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What is the area of the quadrilateral shown above?

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What is the area of the quadrilateral shown above?  [#permalink]

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New post 18 Jul 2018, 20:26
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A
B
C
D
E

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68% (01:00) correct 32% (01:27) wrong based on 34 sessions

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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


Attachment:
2018-07-19_0825.png
2018-07-19_0825.png [ 31.22 KiB | Viewed 548 times ]

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Re: What is the area of the quadrilateral shown above?  [#permalink]

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New post 18 Jul 2018, 20:45
Area of quadrilateral = (a+b/2)×h

Let the vertices be A,B,C,D (From top left in clockwise)

Consider right angle triangle ABE (Drop 2 perpendiculars E,F on CD)

EF=2
ED=CF=1
Apply Pythagoras theorem,
AD^2=ED^2+AE^2
4=1+AE^2
AE=Height of quadrilateral=√3

Area of quadrilateral= ((AB+CD)/2)×H
= ((2+4)/2)×√3
= 3√3

Answer:B

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Re: What is the area of the quadrilateral shown above?  [#permalink]

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New post 18 Jul 2018, 21:07
Bunuel wrote:
Image
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


Attachment:
The attachment 2018-07-19_0825.png is no longer available



the trapezium is isosceles trapezium as it has same angles 'a'. it means that the sides will also be same, also given as 2 ....

look at the attached figure.

drop a perpendicular AB on base..
Now two sides are 1:AB:2.... so it is a 30:60:90 triangle and hence sides are 1:\(\sqrt{3}\):2
Perpendicular = AB = \(\sqrt{3}\)

area of trapezium = \(\frac{1}{2}*\sqrt{3}*(4+2)=3\sqrt{3}\)

B
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What is the area of the quadrilateral shown above?  [#permalink]

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New post 19 Jul 2018, 01:11
Let's extend the trapezium to draw a couple of right-angled triangles.

Attachment:
Diagram_Complete.png
Diagram_Complete.png [ 36.33 KiB | Viewed 371 times ]


In the above diagram, Angle DAB = Angle CFE = 90

Applying Pythagoras Theorem to trangle ABD, \(BD^2 = AB^2 + AD^2\)

Solving for AD, we get \(AD^2 = 2^2 - 1^2\) -> \(AD^2 = 3\) -> \(AD = \sqrt{3}\)

Area of Trapezium(BCDE) = Area of Rectangle(AFDE) - Area of Triangle(ABD) - Area of Triangle(CFE)

Substituting values, we get Area(BCDE) =\(4 * \sqrt{3} - 0.5 * \sqrt{3} * 1 - 0.5* \sqrt{3} * 1 = \sqrt{3}(4 - 0.5 - 0.5) = 3\sqrt{3}\)

Therefore, the area of the trapezium is \(3\sqrt{3}\) (Option B)
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What is the area of the quadrilateral shown above? &nbs [#permalink] 19 Jul 2018, 01:11
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