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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If A is the center of the circle shown above and AB = BC = CD, what is
[#permalink]
29 Nov 2021, 07:28
29 Nov 2021, 07:28
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Question Stats:
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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
If A is the center of the circle shown above and AB = BC = CD, what is
[#permalink]
29 Nov 2021, 07:50
29 Nov 2021, 07:50
1
Consider \(\triangle ABC\)
We are given that \(AB = BC = CD\)
We can infer from the diagram that \(AB\) and \(AC\) are both radii of the circles
So, now we have \(AB = AC = r\) also \(AB = BC\)
Hence, we can easily say that \(AB = BC = AC\), which makes \(\triangle ABC\) an equilateral triangle
Using property - All angles in a equilateral triangle are equal to 60 degrees
Similar case goes for \(\triangle ADC\) which essentially makes \(BD\) an median and angle bisector at \(B\) as well as at \(D\)
Therefore, \(x^o = \frac{∠ABC}{2} = \frac{60^o}{2} = 30^o\)
Hence, Answer is B
We are given that \(AB = BC = CD\)
We can infer from the diagram that \(AB\) and \(AC\) are both radii of the circles
So, now we have \(AB = AC = r\) also \(AB = BC\)
Hence, we can easily say that \(AB = BC = AC\), which makes \(\triangle ABC\) an equilateral triangle
Using property - All angles in a equilateral triangle are equal to 60 degrees
Similar case goes for \(\triangle ADC\) which essentially makes \(BD\) an median and angle bisector at \(B\) as well as at \(D\)
Therefore, \(x^o = \frac{∠ABC}{2} = \frac{60^o}{2} = 30^o\)
Hence, Answer is B
gmatclubot
If A is the center of the circle shown above and AB = BC = CD, what is [#permalink]
29 Nov 2021, 07:50
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