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01 Sep 2008, 04:33
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
25% (02:24) correct
75%
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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) \(\angle ADB > \angle CAD\)
(1) \(\angle ADB\) is acute
(2) \(\angle ADB > \angle CAD\)
20 Apr 2014, 19:59
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arjtryarjtry
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 23926 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 23865 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.
Answer (C)
General Discussion
02 Sep 2008, 09:30
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arjtryarjtry
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.
