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Moderator  Joined: 18 Apr 2015
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What is the area of the quadrilateral shown above? [#permalink]
Expert's post 00:00

Question Stats: 86% (01:03) correct 13% (01:34) wrong based on 36 sessions

Attachment: #GREpracticequestion What is the area of the quadrilateral shown above .jpg [ 6.78 KiB | Viewed 891 times ]

What is the area of the quadrilateral shown above?

A) $$2 \sqrt{3}$$

B) $$3 \sqrt{3}$$

C) $$6$$

D) $$6 \sqrt{3}$$

E) $$8$$
[Reveal] Spoiler: OA

_________________ Intern Joined: 02 Jan 2018
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Re: What is the area of the quadrilateral shown above? [#permalink]
1
KUDOS
Draw a perpendicular forming two right angle triangle, now the figure has two right angle triangle and one rectangle.Both the triangles have same angle so the area will be same.
By using pythagoras theorem find the height of the triangle i.e sqrt(3), now find the area of both the triangle and a rectangle.
sqrt(3)/2+sqrt(3)/2+2*sqrt(3) = 3sqrt(3) Intern Joined: 21 Dec 2017
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Re: What is the area of the quadrilateral shown above? [#permalink]
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Consider three equilateral triangles within quadrilateral,each with side measuring 2.Now Area for Equilateral triangle is (sqrt(3)/4)*(side)^2 and multiply it by 3 to get area for three equilateral triangle i.e whole quadrilateral.
Manager Joined: 02 Jan 2018
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Re: What is the area of the quadrilateral shown above? [#permalink]
Answer for this please. I am getting 3\sqrt{3} as my answer.
Moderator  Joined: 18 Apr 2015
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Kudos [?]: 1138 , given: 5418

Re: What is the area of the quadrilateral shown above? [#permalink]
Expert's post
Added the OA. It is B.

Regards
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Director Joined: 09 Nov 2018
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Re: What is the area of the quadrilateral shown above? [#permalink]
mayurwaghela wrote:
By using pythagoras theorem find the height of the triangle i.e sqrt(3),

How?
Director Joined: 09 Nov 2018
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Re: What is the area of the quadrilateral shown above? [#permalink]
Can we think it as a trapezium?
Intern Joined: 02 Jan 2019
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Re: What is the area of the quadrilateral shown above? [#permalink]
Yes it can be considered as a trapezoid, since the two angles alpha are the same and are connected to the longer of the two bases.

Area: 0.5(Base 1 + Base 2) * height.

How do we get the height? The shorter base (length 2) must have its center where the longer base has its center due to the fact that both angles are equal. Thus, we derive that the longer base just extends the shorter base by (4-2 = 2). Split equally on each side, we can see the longer base composition of lengths 1 + 2 + 1.

If we look at the left part of the figure we have the upward sloping line with length 2. If we let fall a perpendicular from the connection of the upward sloping line and its vertex with the shorter base, we arrive exactly at the first part of the 1 + 2 + 1 composition of the longer base.

Thus we have a created triangle with a base 1 and hypotenuse 2. Since we also know that it has a 90-degree angle, we can deduct it must be a 30 - 60 - 90 triangle. (Remainder: Side lengths of a 30 - 60 - 90 triangle are 1:2:sqrt(3). Thus the height of the triangle is sqrt(3) which equals the height of the trapezoid.

Plug in the formula. Re: What is the area of the quadrilateral shown above?   [#permalink] 23 Jan 2019, 07:43
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