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22 Jul 2015, 02:02
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D
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Difficulty:
35%
(medium)
Question Stats:
75% (02:19) correct
25%
(02:15)
wrong
based on 209
sessions
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In the figure above, lines L and P are parallel. The segment AD is the same length as the segment DC; AD is parallel to BC. If the length of AD is equal to 4, and angle ADC is equal to 60º, what is the area of ABCD?
A. 4√2
B. 4√3
C. 8√2
D. 8√3
E. 16√2
Kudos for a correct solution.
Attachment:
parallel.gif [ 2.38 KiB | Viewed 10365 times ]
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
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Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
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22 Jul 2015, 03:51
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Bunuel
Line L||P and AD||BC ---> \(\angle {ADC} = \angle{BCD} = 60\)
As AD = CD and AD || BC ---> ABCD is a rhombus with sides 4 inches each.
Thus the area = base X height
For height, draw a perpendicular line from A to CD meeting CD at O. Thus in right triangle AOD (rigth angled at O), we can apply the property of 30-60-90 triangle with the hypotenuse = AD = 4
Thus we get the height = \(2\sqrt{3}\)
Finally, the area = base X height = \(4*2\sqrt{3}\) = \(8\sqrt{3}\). D is the correct answer.
22 Jul 2015, 02:39
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Bunuel
Drawing a line from A to C, we find that Triangle ADC is equilateral triangle.
Angle ADP is 120 deg=DCB=120. therefore ACB=60 deg
CBL=120 deg=ABC=60 deg. there fore triangle ABC is equilateral triangle.
Total area=area of 2 equilateral triangle=D
