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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (03:41) correct
69%
(03:21)
wrong
based on 29
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
QS429.png
Triangles AED and BEC are formed using the straight lines AB and CD as shown in the figure above. If \(BE = BC, DE^2 > AD^2 + AE^2\) and the measure of \(∠AED\) is x, which of the following statements must be true?
I. \(CE^2 > 2BE^2\)
II. \(AE < AD\)
III. \(DE > CE\)
A) I only
B) II only
C) III only
D) I, II and III
E) None of the above
Archived Topic
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15 Jan 2019, 11:12
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EXPLANATION:
1) Labeling the figure from given information:
Let's label the figure from the given information:
1.1) \(∠DAE\) = (180-3x) from lower triangle
1.2) Since E is the intersection point of two lines, \(∠BEC\) = \(∠AED\) = x
1.3) Since BE = BC, \(∠ECB\) = \(∠BCE\) = x
1.4) \(∠EBC\) = 180-2x
1.5) Since \(DE^2 > AD^2 + AE^2\), we know that angle opposite to side DE is > 90 degrees and it is an obtuse triangle. So angle opposite to DE is \(∠DAE\) which from 1.1 is 180-3x.
So 180-3x >90, and x <30
2) Evaluating each statement independently:
Now that we have established our pre-thinking, let's evaluate each statement:
I) \(CE^2 > 2BE^2\) which is the same as \(CE^2 > BE^2 + BC^2\) since BE = BC
Now for the above inequality to be true, angle opposing CE should be greater than 90, which means 190-2x should be greater than 90 , we already know that x <30, so 180-2x should be less than 120.
Hence sufficient condition. I is true.
II) AE <AD
we know that angle opposing AD is greater than angle opposing side AE, so the above inequality cannot be true.
III) AE > AD
we know that angles opposing each side is proportional to that side length. So angle opposing AE = 180-3x and angle opposing AD is 180-2x. Under all circumstances. 180-3x < 180-2x,
hence Statement III is not true.
So answer is I only. Hence A is the answer.
----------------------------------------------------------------------------------------------------------------------------------
"Please hit :thumbup: +1 Kudos if you like this post"
----------------------------------------------------------------------------------------------------------------------------------
1) Labeling the figure from given information:
Let's label the figure from the given information:
1.1) \(∠DAE\) = (180-3x) from lower triangle
1.2) Since E is the intersection point of two lines, \(∠BEC\) = \(∠AED\) = x
1.3) Since BE = BC, \(∠ECB\) = \(∠BCE\) = x
1.4) \(∠EBC\) = 180-2x
1.5) Since \(DE^2 > AD^2 + AE^2\), we know that angle opposite to side DE is > 90 degrees and it is an obtuse triangle. So angle opposite to DE is \(∠DAE\) which from 1.1 is 180-3x.
So 180-3x >90, and x <30
2) Evaluating each statement independently:
Now that we have established our pre-thinking, let's evaluate each statement:
I) \(CE^2 > 2BE^2\) which is the same as \(CE^2 > BE^2 + BC^2\) since BE = BC
Now for the above inequality to be true, angle opposing CE should be greater than 90, which means 190-2x should be greater than 90 , we already know that x <30, so 180-2x should be less than 120.
Hence sufficient condition. I is true.
II) AE <AD
we know that angle opposing AD is greater than angle opposing side AE, so the above inequality cannot be true.
III) AE > AD
we know that angles opposing each side is proportional to that side length. So angle opposing AE = 180-3x and angle opposing AD is 180-2x. Under all circumstances. 180-3x < 180-2x,
hence Statement III is not true.
So answer is I only. Hence A is the answer.
----------------------------------------------------------------------------------------------------------------------------------
"Please hit :thumbup: +1 Kudos if you like this post"
----------------------------------------------------------------------------------------------------------------------------------
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
