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# What is the area of the quadrilateral shown above?

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Math Expert
Joined: 02 Sep 2009
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18 Jul 2018, 20:26
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Difficulty:

35% (medium)

Question Stats:

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 [ 31.22 KiB | Viewed 548 times ]

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18 Jul 2018, 20:45

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,
4=1+AE^2

= ((2+4)/2)×√3
= 3√3

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18 Jul 2018, 21:07
Bunuel wrote:

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
Attachments

2018-07-19_0825.png [ 38.33 KiB | Viewed 448 times ]

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

Attachment:

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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