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# If in the figure above AE=AD, FB=6 and the area of the parallelogram D

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If in the figure above AE=AD, FB=6 and the area of the parallelogram D  [#permalink]

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06 Sep 2015, 10:01
2
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45% (medium)

Question Stats:

70% (02:36) correct 30% (02:54) wrong based on 82 sessions

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If in the figure above AE=AD, FB=6 and the area of the parallelogram DEBF is 36, what is the area of the rectangle ABCD?

A. 45
B. 54
C. 72
D. 90
E. 108

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Re: If in the figure above AE=AD, FB=6 and the area of the parallelogram D  [#permalink]

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06 Sep 2015, 17:44
IMO : B

Since DEBF is parallelogram
BF = DE = 6
Using Pythagorean theorem $$AD^2 + AE^2 = DE^2$$
Thus BF = BC = 3√2

The diagram is as follows
Attachment:

1.JPG [ 17.86 KiB | Viewed 1249 times ]

Area of Parallelogram DEBF = 36
b*h = 36 (Where b = DF & h = AD)
b*3√2 = 36
b = 6√2

Thus DC = DF + FC
DC = 6√2 + 3√2
DC = 9√2

Area of ABCD = AD * DC
= 3√2 * 9√2
= 54
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Re: If in the figure above AE=AD, FB=6 and the area of the parallelogram D  [#permalink]

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06 Sep 2015, 18:00
1
reto wrote:
Attachment:
T8925.png

If in the figure above AE=AD, FB=6 and the area of the parallelogram DEBF is 36, what is the area of the rectangle ABCD?

A. 45
B. 54
C. 72
D. 90
E. 108

Area of rectangle ABCD = Area of parallelogram DFBE + 2*Area triangle ADE ...(triangles ADE and BFC are congruent)....(1)

Area of triangle ADE = $$0.5$$*$$x^2$$, where x = AD=AE

For x, in right triangle ADE, right angled at A,

$$AD^2+AE^2=6^2$$ ----> 2$$x^2$$=36 ---> $$x^2$$ = 18

Thus, from 1,

Area of rectangle = 36+18 = 54. B is the correct answer.
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Re: If in the figure above AE=AD, FB=6 and the area of the parallelogram D  [#permalink]

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06 Sep 2015, 21:52
1
reto wrote:
Attachment:
T8925.png

If in the figure above AE=AD, FB=6 and the area of the parallelogram DEBF is 36, what is the area of the rectangle ABCD?

A. 45
B. 54
C. 72
D. 90
E. 108

DEBF is a parallelogram. So if BF = 6, ED = 6 too.
Now, AE = AD so AED forms a right isosceles triangle in which ratio of the sides is $$1:1:\sqrt{2}$$.
Hence, AE = AD = $$6/\sqrt{2}$$

Note that the altitude of parallelogram BEDF is AD and base is BE.
Area of parallelogram = 36 = AD * BE = $$6/\sqrt{2} * BE$$
So $$BE = 6*\sqrt{2}$$

Area of rectangle = AD * (AE + BE) = $$6/\sqrt{2} *(6/\sqrt{2} + 6*\sqrt{2}) = 54$$

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Re: If in the figure above AE=AD, FB=6 and the area of the parallelogram D  [#permalink]

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02 Aug 2017, 14:01
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Re: If in the figure above AE=AD, FB=6 and the area of the parallelogram D   [#permalink] 02 Aug 2017, 14:01
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