not three to conclude that it is a 306090 triangle?
Should we know beforehand that it is a 306090 triangle to use two sides to calculate the third one? When we know two sides in a right triangle the third one is fixed. We have side:hypotenuse=1x:2x > third side = \(\sqrt{(2x)^2x^2}=\sqrt{3}*x\), so the sides are in the ratio: \(1:\sqrt{3}:2\) > 306090 triangle. Does this make sense? Bunuel, If 2 pi r = 2 pi r^2 then either r=0 or r=1 Since radius is always +ve, its safe to assume that r=1. Is that correct?
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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03 Sep 2013, 05:23
maaadhu wrote: Bunuel wrote: keenys wrote: Thanks for you reply Bunuel.
However, I still did not understand. Here we have angle XZP=90, XP=r and ZP=r/2.
We do not know that the other angles are 60 and 30 respectively. How can we use the ratio of two sides not three to conclude that it is a 306090 triangle?
Should we know beforehand that it is a 306090 triangle to use two sides to calculate the third one? When we know two sides in a right triangle the third one is fixed. We have side:hypotenuse=1x:2x > third side = \(\sqrt{(2x)^2x^2}=\sqrt{3}*x\), so the sides are in the ratio: \(1:\sqrt{3}:2\) > 306090 triangle. Does this make sense? Bunuel, If 2 pi r = 2 pi r^2 then either r=0 or r=1 Since radius is always +ve, its safe to assume that r=1. Is that correct? Yes, because we obviously have a circle.
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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28 Feb 2014, 08:19
Hey! Could someone please check my calculations? I keep getting a wrong answer (I tried to solve it in a slightly different way but nonetheless the solution should be the same) for statement 1: so, according to statement 1 > 2pir=2pir² <=> r=1 ; for the following calculations, please see the attached image below. (1) m°+n°=90° > n°=90°m° (2) w°+p°=90° (3)n°+p°=90° > (1) in (2): 90°m°+p°=90° > m°=p°, similarly: n°=w° > AXZ and XZB are similar triangles, hence, their side ratios will be equal. > (XZ/0.5)=(0.75/XZ) <=>2XZ=(3/4XZ) <=> XZ²=3/8 <=> XZ=0.5(3/2)^(1/2) > XZ=2XZ=(3/2)^(1/2) I tried the calculations again and again, but i keep getting the same wrong answer and not 3^(1/2). What did I do wrong? I know that the 306090 approach is easier and probably quicker but I am still confused about what error I made in my calculations/ approach. If someone can help, please do so Max
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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28 Feb 2014, 08:42
damamikus wrote: Hey! Could someone please check my calculations? I keep getting a wrong answer (I tried to solve it in a slightly different way but nonetheless the solution should be the same) for statement 1: so, according to statement 1 > 2pir=2pir² <=> r=1 ; for the following calculations, please see the attached image below. (1) m°+n°=90° > n°=90°m° (2) w°+p°=90° (3)n°+p°=90° > (1) in (2): 90°m°+p°=90° > m°=p°, similarly: n°=w° > AXZ and XZB are similar triangles, hence, their side ratios will be equal. > (XZ/0.5)=( 0.75/XZ) <=>2XZ=(3/4XZ) <=> XZ²=3/8 <=> XZ=0.5(3/2)^(1/2) > XZ=2XZ=(3/2)^(1/2) I tried the calculations again and again, but i keep getting the same wrong answer and not 3^(1/2). What did I do wrong? I know that the 306090 approach is easier and probably quicker but I am still confused about what error I made in my calculations/ approach. If someone can help, please do so Max \(\frac{XZ}{AZ} = \frac{ZB}{XZ}\) > \(AZ = 0.5\) and \(ZB = 1.5\), not 0.75. \(\frac{XZ}{0.5} = \frac{1.5}{XZ}\) > \(XZ^2 = \frac{3}{4}\) > \(XZ=\frac{\sqrt{3}}{2}\). Hope it helps.
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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28 Feb 2014, 08:57
Thanks a lot Bunuel! I totally missed that numbererror!



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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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01 Mar 2014, 05:17
enigma123 wrote: Attachment: Untitled.png For a circle with center point P, cord XY is the perpendicular bisector of radius AP (A is a point on the edge of the circle). What is the length of cord XY? (1) The circumference of circle P is twice the area of circle P. (2) The length of Arc XAY = \(\frac{2\pi}{3}\). How come the answer is D? I have drawn these pictures as they were not provided with the questions. Even though with my guess work I have selected A which is incorrect. Can someone please let me know how to solve this? Also, I understand this will include a concept of 306090 degree triangle  any idea which angles to assign 30 and 60 degrees? hi bunuel ! can u please explain the 2nd condition how did we get 120 degree ?



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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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01 Mar 2014, 07:04
adymehta29 wrote: enigma123 wrote: Attachment: Untitled.png For a circle with center point P, cord XY is the perpendicular bisector of radius AP (A is a point on the edge of the circle). What is the length of cord XY? (1) The circumference of circle P is twice the area of circle P. (2) The length of Arc XAY = \(\frac{2\pi}{3}\). How come the answer is D? I have drawn these pictures as they were not provided with the questions. Even though with my guess work I have selected A which is incorrect. Can someone please let me know how to solve this? Also, I understand this will include a concept of 306090 degree triangle  any idea which angles to assign 30 and 60 degrees? hi bunuel ! can u please explain the 2nd condition how did we get 120 degree ? The central angle which subtends arc XAY is angle XPY, which is 60+60=120 degrees. Hope it's clear.
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Re: For a circle with center point P cord XY is the perpendicular bisector [#permalink]
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02 Mar 2014, 10:17
Hi Bunuel,
What formula have you applied to make statement 2 sufficient. Could you please explain. thanks!
(2) The length of Arc XAY = 2pi/3 > \frac{2\pi}{3}=\frac{120}{360}*2\pi{r} > r=1, the same as above. Sufficient.




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