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In the circle shown in the figure, the length of the arc ACB [#permalink]
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18 Jan 2012, 13:01
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well it is tricky to me... lol... Nova Gmat Bible (example problem) In the circle shown in the figure, the length of the arc ACB is 3 times the length of the arc AB. What is the length of the line segment AB? A) 3 (B) 4 (C) 5 (D) 2 √3 (E) 3 √2 anyone care to explain the solution in detail?
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Re: triangle inscribed in circle tricky question [#permalink]
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18 Jan 2012, 13:11
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shawndx wrote: well it is tricky to me... lol... Nova Gmat Bible (example problem)
In the circle shown in the figure, the length of the arc ACB is 3 times the length of the arc AB. What is the length of the line segment AB ?
A) 3 (B) 4 (C) 5 (D) 2 √3 (E) 3 √2 Assuming O is the center of the circle. Arc(ACB)=3*arc(AB). Now as arc(ACB)+arc(AB)=3*arc(AB)+arc(AB)=circumference then arc(AB)=circumference/4 > angle AOB=360/4=90 degrees. So, triangle AOB is a right isosceles triangle with OA=OB=r=3 > hypotenuse \(AB=\sqrt{3^2+3^2}=3\sqrt{2}\) (or you can find the length of AB by applying the property of 454590 right triangle where the sides are in the ratio \(1:1:\sqrt{2}\)). Answer: E. Check Triangles chapter of Math Book for more on this topic: mathtriangles87197.htmlHope it helps.
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Re: In the circle shown in the figure, the length of the arc ACB [#permalink]
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18 Jan 2012, 13:24
one question that I have is this... in mgmat guide it says that any right triangle inscribed in a circle must have the diameter of the circle as one of its sides... so I am assuming then AB would be the diameter... because AO and BO cannot be? however, if 3 is the radius and diameter is 2 times the radius, then wont the diameter then be 6? just confused really... also Bunuel how did you know to assume from the diagram that O is the center without the question telling you as such? I was not able to do that, so is there some way of knowing? and also how did you know it was an isosceles triangle as opposed to half of an equilateral triangle? (is it because both points stem from center O and so must be equal?) thanks a lot!



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Re: In the circle shown in the figure, the length of the arc ACB [#permalink]
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18 Jan 2012, 13:39
shawndx wrote: one question that I have is this... in mgmat guide it says that any right triangle inscribed in a circle must have the diameter of the circle as one of its sides... so I am assuming then AB would be the diameter... because AO and BO cannot be? however, if 3 is the radius and diameter is 2 times the radius, then wont the diameter then be 6? just confused really... also Bunuel how did you know to assume from the diagram that O is the center without the question telling you as such? I was not able to do that, so is there some way of knowing? thanks a lot! First of all the property you refer to is correct: a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle. BUT: triangle AOB is NOT inscribed in a circle. In order a triangle to be inscribed in a circle ALL three vertices must be on the circumference. Check Circles chapter of Math Book for more on this topic: mathcircles87957.htmlNext, even though the letter O is commonly used for the center of a circle, the real GMAT question will explicitly specify this information (so in a sense it's not 100% accurate question). But if we don't assume that O is the center then the question can not be answered at all. Hope it's clear.
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Re: In the circle shown in the figure, the length of the arc ACB [#permalink]
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18 Jan 2012, 13:45
thank you so much! your the best Bunnel!



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Re: In the circle shown in the figure, the length of the arc ACB [#permalink]
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18 Jan 2012, 15:15
In geometry problems, as in all other problems, it's crucial to really look at all the given information and think about what it wants to tell you.
Start with the most obvious information, the length of CO, which is the radius. You're also given information about the ratios of the arcs. Since the length of an arc depends on it's angle, you're also given information about the angle AOB. Because arc AB is 1/4 of the whole circumference, you know that the angle is 1/4 of 360°, so the triangle AOB is a right triangle. Two of the sides are radii, which lets you know that the hypothenuse is 3*2^1/2



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Re: In the circle shown in the figure, the length of the arc ACB [#permalink]
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19 Jan 2012, 05:26
length of the arc ACB /length of the arc AB=3/1 it means that length of the arc AB =1/4*360=90degree so AOB is a right isosceles triangle (454590 degree) with hypotenuse AB=3sqroot2
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Re: In the circle shown in the figure, the length of the arc ACB [#permalink]
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19 Jan 2012, 10:07
OK, I will be glad to help. since CO is 3, and O is the centre of the circle, then the line extended from O to any point on the edge of the circle is also 3. from this we know that OA and OB are 3 each. and any line extends from the centre of the circle to the any point on the the edge forms a right triangle, so we now we know that <Aob is right triangle. Thus, we can apply the Pythagorean theorm that states that AO^2 + OB^2=AB^2 so, 3^2+3^2=sqrt 18.......> 3sqrt2 hope this is clear enough
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