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# Points A, B, C and D lie on a circle of radius 1. Let x be

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Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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01 Sep 2008, 04:33
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Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ?

(1) $$\angle ADB$$ is acute
(2) $$\angle ADB > \angle CAD$$
[Reveal] Spoiler: OA

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Re: CIRCLE ... NICE ONE [#permalink]

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02 Sep 2008, 09:30
arjtryarjtry wrote:
Points A, B, C and D lie on a circle of radius 1. Let $$x$$ be the length of arc AB and $$y$$ the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ?

1. $$\angle$$ ADB is acute
2. $$\angle$$ ADB > $$\angle$$ CAD

The answer is not B. If we use only that statement, we can draw the following (no need to use co-ordinate geometry, but it's easier to explain if I do, since I can't draw a picture):

-draw the circle in the co-ordinate plane, with centre at (0,0). The radius is 1.
-make AC a diameter: put A at (-1, 0), and C at (1,0).
-draw D and B in the third quadrant (i.e. where x and y are both negative), and so that when read counterclockwise, the points are in the sequence ADBC. That is, minor arc AD should be shorter than minor arc AB.

Drawn this way, ADB should be (very) obtuse, and much larger than CAD, which must be less than 90 degrees (angle CDA must be 90, since AC is a diameter so CAD and DCA are both less than 90). The length of minor arc AB is clearly less than one quarter of the circumference, while minor arc CD is clearly greater than one quarter of the circumference, so it is possible for x to be less than y.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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27 Sep 2012, 11:18
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I also fell for trap B, then realized.

I think answer must be C only, because order of the points on the circle were not mentioned, only if you consider that ADB<90 then only we know B is between A and D.

Correct me If any other explanation is possible.

Thanks,
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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30 Sep 2012, 17:36
If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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voodoochild wrote:
If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?

(2) alone is not sufficient.
Remember the condition in (1) and check the case when that condition doesn't hold.
In my attached drawing, in the isosceles trapezoid, $$x=y.$$
Attachments

ArcsComparison.jpg [ 14.92 KiB | Viewed 8229 times ]

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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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06 Oct 2012, 01:24
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ?

1. $$\angle ADB$$ is acute
2. $$\angle ADB > \angle CAD$$

Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y.
Statement 2 says one angle is greater than the other. So this must be sufficient

Is there an explanation for Y it is C.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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08 Oct 2012, 01:25
154238 wrote:
OA is B for me . Kindly Confirm Bunuel.

cheers

No, B is not the answer. See my previous post above points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1126726
In the drawing, an isosceles trapezoid is depicted, and obviously $$x=y.$$
For example, by moving point B, you can increase or decrease the arc AB, while arc DC stays the same. So, you can have both $$x>y$$ and $$x<y,$$ after you already had $$x=y.$$
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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08 Oct 2012, 03:14
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rajathpanta wrote:
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ?

1. $$\angle ADB$$ is acute
2. $$\angle ADB > \angle CAD$$

Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y.
Statement 2 says one angle is greater than the other. So this must be sufficient

Is there an explanation for Y it is C.

The answer is not B, (2) alone is not sufficient. See my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is $$2\pi$$, therefore $$x$$ and $$y$$ being less than $$\pi$$ are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that $$\angle{ADB}$$ is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that $$\angle{ADB}$$ is acute. In the attached drawing, D can be only above the chord AB. Therefore to $$\angle{ADB}$$ corresponds the shorter arc $$x.$$
From (2) we have that $$\angle{ADB}>\angle{CAD},$$ meaning that $$\angle{CAD}$$ is also acute. Therefore the corresponding arc to $$\angle{CAD}$$ is the shorter arc $$y.$$ In the attached drawing C can be only on the right of the diameter $$DC_1.$$
Using (2), now we can conclude that $$x>y.$$
So, (1) and (2) together is sufficient.

Attachments

ArcsComparison-more.jpg [ 16.77 KiB | Viewed 8021 times ]

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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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09 Oct 2012, 01:28
EvaJager wrote:
rajathpanta wrote:
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ?

1. $$\angle ADB$$ is acute
2. $$\angle ADB > \angle CAD$$

Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y.
Statement 2 says one angle is greater than the other. So this must be sufficient

Is there an explanation for Y it is C.

The answer is not B, (2) alone is not sufficient. See my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is $$2\pi$$, therefore $$x$$ and $$y$$ being less than $$\pi$$ are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that $$\angle{ADB}$$ is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that $$\angle{ADB}$$ is acute. In the attached drawing, D can be only above the chord AB. Therefore to $$\angle{ADB}$$ corresponds the shorter arc $$x.$$
From (2) we have that $$\angle{ADB}>\angle{CAD},$$ meaning that $$\angle{CAD}$$ is also acute. Therefore the corresponding arc to $$\angle{CAD}$$ is the shorter arc $$y.$$ In the attached drawing C can be only on the right of the diameter $$DC_1.$$
Using (2), now we can conclude that $$x>y.$$
So, (1) and (2) together is sufficient.

Hi Eva,

we have that $$\angle{ADB}>\angle{CAD}, how can you conclude that that [m]\angle{CAD}$$ is also acute.
Let say you make a square inside the circle
a---------d
| |
| |
c---------b

Its still not clear to me, why B is wrong ?

Thanks

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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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09 Oct 2012, 02:03
EvaJager wrote:
rajathpanta wrote:
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ?

1. $$\angle ADB$$ is acute
2. $$\angle ADB > \angle CAD$$

The answer is not B, (2) alone is not sufficient. See my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is $$2\pi$$, therefore $$x$$ and $$y$$ being less than $$\pi$$ are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that $$\angle{ADB}$$ is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that $$\angle{ADB}$$ is acute. In the attached drawing, D can be only above the chord AB. Therefore to $$\angle{ADB}$$ corresponds the shorter arc $$x.$$
From (2) we have that $$\angle{ADB}>\angle{CAD},$$ meaning that $$\angle{CAD}$$ is also acute. Therefore the corresponding arc to $$\angle{CAD}$$ is the shorter arc $$y.$$ In the attached drawing C can be only on the right of the diameter $$DC_1.$$
Using (2), now we can conclude that $$x>y.$$
So, (1) and (2) together is sufficient.

Hi Eva,

we have that $$\angle{ADB}>\angle{CAD}, how can you conclude that that [m]\angle{CAD}$$ is also acute.
Let say you make a square inside the circle
a---------d
| |
| |
c---------b

Here, in fact all angles are equal and are right angles. This cannot be the situation in the present question. In addition, if both angles ADB and CAD are right angles, then the inequality $$\angle{ADB}>\angle{CAD}$$ doesn't hold!
See below.

Its still not clear to me, why B is wrong ?

Thanks

For why B is not sufficient see my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047

When taking (1) and (2) together, from (1) we have that $$\angle{ADB}$$ is acute, meaning its measure is less than 90 degrees. Then using (2) we get that $$\angle{CAD}<\angle{ADB}<90^o$$, so necessarily $$\angle{CAD}$$ is also acute.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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06 Oct 2013, 04:22
EvaJager wrote:
voodoochild wrote:
If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?

(2) alone is not sufficient.
Remember the condition in (1) and check the case when that condition doesn't hold.
In my attached drawing, in the isosceles trapezoid, $$x=y.$$

In your diagram Angel ADB is not acute. How can we use that to prove that we can have x = y with both (1) & (2) statements.

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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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arjtryarjtry wrote:
Points A, B, C and D lie on a circle of radius 1. Let $$x$$ be the length of arc AB and $$y$$ the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ?

1. $$\angle$$ ADB is acute
2. $$\angle$$ ADB > $$\angle$$ CAD

Responding to a pm:

A, B, C and D could lie anywhere on the circle. There is nothing given to say that they must lie in that order.
What is the relevance of $$x \lt \pi$$ and $$y \lt \pi$$? The total circumference of the circle of radius 1 is $$2\pi$$. So this is to tell you that both arcs are less than semi circles so we are talking about the minor arcs of AB and CD.

Question: Is x > y?

1. $$\angle$$ ADB is acute

This tells us that D lies on the major arc of AB, not on minor arc i.e. it doesn't lie on the red part which is x. Had D been on x, the angle ADB would have been obtuse.

Attachment:

Ques3.jpg [ 9.91 KiB | Viewed 6816 times ]

But y may be smaller than x or greater depending on where C is. Hence this statement alone is not sufficient.

2. $$\angle$$ ADB > $$\angle$$ CAD
Angles ADB and CAD could be inscribed angles of arcs x and y on their major arcs in which case this implies that x > y. Or angle ADB could be obtuse and arc x could be less than arc y.

Attachment:

Ques4.jpg [ 24.37 KiB | Viewed 6816 times ]

Note that in both the cases above, angle ADB > angle CAD. Angle ADB is obtuse since the inscribed angle is in the minor arc. Angle CAD is acute since the angle is in the major arc. But in the first case x > y and in the second case x < y. Hence this statement alone is not sufficient.

Using both statements together, angle ADB is acute and greater than angle CAD so angle CAD is also acute.
This means we are talking about inscribed angles in the major arc in both cases i.e. the first case in the figure given above. Then, if inscribed angle of x is greater than inscribed angle of y, it means central angle of x is greater than central angle of y (Central angle = 2*Inscribed angle in major arc) and hence arc x is greater than arc y.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17357 [8], given: 232 Senior Manager Joined: 08 Apr 2012 Posts: 446 Kudos [?]: 79 [0], given: 58 Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] ### Show Tags 21 Sep 2014, 08:51 VeritasPrepKarishma wrote: arjtryarjtry wrote: Points A, B, C and D lie on a circle of radius 1. Let $$x$$ be the length of arc AB and $$y$$ the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ? 1. $$\angle$$ ADB is acute 2. $$\angle$$ ADB > $$\angle$$ CAD Responding to a pm: A, B, C and D could lie anywhere on the circle. There is nothing given to say that they must lie in that order. What is the relevance of $$x \lt \pi$$ and $$y \lt \pi$$? The total circumference of the circle of radius 1 is $$2\pi$$. So this is to tell you that both arcs are less than semi circles so we are talking about the minor arcs of AB and CD. Question: Is x > y? 1. $$\angle$$ ADB is acute This tells us that D lies on the major arc of AB, not on minor arc i.e. it doesn't lie on the red part which is x. Had D been on x, the angle ADB would have been obtuse. Attachment: Ques3.jpg But y may be smaller than x or greater depending on where C is. Hence this statement alone is not sufficient. 2. $$\angle$$ ADB > $$\angle$$ CAD Angles ADB and CAD could be inscribed angles of arcs x and y on their major arcs in which case this implies that x > y. Or angle ADB could be obtuse and arc x could be less than arc y. Attachment: Ques4.jpg Note that in both the cases above, angle ADB > angle CAD. Angle ADB is obtuse since the inscribed angle is in the minor arc. Angle CAD is acute since the angle is in the major arc. But in the first case x > y and in the second case x < y. Hence this statement alone is not sufficient. Using both statements together, angle ADB is acute and greater than angle CAD so angle CAD is also acute. This means we are talking about inscribed angles in the major arc in both cases i.e. the first case in the figure given above. Then, if inscribed angle of x is greater than inscribed angle of y, it means central angle of x is greater than central angle of y (Central angle = 2*Inscribed angle in major arc) and hence arc x is greater than arc y. Answer (C) Hi Karishma, Do you have any tips on how to approach these kinds of question which can't be solved algebraically? This question or for example "to how many "slices" can a circle be divided using 4 straight lines"? Both of these aren't solvable using algebra, but I don't really know how to approach them either.... Kudos [?]: 79 [0], given: 58 GMAT Club Legend Joined: 09 Sep 2013 Posts: 16678 Kudos [?]: 273 [0], given: 0 Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] ### Show Tags 22 Mar 2016, 03:19 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 273 [0], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17357 [0], given: 232 Location: Pune, India Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] ### Show Tags 01 Aug 2016, 22:13 VeritasPrepKarishma wrote: arjtryarjtry wrote: Points A, B, C and D lie on a circle of radius 1. Let $$x$$ be the length of arc AB and $$y$$ the length of arc CD respectively, such that $$x \lt \pi$$ and $$y \lt \pi$$ . Is $$x \gt y$$ ? 1. $$\angle$$ ADB is acute 2. $$\angle$$ ADB > $$\angle$$ CAD Responding to a pm: A, B, C and D could lie anywhere on the circle. There is nothing given to say that they must lie in that order. What is the relevance of $$x \lt \pi$$ and $$y \lt \pi$$? The total circumference of the circle of radius 1 is $$2\pi$$. So this is to tell you that both arcs are less than semi circles so we are talking about the minor arcs of AB and CD. Question: Is x > y? 1. $$\angle$$ ADB is acute This tells us that D lies on the major arc of AB, not on minor arc i.e. it doesn't lie on the red part which is x. Had D been on x, the angle ADB would have been obtuse. Attachment: Ques3.jpg But y may be smaller than x or greater depending on where C is. Hence this statement alone is not sufficient. 2. $$\angle$$ ADB > $$\angle$$ CAD Angles ADB and CAD could be inscribed angles of arcs x and y on their major arcs in which case this implies that x > y. Or angle ADB could be obtuse and arc x could be less than arc y. Attachment: Ques4.jpg Note that in both the cases above, angle ADB > angle CAD. Angle ADB is obtuse since the inscribed angle is in the minor arc. Angle CAD is acute since the angle is in the major arc. But in the first case x > y and in the second case x < y. Hence this statement alone is not sufficient. Using both statements together, angle ADB is acute and greater than angle CAD so angle CAD is also acute. This means we are talking about inscribed angles in the major arc in both cases i.e. the first case in the figure given above. Then, if inscribed angle of x is greater than inscribed angle of y, it means central angle of x is greater than central angle of y (Central angle = 2*Inscribed angle in major arc) and hence arc x is greater than arc y. Answer (C) Responding to a pm: Quote: for the question above ..can we consider a case where intersection point of AC and DB is the centre of circle then in that case both x=y No. This is not possible. Note that if AC and DB pass through the centre of the circle, they are diameters. They will split the circle into half. Then arcs x and y will be of lengths $$\pi$$ each (because circumference is $$2\pi$$). But you are given that both x and y are less than $$\pi$$. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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Hello from the GMAT Club BumpBot!

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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink]

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31 Aug 2017, 03:47
EvaJager wrote:
voodoochild wrote:
If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?

(2) alone is not sufficient.
Remember the condition in (1) and check the case when that condition doesn't hold.
In my attached drawing, in the isosceles trapezoid, $$x=y.$$

How to approach these type of sums? People please explain.

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Re: Points A, B, C and D lie on a circle of radius 1. Let x be   [#permalink] 31 Aug 2017, 03:47
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