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# In the figure above, lines L and P are parallel. The segment AD is the

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Math Expert
Joined: 02 Sep 2009
Posts: 58434
In the figure above, lines L and P are parallel. The segment AD is the  [#permalink]

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22 Jul 2015, 02:02
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Difficulty:

45% (medium)

Question Stats:

74% (02:25) correct 26% (02:14) wrong based on 182 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 5231 times ]

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Re: In the figure above, lines L and P are parallel. The segment AD is the  [#permalink]

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22 Jul 2015, 02:39
2
Bunuel wrote:

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

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
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Re: In the figure above, lines L and P are parallel. The segment AD is the  [#permalink]

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22 Jul 2015, 03:51
3
Bunuel wrote:

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

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58434
Re: In the figure above, lines L and P are parallel. The segment AD is the  [#permalink]

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26 Jul 2015, 11:51
2
1
Bunuel wrote:

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

800score Official Solution:

AB is parallel to DC and AD is parallel to BC, so ABCD is parallelogram. We know how to find the area of a parallelogram we multiply the height of parallelogram by its base length.

We are told that the angle ADC is a 60º angle. If we draw an altitude from A straight down and perpendicular to line P, the length of that altitude will be equal to the height of the parallelogram. Moreover, it forms a 60-30-90 triangle, so we can easily find its length. Since the hypotenuse of that triangle is equal to 4, the second longest side must be equal to 2√3 (since the proportions for the triangle run x, x√3, and 2x).

To find the area of a parallelogram , multiply the height of the parallelogram by its base length: 4 × 2√3 = 8√3, or answer choice (D).
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Re: In the figure above, lines L and P are parallel. The segment AD is the  [#permalink]

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12 Apr 2017, 21:24
Hello Bunuel,

Is below approach correct?

$$\angle$$ ADC = 60, AD=DC; so ADC will be an equilateral triangle with sides 4 each.

Thus height of the triangle will be the height of parallelogram = ($$\sqrt{3}$$ /2 )* side 4

Thus it will 8 * $$\sqrt{3}$$
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Re: In the figure above, lines L and P are parallel. The segment AD is the  [#permalink]

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02 Nov 2018, 02:09
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Re: In the figure above, lines L and P are parallel. The segment AD is the   [#permalink] 02 Nov 2018, 02:09
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