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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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18 Jan 2018, 03:03
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Difficulty:

25% (medium)

Question Stats:

88% (02:03) correct 12% (02:24) wrong based on 61 sessions

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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. $$27\sqrt{3}$$

Attachment:

2018-01-18_1400.png [ 6.24 KiB | Viewed 2408 times ]

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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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18 Jan 2018, 05:42
Bunuel wrote:

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. $$27\sqrt{3}$$

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

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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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18 Jan 2018, 07:26
Bunuel wrote:

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. $$27\sqrt{3}$$

Attachment:
The attachment 2018-01-18_1400.png is no longer available
Attachment:

2018-01-18_1400.png [ 7.56 KiB | Viewed 1917 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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20 Jan 2018, 07:35
Bunuel wrote:

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. $$27\sqrt{3}$$

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

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Re: What is the area of the trapezoid above if AB = BC = CD = 6, and AD =   [#permalink] 20 Jan 2018, 07:35
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