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29 Jul 2019, 01:23
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D
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Difficulty:
65%
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Question Stats:
55% (02:27) correct
45%
(02:35)
wrong
based on 78
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[GMAT math practice question]
\(ABCD\) is a square and \(ADE\) is an equilateral triangle. What is the value of \(x\)?
7.29(ps).png [ 6.57 KiB | Viewed 3302 times ]
\(A. 120^o\)
\(B. 130^o\)
\(C. 140^o\)
\(D. 150^o\)
\(E. 160^o\)
\(ABCD\) is a square and \(ADE\) is an equilateral triangle. What is the value of \(x\)?
Attachment:
7.29(ps).png [ 6.57 KiB | Viewed 3302 times ]
\(A. 120^o\)
\(B. 130^o\)
\(C. 140^o\)
\(D. 150^o\)
\(E. 160^o\)
29 Jul 2019, 01:46
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Please, find the attached file.
Posted from my mobile device
Posted from my mobile device
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File comment: PFA
96A066B9-5425-4FF7-87DA-38F927817F10.jpeg [ 2.61 MiB | Viewed 3251 times ]
firas92
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29 Jul 2019, 03:08
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Since ADE is an equilateral triange,
\(AE=DE=AD=\)Side length of square and \(\angle DAE=\angle ADE=\angle AED=60^{\circ}\)
So, \(\angle BAE=90-\angle DAE=90-60=30^{\circ}\)
We know that \(AE=AD\) and \(AD=AB\)(side length of square) and so \(AE=AB\)
Therefore, triangle ABE is isoceles and \(\angle ABE=\angle AEB\)
So in triangle ABE, \(30+\angle ABE+\angle AEB=180\) or
\(30+2*\angle AEB=180\)
\(\angle AEB=75^{\circ}\)
Similarly, \(\angle DEC=75^{\circ}\)
Now \(\angle AEB+x+\angle DEC+\angle AED= 360\)
\(75+x+75+60=360\)
\(x=150^{\circ}\)
Answer is (D)
Hit Kudos if this helped!
Posted from my mobile device
\(AE=DE=AD=\)Side length of square and \(\angle DAE=\angle ADE=\angle AED=60^{\circ}\)
So, \(\angle BAE=90-\angle DAE=90-60=30^{\circ}\)
We know that \(AE=AD\) and \(AD=AB\)(side length of square) and so \(AE=AB\)
Therefore, triangle ABE is isoceles and \(\angle ABE=\angle AEB\)
So in triangle ABE, \(30+\angle ABE+\angle AEB=180\) or
\(30+2*\angle AEB=180\)
\(\angle AEB=75^{\circ}\)
Similarly, \(\angle DEC=75^{\circ}\)
Now \(\angle AEB+x+\angle DEC+\angle AED= 360\)
\(75+x+75+60=360\)
\(x=150^{\circ}\)
Answer is (D)
Hit Kudos if this helped!
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