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In square ABCD above, if DE = EB and DF = FC, then the area of the sha

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In square ABCD above, if DE = EB and DF = FC, then the area of the sha [#permalink]

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New post 06 Jul 2017, 23:36
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In square ABCD above, if DE = EB and DF = FC, then the area of the shaded region is what fraction of the area of square region ABCD?

(A) 1/16

(B) 1/8

(C) 1/6

(D) 1/4

(E) 1/3


[Reveal] Spoiler:
Attachment:
2017-07-07_1134.png
2017-07-07_1134.png [ 6.71 KiB | Viewed 1628 times ]
[Reveal] Spoiler: OA

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Re: In square ABCD above, if DE = EB and DF = FC, then the area of the sha [#permalink]

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New post 07 Jul 2017, 00:13
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Consider any square. Draw the diagonals. Join the midpoints of the opposite sides.

You will see that there are eight smaller triangles each similar to the other. The area of all these smaller triangles is same.
Lets say the area of one small triangle is x. Thus the sum of the areas of all 8 triangles would be 8x which would also be the area of the square itself .

Triangle EFC's area shown in the figure above is thus 1/8th of the area of the square.

Thus option B would be correct.

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Re: In square ABCD above, if DE = EB and DF = FC, then the area of the sha [#permalink]

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New post 07 Jul 2017, 07:05
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Bunuel wrote:
Image
In square ABCD above, if DE = EB and DF = FC, then the area of the shaded region is what fraction of the area of square region ABCD?

(A) 1/16

(B) 1/8

(C) 1/6

(D) 1/4

(E) 1/3


[Reveal] Spoiler:
Attachment:
2017-07-07_1134.png


Let the side of square be \(= 2\). Therefore, \(AB = BC = CD = AD = 2\)

Area of square \(ABCD = 2^2 = 4\)

Given; \(DF = FC\). Therefore; \(DF = FC = 1\)

Diagonal of square \(= (side) \sqrt{2}\). Therefore, \(BD = 2\sqrt{2}\)

\(DE = EB = \frac{(2\sqrt{2})}{ 2}\) \(=> DE = EB = \sqrt{2}\)

Joining \(EC\) to \(A\), we get \(AC\) as another diagonal. Therefore, \(EC = DE = EB\). Hence \(EC = \sqrt{2}\)

Using Pythagorean theorem in \(\triangle EFC\), we get;

\(EF^2 + FC^2 = EC^2\)

\(EF^2 = EC^2 - FC^2\)

\(EF^2 =\)\(\sqrt{2^2}\) \(- 1^2\)

\(EF^2 = 2 - 1 = 1\)

\(EF = \sqrt{1} = 1\)

Shaded region is \(\triangle EFC\), Which is right angled triangle.

Area of shaded region \(= \frac{1}{2} EF * FC = \frac{1}{2} * 1*1 = \frac{1}{2}\)

Required fraction = Area of the shaded region / Area of square

Required fraction \(= \frac{(1/2)}{4} = \frac{1}{8}\)

Answer (B)...

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Re: In square ABCD above, if DE = EB and DF = FC, then the area of the sha [#permalink]

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New post 09 Jul 2017, 19:49
B is the correct answer as Area of square is Xsquare and area of triangle is xsquare/8 so fraction will be 1/8.


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Re: In square ABCD above, if DE = EB and DF = FC, then the area of the sha [#permalink]

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New post 09 Jul 2017, 20:11
We are given that E is the mid point of diagonal DB.

Now lets say area of square ABCD is 1.

Then, area of triangle DCB = 1/2 of full square = 1/2
Now, since E is mid point of DB, area of triangle CED = 1/2 of triangle DCB = 1/4
Also, since F is mid point of CD, area of triangle EFC (Shaded region) = 1/2 of triangle CED = 1/8

Thus shaded area is 1/8 of total square's area. Hence B answer.

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Re: In square ABCD above, if DE = EB and DF = FC, then the area of the sha [#permalink]

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New post 13 Nov 2017, 05:28
Bunuel wrote:
Image
In square ABCD above, if DE = EB and DF = FC, then the area of the shaded region is what fraction of the area of square region ABCD?

(A) 1/16

(B) 1/8

(C) 1/6

(D) 1/4

(E) 1/3


[Reveal] Spoiler:
Attachment:
2017-07-07_1134.png


I solved this question using:
- 45-45-90 triangle

EF is perpendicular to DC and thus angle DFC is 90.

In triangle BDC, BC= DC (sides of a square). Therefore, angle DBC is 45.

Now triangle DFE is an isosceles triangle in which DF=FE. So sides are in ratio 1:1:root 2

Area of triangle (shaded)= 1/2 *x^2
Area of square= 4x^2

The answer is 1/8

Kudos [?]: 21 [0], given: 275

Re: In square ABCD above, if DE = EB and DF = FC, then the area of the sha   [#permalink] 13 Nov 2017, 05:28
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In square ABCD above, if DE = EB and DF = FC, then the area of the sha

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