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27 Jul 2020, 01:51
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[GMAT math practice question]
\(△ABC\) and \(△CDE\) are equilateral triangles, as the figure shows. What is the measure of the \(∠x\)?
7.27PS.png [ 9.36 KiB | Viewed 6826 times ]
A. \(45°\)
B. \(50°\)
C. \(55°\)
D. \(60°\)
E. \(65°\)
\(△ABC\) and \(△CDE\) are equilateral triangles, as the figure shows. What is the measure of the \(∠x\)?
Attachment:
7.27PS.png [ 9.36 KiB | Viewed 6826 times ]
A. \(45°\)
B. \(50°\)
C. \(55°\)
D. \(60°\)
E. \(65°\)
yashikaaggarwal
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27 Jul 2020, 03:05
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MathRevolution I think we do need another constraint to prove BE bisects angle B.
Proving all triangles congruent is a long process, and do require a lot of time if one go by logic. Can you check that. Regards
Posted from my mobile device
Proving all triangles congruent is a long process, and do require a lot of time if one go by logic. Can you check that. Regards
Posted from my mobile device
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
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Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
27 Jul 2020, 04:24
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Angle DEC = 60
Angle DEA = 120 (180-DEC = supplementary angle)
Also CEF = 60+Y = Angle EBC + Angle ECB
60+Y = 60+EBC
EBC = Y
Hence we can say,
ED is parallel to BC (alternate angles)
Extend ED till it cut AB at H
Such that angle HEA = angle CED = 60° (opposite angle)
Now. AHE is an equilateral triangle too.
AE = ED
In triangle AED
Angle AED + EAD + EDA = 180
120 + 2EAD = 180
2EAD = 60
EAD = 30°
Draw an altitude EG such that it meet AD at G and
Extend EG such that it meet BC at I.
Now HBEI is a parallelogram, whose opposite angles are equal.
Also parallelogram diagonals bisect angle hence Angle HBE = angle IBE = 30° = Angle DEF = FEG.
In triangle FGE, angle EGF = 90° , angle GEF = 30° and
Remaining angle X = angle EFG = 180-90-30 = 60° .
Answer is D
Posted from my mobile device
Angle DEA = 120 (180-DEC = supplementary angle)
Also CEF = 60+Y = Angle EBC + Angle ECB
60+Y = 60+EBC
EBC = Y
Hence we can say,
ED is parallel to BC (alternate angles)
Extend ED till it cut AB at H
Such that angle HEA = angle CED = 60° (opposite angle)
Now. AHE is an equilateral triangle too.
AE = ED
In triangle AED
Angle AED + EAD + EDA = 180
120 + 2EAD = 180
2EAD = 60
EAD = 30°
Draw an altitude EG such that it meet AD at G and
Extend EG such that it meet BC at I.
Now HBEI is a parallelogram, whose opposite angles are equal.
Also parallelogram diagonals bisect angle hence Angle HBE = angle IBE = 30° = Angle DEF = FEG.
In triangle FGE, angle EGF = 90° , angle GEF = 30° and
Remaining angle X = angle EFG = 180-90-30 = 60° .
Answer is D
Posted from my mobile device
Attachments
IMG_20200727_165231.jpg [ 1.45 MiB | Viewed 6491 times ]
