H,G,F and E are midpoints of the sides of square ABCD : GMAT Problem Solving (PS)
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# H,G,F and E are midpoints of the sides of square ABCD

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H,G,F and E are midpoints of the sides of square ABCD [#permalink]

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10 Feb 2012, 23:48
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H,G,F and E are midpoints of the sides of square ABCD. Arcs FG and EH are centered at B and D respectively, as shown above. If the side of the square ABCD is 4, what is the area of the shaded region HEFG?

A) 4(3-$$\pi$$)
B) 2(4-$$\pi$$)
C) 4(4-$$\pi$$)
D) 2(6-$$\pi$$)
E) 8(1+$$\pi$$)
[Reveal] Spoiler: OA
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Re: H,G,F and E are midpoints of the sides of square ABCD [#permalink]

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11 Feb 2012, 00:14
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nafishasan60 wrote:

H,G,F and E are midpoints of the sides of square ABCD. Arcs FG and EH are centered at B and D respectively, as shown above. If the side of the square ABCD is 4, what is the area of the shaded region HEFG?

A) 4(3-$$\pi$$)
B) 2(4-$$\pi$$)
C) 4(4-$$\pi$$)
D) 2(6-$$\pi$$)
E) 8(1+$$\pi$$)

The legs of right isosceles triangles FAE and GCH equal to half of the side of the square, so to 4/2=2, hence their combined area is $$2*(\frac{1}{2}*2*2)=4$$;

The same way the radii of arcs FG and EH equal to 4/2=2. Since both arcs are 90 degrees, then their commbined area is $$2*\frac{90}{360}*\pi*{r^2}=2\pi$$;

The are of the square is 4^2=16, thus the area of the shaded region is $$16-(4+2\p)=2(6-\pi)$$.

Hope it's clear.
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Re: H,G,F and E are midpoints of the sides of square ABCD [#permalink]

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06 Mar 2014, 01:34
Bumping for review and further discussion.

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Re: H,G,F and E are midpoints of the sides of square ABCD [#permalink]

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07 Mar 2014, 00:15
Unshaded region has 2 equal quartercircles (means 1 semicircle) & 2 equal right triangles (Means 1 square)

Area of Semicircle = pie 2^2/2 = 2pi
Area of small Square = 4

Area of shaded region = 16 - 4 - 2pi

= 12 - 2pi
= 2(6-pi) = D
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H,G,F and E are midpoints of the sides of square ABCD [#permalink]

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26 Sep 2015, 04:52
I would ballpark for the right solution. Divide the square into 4 small squares (blue line) and call the midpoint M. From this you see that the shaded part within AFME and MGCH is always half. From this you have each 1/8 = 2/8 of the total square. Lastly also divide EMHD and FBGM further into halfes (red lines). There you see that the shaded region is half of 1/8 and therefore 1/16. So add 2/16 to 2/8 to get 3/8.

Attachment:

Unbenannt.png [ 7.56 KiB | Viewed 1633 times ]

So we are looking for 3/8 of 16 = 6 in the answer choices. D is closest.
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H,G,F and E are midpoints of the sides of square ABCD   [#permalink] 26 Sep 2015, 04:52
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