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# M22-02

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:15
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Difficulty:

25% (medium)

Question Stats:

73% (01:04) correct 27% (01:11) wrong based on 124 sessions

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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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Math Expert
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16 Sep 2014, 01:15
Official Solution:

Statement (1) by itself is sufficient. $$\angle AEB = 180 - 2*\angle EDC = 150^o$$.

Statement (2) by itself is sufficient. $$\angle AEB = 180 - 2*\angle BAC = 180 - 2*(180 - 90 - \angle ACB) = 150^o$$.

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Intern
Joined: 12 Sep 2013
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19 Aug 2015, 05:53
Hi Bunuel,

What properties are applied here ? Is there any alternative here ?
Intern
Joined: 17 Jul 2015
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Schools: Booth '18 (II)

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25 Aug 2015, 13:26
2
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.
Intern
Joined: 01 Apr 2014
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Schools: ISB '17
GMAT 1: 530 Q35 V28
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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 ....
Director
Joined: 08 Jun 2015
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Location: India
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GMAT 2: 700 Q48 V38

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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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Manager
Joined: 10 Sep 2014
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27 Mar 2018, 20:53
Bunuel. Please post a pictorial explanation. Thanks.
Math Expert
Joined: 02 Sep 2009
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27 Mar 2018, 21:29
1
Bunuel. Please post a pictorial explanation. Thanks.

Here it is:

Does it help?

Attachment:
Untitled.png

>> !!!

You do not have the required permissions to view the files attached to this post.

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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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28 Mar 2018, 02:30
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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 $$\angle AEB = 180 - 2*\angle BAC = 180 - 2*(180 - 90 - \angle ACB) = 150^o$$?

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

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