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What is the area of parallelogram ABCD?

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What is the area of parallelogram ABCD?  [#permalink]

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New post 11 Sep 2018, 01:40
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A
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D
E

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What is the area of parallelogram ABCD?


A. 20
B. \(20\sqrt{3}\)
C. 40
D. \(40\sqrt{3}\)
E. 80


Attachment:
image005.jpg
image005.jpg [ 2.22 KiB | Viewed 756 times ]

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Re: What is the area of parallelogram ABCD?  [#permalink]

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New post 11 Sep 2018, 03:34
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Bunuel wrote:
Image
What is the area of parallelogram ABCD?

A. 20
B. \(20\sqrt{3}\)
C. 40
D. \(40\sqrt{3}\)
E. 80
Attachment:
image005.jpg
[/spoiler]


Opposite angles of the parallelograms are equal.
That means we have 2 150 degrees and 1 30 degrees angles.

If we drop a perpendicular, from one of 150degree angle vertex to the opposite side
We will have a 30-60-90 triangle.
We can calculate the height from there, which is 4 cm.
We have the base. we have the height.

Area=40
Answer: C
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Re: What is the area of parallelogram ABCD?  [#permalink]

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New post 11 Sep 2018, 10:50
Using trigo , area of parallelogram = side 1*side2 *sin of included angle
= 10 * 8 * 1/2 = 40
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Re: What is the area of parallelogram ABCD?  [#permalink]

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New post 12 Sep 2018, 17:25
1
1
Bunuel wrote:
Image
What is the area of parallelogram ABCD?


A. 20
B. \(20\sqrt{3}\)
C. 40
D. \(40\sqrt{3}\)
E. 80

Attachment:
image005.jpg


Since angle B = 150, angle A = 30.

Thus, when we drop a height from angle D to side AB and call the intersection E, we will have triangle ADE as a 30-60-90 right triangle. Since AD = BC = 8 is the hypotenuse of the ADE and DE is the side opposite 30-degree angle, DE = 4. However, DE is also the height of the parallelogram. Therefore, the area of the parallelogram is 10 x 4 = 40.

Answer: C
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Re: What is the area of parallelogram ABCD?  [#permalink]

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New post 06 Oct 2018, 23:12
Area of the parallelogram ABCD = AB.CD. sin150 deg. = 40 sq. units.(C)
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Re: What is the area of parallelogram ABCD? &nbs [#permalink] 06 Oct 2018, 23:12
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What is the area of parallelogram ABCD?

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