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What is the area of the trapezoid above if AB = BC = CD = 6, and AD =

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What is the area of the trapezoid above if AB = BC = CD = 6, and AD = [#permalink]

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New post 18 Jan 2018, 02:03
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What is the area of the trapezoid above if AB = BC = CD = 6, and AD = 12?

A. \(6\sqrt{6}\)

B. \(18\sqrt{2}\)

C. \(18\sqrt{3}\)

D. \(27\sqrt{2}\)

E. \(27\sqrt{3}\)


[Reveal] Spoiler:
Attachment:
2018-01-18_1400.png
2018-01-18_1400.png [ 6.24 KiB | Viewed 454 times ]
[Reveal] Spoiler: OA

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Re: What is the area of the trapezoid above if AB = BC = CD = 6, and AD = [#permalink]

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New post 18 Jan 2018, 04:42
Bunuel wrote:
Image
What is the area of the trapezoid above if AB = BC = CD = 6, and AD = 12?

A. \(6\sqrt{6}\)

B. \(18\sqrt{2}\)

C. \(18\sqrt{3}\)

D. \(27\sqrt{2}\)

E. \(27\sqrt{3}\)


[Reveal] Spoiler:
Attachment:
2018-01-18_1400.png


Area of Trapezium = SUm of the two parallel sides * Distance between parallel sides / 2

i.e. Area of Trapezium = = (BC+AD)*Distance /2

Distance between two sides = Perpendicular dropped from B on AD = h

The triangle will have side 3-6-h

Now using pythagorus theorem, h = 3√3

Hence Area of Trapezium = (6+12)*3√3 /2 = 27√3

Answer: option E
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Re: What is the area of the trapezoid above if AB = BC = CD = 6, and AD = [#permalink]

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New post 18 Jan 2018, 06:26
Bunuel wrote:
Image
What is the area of the trapezoid above if AB = BC = CD = 6, and AD = 12?

A. \(6\sqrt{6}\)

B. \(18\sqrt{2}\)

C. \(18\sqrt{3}\)

D. \(27\sqrt{2}\)

E. \(27\sqrt{3}\)


[Reveal] Spoiler:
Attachment:
The attachment 2018-01-18_1400.png is no longer available
Attachment:
2018-01-18_1400.png
2018-01-18_1400.png [ 7.56 KiB | Viewed 258 times ]


Here BE = CF =\(3\sqrt{3}\)

Area of Triangle ABE + Triangle CFD = \(2*\frac{1}{2}*3*3\sqrt{3}\) = \(9\sqrt{3}\)

Area of rectangle BEFC = \(6*3\sqrt{3}\) = \(18\sqrt{3}\)

So, The total area is \(27\sqrt{3}\), answer will be (E)
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Re: What is the area of the trapezoid above if AB = BC = CD = 6, and AD = [#permalink]

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New post 20 Jan 2018, 06:35
Bunuel wrote:
Image
What is the area of the trapezoid above if AB = BC = CD = 6, and AD = 12?

A. \(6\sqrt{6}\)

B. \(18\sqrt{2}\)

C. \(18\sqrt{3}\)

D. \(27\sqrt{2}\)

E. \(27\sqrt{3}\)


[Reveal] Spoiler:
Attachment:
2018-01-18_1400.png


Notice that trapezoid ABCD is an isosceles trapezoid since AB = CD. By dropping a perpendicular from B to AD and C to AD we have created two right triangles and a rectangle in the middle. Let E and F be points on AD where the perpendiculars are dropped from B and C, respectively. We see that AE = FD and BE = CF. We can let n = AE = FD and h = BE = CF.

Since BC = EF, we can say that:

EF + n + n = AD

6 + 2n = 12

2n = 6

n = 3 = AE

Since AE^2 + BE^2 = AB^2, we have

3^2 + h^2 = 6^2

h^2 = 27

h = 3√3

We now can determine the area of the trapezoid.

area = (base 1 + base 2)/2 x height

area = (6 + 12)/2 x 3√3 = 9 x 3√3 = 27√3

Answer: E
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Re: What is the area of the trapezoid above if AB = BC = CD = 6, and AD =   [#permalink] 20 Jan 2018, 06:35
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