If A is the center of the circle shown above (see attachment : Problem Solving (PS)

TheRob

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If A is the center of the circle shown above (see attachment [#permalink]

Updated on: 18 Jul 2022, 08:11

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A

B

C

D

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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

PS21278

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Circle - original.PNG [ 3.31 KiB | Viewed 47763 times ]

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21278-0.png [ 32.38 KiB | Viewed 19399 times ]

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Originally posted by TheRob on 07 Sep 2009, 07:19.
Last edited by Bunuel on 18 Jul 2022, 08:11, edited 4 times in total.

Edited the question, added the diagram and added the OA

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Bunuel

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Re: If A is the center of the circle shown above (see attachment [#permalink]

22 May 2020, 07:15

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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

PS21278

Show Spoiler

Attachment:

1.png [ 7.92 KiB | Viewed 13576 times ]

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Bunuel

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Re: If A is the center of the circle shown above (see attachment [#permalink]

25 Feb 2012, 12:41

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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

Look at the diagram below:

Attachment:

Circle.PNG [ 26.39 KiB | Viewed 49459 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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Jivana

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Re: If A is the center of the circle shown above (see attachment [#permalink]

07 Sep 2009, 18:49

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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

MBAhereIcome

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Re: If A is the center of the circle shown above (see attachment [#permalink]

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

pbull78

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Re: If A is the center of the circle shown above (see attachment [#permalink]

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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Bunuel

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Re: If A is the center of the circle shown above (see attachment [#permalink]

27 Feb 2012, 08:42

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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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aakrity

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Re: If A is the center of the circle shown above (see attachment [#permalink]

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?

Attachments

Circle.PNG [ 26.39 KiB | Viewed 42752 times ]

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Bunuel

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Re: If A is the center of the circle shown above (see attachment [#permalink]

14 Nov 2013, 02:49

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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).
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Re: If A is the center of the circle shown above (see attachment [#permalink]

16 Feb 2020, 23:09

Expert Reply

TheRob wrote:

Attachment:

Circle - original.PNG

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

Adding teh video solution to teh thread.

B

Batman117

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Re: If A is the center of the circle shown above (see attachment [#permalink]

22 May 2020, 07:47

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Here AB=BC
Where AB is radius.
From figure AC is also radius.
So AB=BC=AC
So ABC is equilateral triangle.
So each side is of 60
So X=60/2
X=30

Posted from my mobile device

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BrentGMATPrepNow

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Re: If A is the center of the circle shown above (see attachment [#permalink]

23 May 2020, 15:52

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Top Contributor

Bunuel wrote:

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
PS21278

Show Spoiler

Attachment:

1.png

Since all of the radii in a circle have the same length, we know that the 3 red lines below all have the same length.

Since we are told that AB = BC = CD, we can say that all 5 red lines below have the same length

This means the two red triangles must be equilateral triangles, which means all of their angles are 60°

Since all 4 sides of the red quadrilateral below have the same length, we know that the quadrilateral is a rhombus

In a rhombus, the diagonals bisect the angles.
So the blue diagonal above, bisects the 60° angle into two equal angles of 30° each, which means x = 30

Answer: B

Cheers,
Brent

B

Rainman91

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Re: If A is the center of the circle shown above (see attachment [#permalink]

31 Jul 2022, 04:05

Do we assume that the lines are touching the circle's circumference? Just based on the image?

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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]

11 Jan 2024, 10:09

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