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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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A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

70% (02:38) correct
30% (01:36) wrong based on 193 sessions

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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+\(\pi\))

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 \(16-(4+2\p)=2(6-\pi)\).

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 1689 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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