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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] New post 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] New post 11 Feb 2012, 00:14
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nafishasan60 wrote:
Image

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).

Answer: D.

Hope it's clear.
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Re: H,G,F and E are midpoints of the sides of square ABCD [#permalink] New post 06 Mar 2014, 01:34
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Re: H,G,F and E are midpoints of the sides of square ABCD [#permalink] New post 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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Re: H,G,F and E are midpoints of the sides of square ABCD   [#permalink] 07 Mar 2014, 00:15
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