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13 Feb 2020, 01:29
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:23) correct
38%
(02:07)
wrong
based on 88
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Not Attempted Yet
[GMAT math practice question]
The figure below shows parallelogram \(ABCD\) and point \(E\) is on line \(BC.\) Line DE bisects \(∠D\). Moreover, \(BE = DE\) and \(∠A = 120^o\). What is the measure of \(∠BDE\)?
2.13PS.png [ 9.96 KiB | Viewed 2180 times ]
A. \(20^o\)
B. \(30^o\)
C. \(33^o\)
D. \(40^o\)
E. \(43^o\)
The figure below shows parallelogram \(ABCD\) and point \(E\) is on line \(BC.\) Line DE bisects \(∠D\). Moreover, \(BE = DE\) and \(∠A = 120^o\). What is the measure of \(∠BDE\)?
Attachment:
2.13PS.png [ 9.96 KiB | Viewed 2180 times ]
A. \(20^o\)
B. \(30^o\)
C. \(33^o\)
D. \(40^o\)
E. \(43^o\)
16 Feb 2020, 18:51
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=>
Since \(BE = DE\), we have \(∠DBE = ∠BDE\) and \(∠DEC = 2*(∠BDE) = 2*(∠EDC).\)
Since \(∠EDC + ∠DEC + ∠C = 180°, ∠EDC + 2*(∠EDC) + 120° = 180°, 3*(∠EDC) + 120° = 180°, 3*(∠EDC) = 60°, ∠EDC = 20°.\)
Then we have \(∠BDE = ∠EDC = 20°.\)
Therefore, the answer is A.
Answer: A
Since \(BE = DE\), we have \(∠DBE = ∠BDE\) and \(∠DEC = 2*(∠BDE) = 2*(∠EDC).\)
Since \(∠EDC + ∠DEC + ∠C = 180°, ∠EDC + 2*(∠EDC) + 120° = 180°, 3*(∠EDC) + 120° = 180°, 3*(∠EDC) = 60°, ∠EDC = 20°.\)
Then we have \(∠BDE = ∠EDC = 20°.\)
Therefore, the answer is A.
Answer: A
22 Feb 2020, 03:47
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MathRevolution
CAn be solved by looking at answer choices
angle ADC is 60Deg.
Also angle BDE = EDC = 2x(say)
we look for value x such that,it is less than 60 Deg when we put it in 2x
only 20 fits
Hence it is our answer.
Ans A
