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Re M2120 [#permalink]
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16 Sep 2014, 01:11
Official Solution:If a circle passes through points \((1, 2)\), \((2, 5)\), and \((5, 4)\), what is the diameter of the circle? A. \(\sqrt{18}\) B. \(\sqrt{20}\) C. \(\sqrt{22}\) D. \(\sqrt{26}\) E. \(\sqrt{30}\) Look at the diagram below: Calculate the lengths of the sides of triangle \(ABC\): \(AB=\sqrt{10}\); \(BC=\sqrt{10}\); \(AC=\sqrt{20}=\sqrt{2}*\sqrt{10}\); As we see the ratio of the sides of triangle \(ABC\) is \(1:1:\sqrt{2}\), so \(ABC\) is 45°45°90° right triangle (in 45°45°90° right triangle the sides are always in the ratio \(1:1:\sqrt{2}\)). So, we have right triangle \(ABC\) inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so \(AC=diameter=\sqrt{20}\). Answer: B
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Re: M2120 [#permalink]
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12 Oct 2014, 20:55
I am having difficulties with this problem I actually got it right but I might have been lucky I tried to find the slope which i believe the equation for this is y=3x1... would this be helpful in finding the answer and second how did you get the legnths of each side of the triangle. I am having problems trying to find each length. Could someone please go over this problem and answer thank you



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Re: M2120 [#permalink]
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13 Oct 2014, 00:10



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Re: M2120 [#permalink]
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06 Jan 2015, 12:24
Bunuel wrote: Official Solution:If a circle passes through points \((1, 2)\), \((2, 5)\), and \((5, 4)\), what is the diameter of the circle? A. \(\sqrt{18}\) B. \(\sqrt{20}\) C. \(\sqrt{22}\) D. \(\sqrt{26}\) E. \(\sqrt{30}\) Look at the diagram below: Calculate the lengths of the sides of triangle \(ABC\): \(AB=\sqrt{10}\); \(BC=\sqrt{10}\); \(AC=\sqrt{20}=\sqrt{2}*\sqrt{10}\); As we see the ratio of the sides of triangle \(ABC\) is \(1:1:\sqrt{2}\), so \(ABC\) is 45°45°90° right triangle (in 45°45°90° right triangle the sides are always in the ratio \(1:1:\sqrt{2}\)). So, we have right triangle \(ABC\) inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so \(AC=diameter=\sqrt{20}\). Answer: B Hi Bunuel I did not understand how did you calculate AB & BC (highlighted above). Request you to elaborate if possible Thanks a ton Buddy



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Re: M2120 [#permalink]
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07 Jan 2015, 07:17
buddyisraelgmat wrote: Bunuel wrote: Official Solution:If a circle passes through points \((1, 2)\), \((2, 5)\), and \((5, 4)\), what is the diameter of the circle? A. \(\sqrt{18}\) B. \(\sqrt{20}\) C. \(\sqrt{22}\) D. \(\sqrt{26}\) E. \(\sqrt{30}\) Look at the diagram below: Calculate the lengths of the sides of triangle \(ABC\): \(AB=\sqrt{10}\); \(BC=\sqrt{10}\); \(AC=\sqrt{20}=\sqrt{2}*\sqrt{10}\); As we see the ratio of the sides of triangle \(ABC\) is \(1:1:\sqrt{2}\), so \(ABC\) is 45°45°90° right triangle (in 45°45°90° right triangle the sides are always in the ratio \(1:1:\sqrt{2}\)). So, we have right triangle \(ABC\) inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so \(AC=diameter=\sqrt{20}\). Answer: B Hi Bunuel I did not understand how did you calculate AB & BC (highlighted above). Request you to elaborate if possible Thanks a ton Buddy Check The Distance Between Two Points here: mathcoordinategeometry87652.htmlHope it helps.
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Re: M2120 [#permalink]
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07 Jan 2015, 09:07
Bunuel wrote: buddyisraelgmat wrote: Bunuel wrote: Official Solution:If a circle passes through points \((1, 2)\), \((2, 5)\), and \((5, 4)\), what is the diameter of the circle? A. \(\sqrt{18}\) B. \(\sqrt{20}\) C. \(\sqrt{22}\) D. \(\sqrt{26}\) E. \(\sqrt{30}\) Look at the diagram below: Calculate the lengths of the sides of triangle \(ABC\): \(AB=\sqrt{10}\); \(BC=\sqrt{10}\); \(AC=\sqrt{20}=\sqrt{2}*\sqrt{10}\); As we see the ratio of the sides of triangle \(ABC\) is \(1:1:\sqrt{2}\), so \(ABC\) is 45°45°90° right triangle (in 45°45°90° right triangle the sides are always in the ratio \(1:1:\sqrt{2}\)). So, we have right triangle \(ABC\) inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so \(AC=diameter=\sqrt{20}\). Answer: B Hi Bunuel I did not understand how did you calculate AB & BC (highlighted above). Request you to elaborate if possible Thanks a ton Buddy Check The Distance Between Two Points here: mathcoordinategeometry87652.htmlHope it helps. Yup. Got it  Thanks



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Re M2120 [#permalink]
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25 Jul 2015, 13:57
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. How to find the length of each side? Which formula/concept to apply?



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Re: M2120 [#permalink]
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25 Jul 2015, 14:04



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Re: M2120 [#permalink]
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29 Oct 2016, 12:21
BunuelI used distance formula using 2 points (1,2) and (5,4) which gave me correct ans. (root {(42)^2 + (51)^2} ) is this approach fine?



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Re: M2120 [#permalink]
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30 Oct 2016, 01:16



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Re: M2120 [#permalink]
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30 Dec 2016, 03:25
Bunuel,
Tell me if this thought process is ok
Equation of a Circle is x^2+y^2= r^2 ...... the points mentioned shud satisfy this equation..we get r= sqrt(5)...diameter= 2*sqrt(5)...if 2 goes inside the sqrt sign, becomes sqrt(20)..thats my answer



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Re: M2120 [#permalink]
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30 Dec 2016, 03:40



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Re: M2120 [#permalink]
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30 Dec 2016, 03:54
Bunuel wrote: Omkar.kamat wrote: Bunuel,
Tell me if this thought process is ok
Equation of a Circle is x^2+y^2= r^2 ...... the points mentioned shud satisfy this equation..we get r= sqrt(5)...diameter= 2*sqrt(5)...if 2 goes inside the sqrt sign, becomes sqrt(20)..thats my answer x^2 + y^2 = r^2 is the equation of a circle centred at the origin. The given circle is not centred ant the origin. How did you get that \(r = \sqrt{5}\)? P.S. The correct way is given in the solution above. Oops...I missed the Origin part of it. Sorry !! Omkar Kamat When The Going Gets Tough, The Tough Gets Going !!



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Re: M2120 [#permalink]
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14 May 2017, 04:40
Hi Bunuel, Great solution. However I would like to know how did you realize that finding the lengths of the sides of the triangle formed by the three points would most certainly give you a clue whether this triangle was a right triangle. Frankly, when I started this problem I felt that the only way was to choose an arbitrary point as the centre of circle and equate the distances from the centre to the three points (since they would be radii). This took me a some time. Please share your thoughts. Regards



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Can also be solved by calculating the slopes of the 2 lines from these 3 points . One comes to be 3 ie ( 52/21 = 3) and other as 1/3 ie (45/52 = 1/3) so there is a right angle and the line connecting the 2 end points will be diameter.



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Re: M2120 [#permalink]
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12 Jul 2017, 13:21
Isn't solving it by the equation of circle a faster and certain method?
if we hadn't calculated the distances specifically (it doesn't strike naturally to use the distance formula here), we couldn't know then that it is in fact a right angle.



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Re: M2120 [#permalink]
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01 Dec 2017, 07:53
Hi
can anyone explain why triangle ABC is 454590 degree ? i know that one angle must be 90 degrees but the other two angles could be different and all three angles sum is 180. Are sides AB and BC similar ? If so, how ?
Thanks



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01 Dec 2017, 07:56



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Re: M2120 [#permalink]
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09 Apr 2018, 15:23
Why go through the trouble of pythag. theorem and finding separate leg lengths when you can just use distance formula between AC (easy to surmise that distance AC is greater than AB or BC). Is there a possibility that the GMAT just throws you three arbitrary points on the circle as opposed to this equation that included the endpoints of the diameter?










