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In the figure below, ADE
[#permalink]
19 May 2026, 13:11
19 May 2026, 13:11
Expert Reply
00:00
Question Stats:
58% (02:20) correct
41% (02:52) wrong based on 17 sessions
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In the figure below, \(\angle \mathrm{ADE}=60^{\circ}, \angle \mathrm{EFC}=40^{\circ}\), and \(\angle \mathrm{DAE}= 55^{\circ}\). If \(A B |\mid C D\), what is the value of x ?
Note: Figure not drawn to scale
GRE in the figure below.png [ 183.63 KiB | Viewed 439 times ]
\(\begin{aligned}
& A. 75 \\
& B. 85 \\
& C. 95 \\
& D. 105 \\
& E. 115
\end{aligned}\)
Note: Figure not drawn to scale
Attachment:
GRE in the figure below.png [ 183.63 KiB | Viewed 439 times ]
\(\begin{aligned}
& A. 75 \\
& B. 85 \\
& C. 95 \\
& D. 105 \\
& E. 115
\end{aligned}\)
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Joined: 20 Feb 2017
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In the figure below, ADE
[#permalink]
23 May 2026, 02:10
23 May 2026, 02:10
1
Expert Reply
Explanation
ADE=60, DAE=55, So
AED=180
Use the straight lines
Points A,E,F are collinear, and D,E,C are collinear. Therefore the angle at E in ECF is vertically opposite to AED, so
CEF=65
Also given: EFC=40
3) Find the angle at C
In ECF, ECF=180-65-40=ECF=75
Use AB∥CD
Since AB∥CD, and E,C,D lie on the same straight line, we have
AB∥EC
Also B,C,F are collinear, so BC is the same line as CF. Hence the angle between AB and BC equals the angle between EC and CF:
x=ECF=75
Answer: A
ADE=60, DAE=55, So
AED=180
Use the straight lines
Points A,E,F are collinear, and D,E,C are collinear. Therefore the angle at E in ECF is vertically opposite to AED, so
CEF=65
Also given: EFC=40
3) Find the angle at C
In ECF, ECF=180-65-40=ECF=75
Use AB∥CD
Since AB∥CD, and E,C,D lie on the same straight line, we have
AB∥EC
Also B,C,F are collinear, so BC is the same line as CF. Hence the angle between AB and BC equals the angle between EC and CF:
x=ECF=75
Answer: A
gmatclubot
In the figure below, ADE [#permalink]
23 May 2026, 02:10
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