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M22-02

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New post 16 Sep 2014, 01:15
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If the diagonals of rectangle \(ABCD\) (\(AB \gt BC\)) cross at point \(E\), what is the value of \(\angle AEB\) ?


(1) The value of \(\angle EDC\) is 15 degrees

(2) The value of \(\angle ACB\) is 75 degrees

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New post 16 Sep 2014, 01:15
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New post 19 Aug 2015, 05:53
Hi Bunuel,

What properties are applied here ? Is there any alternative here ?
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New post 25 Aug 2015, 13:26
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The problem says it's a rectangle. A rectangle's diagonals bisect each other and are equal in length, so if you draw a rectangle and draw in the diagonals, you'll get 4 isosceles triangles. Both statements give one angle of the two angles between the base and the equal sides, so you can calculate the third angle from both statements and get to AEB.
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New post 29 Nov 2015, 01:01
i know its real late to answer... diagonals of rectangle bisect each other ... so both statements give us the 2 angles of the smaller triangles ....
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New post 26 Dec 2017, 09:09
+1 for D. Use the property that diagonals of a rectangle bisect one another. Both statements are sufficient individually. Hence option D.
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New post 27 Mar 2018, 20:53
Bunuel. Please post a pictorial explanation. Thanks.
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New post 28 Mar 2018, 00:37
No Bunuel I could not relate the picture with your solution given earlier. Could you please explain it with picture how \(\angle AEB = 180 - 2*\angle EDC = 150^o\)?

and \(\angle AEB = 180 - 2*\angle BAC = 180 - 2*(180 - 90 - \angle ACB) = 150^o\)?
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New post 28 Mar 2018, 02:30
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sadikabid27 wrote:
No Bunuel I could not relate the picture with your solution given earlier. Could you please explain it with picture how \(\angle AEB = 180 - 2*\angle EDC = 150^o\)?

and \(\angle AEB = 180 - 2*\angle BAC = 180 - 2*(180 - 90 - \angle ACB) = 150^o\)?


Image


For (1). First of all, \(\angle AEB = \angle DEC\). Next, DEC is isosceles, thus \(\angle EDC = \angle ECD=15^o\). The sum of the angles in a triangle is 180 degrees:

\(\angle EDC + \angle ECD + \angle DEC = 180^o\);

\(\angle EDC + \angle EDC + \angle AEB = 180^o\);

\(\angle AEB = 180 - 2*\angle EDC = 150^o\).


For (2). First of all, \(\angle BAC + \angle ACB= 90^o\) --> \(\angle BAC =90^o-\angle ACB\). Next, AEB is isosceles, thus \(\angle BAC = \angle ABE\). The sum of the angles in a triangle is 180 degrees. \(\angle BAC + \angle ABE + \angle AEB = 180^o\);

\(\angle BAC + \angle BAC + \angle AEB = 180^o\);

\(\angle AEB =180^o-2*\angle BAC = 180^o-2*(90^o-\angle ACB)\)
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M22-02 &nbs [#permalink] 28 Mar 2018, 02:30
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