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If A is the center of the circle shown above (see attachment [#permalink]
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 07 Sep 2009, 07:19
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If A is the center of the circle shown above and AB = BC = CD, What is the value of x? A. 15 B. 30 C. 45 D. 60 E. 75
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Last edited by Bunuel on 21 Sep 2013, 13:18, edited 2 times in total.
Edited the question, added the diagram and added the OA
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Re: Weird cricle [#permalink]
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07 Sep 2009, 18:10
A is the center of the circle => AB= AD = AC = BC = CD => ABC is a regular triangle => x = 30° The answer is B TheRob wrote: If A is the center of the circle shown above (see attachment) and AB = BC = CD, What is the value of x?
A) 15
B) 30
C) 45
D) 60
E) 75
(I tried to draw it the best I could sorry for the inconvenience)
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Re: Weird cricle [#permalink]
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Triangle abc is equilateral, therefore all angles are 60. abc = acb = bac = 60.
Similarly, Triangle adc is equilateral, therefore all angles are 60. adc = acd = dac = 60.
Now, angle bad = bac + dac = 120, since each triangle is a reverse image of the other.
Now triangle bad is isosceles, since ab = ad.
Therefore abd = adb = 30, since bad =120.
Hence, B
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Re: Weird cricle [#permalink]
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25 Feb 2012, 11:55
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you can solve the question easily by knowing two important properties of square and rhombus. 1. diagonals bisect each other. 2. each diagonal forms a 90 degree angle with the other diagonal
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Re: If A is the center of the circle shown above [#permalink]
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TheRob wrote: If A is the center of the circle shown above (see attachment) and AB = BC = CD, What is the value of x? A. 15 B. 30 C. 45 D. 60 E. 75 Attachment: The attachment Circle - original.PNG is no longer available (I tried to draw it the best I could sorry for the inconvenience) Look at the diagram below: Attachment: 
Circle.PNG [ 26.39 KiB | Viewed 17493 times ]
Given that AB = BC = CD, also since AB is the radius then AB = AC = AD = radius, so we have that: AB = BC = CD = AC = AD, so basically we have two equilateral triangles ABC and ACD with common base of AC (ABC and ACD are mirror images of each other). Line segment BD cuts the angle ABC in half and since all angles in equilateral triangle equal to 60 degrees then x=60/2=30 degrees. Answer: B.
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Re: If A is the center of the circle shown above [#permalink]
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27 Feb 2012, 08:25
bunuel ,
how can we say that line segment BD cuts the angle ABC in half which property is this , can u expalin me?
thanks
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Re: If A is the center of the circle shown above [#permalink]
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27 Feb 2012, 08:42
pbull78 wrote: bunuel ,
how can we say that line segment BD cuts the angle ABC in half which property is this , can u expalin me?
thanks ABC and ACD are two equilateral triangles, which are mirror images of each other if you join vertices B and D, the segment BD must cut AC, the common base, in half. Which makes BD perpendicular bisector of AC --> so BO is a hight, median, and bisector of angle ABC (O being intersection point of BD and AC). Hope it's clear.
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Re: If A is the center of the circle shown above [#permalink]
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27 Feb 2012, 08:43
This is a proper rhombus... and this is one of the property of rhombus... that they bisect the diagonals in half.. and the diagonals bisect the angles in half...
also Angle (DAC is 120 degree )
and Triangle DAC is an isosceles triangle.. with base BD... hence angle ABD is 30
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Re: If A is the center of the circle shown above [#permalink]
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28 Feb 2012, 11:10
thanks bunuel , i also appreciate the explanation given by jivana.
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Re: Find X [#permalink]
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07 Apr 2012, 22:32
In triangle ABC, AB = BC = AC meaning all angles = 60 deg each. Similarly, in Triangle CDA, AD = CD = AC meaning all angles = 60 deg each. thus, in quadrilateral ABCD, Angle A = 120 = C and Angle B = 60 deg = Angle D. Thus in Triangle, ABC A + Angle ABD + Angle ADC = 180 120 + 2 * Angle ABD = 180 x = 30 deg.
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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14 Nov 2013, 02:44
we are given that AB=BC=CD AB is the radius of the circle, so AB=AD This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45 Why is this not considered that ABCD is a square?
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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14 Nov 2013, 02:49
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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14 Nov 2013, 02:55
Bunuel wrote: aakrity wrote: we are given that AB=BC=CD
AB is the radius of the circle, so AB=AD
This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45
Why is this not considered that ABCD is a square? ABCD is not a square it's a rhombus (the diagonals are not equal). O right how silly of me I assumed if a quadrilateral has four equal sides, it is a square. Now the solution makes sense to me. Thank you.
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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22 Jun 2014, 09:38
Bunuel wrote: aakrity wrote: we are given that AB=BC=CD
AB is the radius of the circle, so AB=AD
This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45
Why is this not considered that ABCD is a square? ABCD is not a square it's a rhombus (the diagonals are not equal). Bunnel, So Diagonals are 'angle bisectors' both in a Rhombus (not congruent here) and a Square. But it's not the case in a rectangle where they are congruent and perpendicular bosectors alone. Is my understanding accurate? I've read that every point on an angle bisector is equidistant from both the adjacent sides. Is there any other way to identify whether a diagonal is also an angle bisector? Thank you
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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22 Jun 2014, 15:55
Kconfused wrote: Bunuel wrote: aakrity wrote: we are given that AB=BC=CD
AB is the radius of the circle, so AB=AD
This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45
Why is this not considered that ABCD is a square? ABCD is not a square it's a rhombus (the diagonals are not equal). Bunnel, So Diagonals are 'angle bisectors' both in a Rhombus (not congruent here) and a Square. But it's not the case in a rectangle where they are congruent and perpendicular bosectors alone. Is my understanding accurate? I've read that every point on an angle bisector is equidistant from both the adjacent sides. Is there any other way to identify whether a diagonal is also an angle bisector? Thank you All is true, except that the diagonals in rectangle are no perpendicular to each other. They bisect each other but not at a right angle.
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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13 Jan 2016, 04:12
Thought BC and CD were the arcs, not straight lines.
Please provide accurate diagrams of clarify in the statement.
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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Re: If A is the center of the circle shown above (see attachment [#permalink]
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22 Jun 2016, 22:37
What we know:
1) The figure is a rhombus (Diagonals are angular bisectors)
2) All lines extended from A are radii of the circle.
Therefore, AB = BC = AC. This in turn makes ABC an equilateral triangle. Now you know angleABC is 60*.
As the figure is a rhombus and its diagonals bisect the angleABC, x = 30
Answer : B
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Re: If A is the center of the circle shown above (see attachment
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22 Jun 2016, 22:37
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