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In the figure, AB is parallel to CD and AB = CD =2*2^(1/2). If the

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In the figure, AB is parallel to CD and AB = CD =2*2^(1/2). If the [#permalink]

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In the figure, AB is parallel to CD and \(AB=CD=2 \sqrt{2}\). If the distance between AB and CD is \(2\sqrt{2}\), what is the area of the shaded region?

(A) \(2+\pi\)

(B) \(2+2\pi\)

(C) \(4+\pi\)

(D) \(4+2\pi\)

(E) \(4+4\pi\)


[Reveal] Spoiler:
Attachment:
Geo.JPG
Geo.JPG [ 6.7 KiB | Viewed 723 times ]
[Reveal] Spoiler: OA

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In the figure, AB is parallel to CD and AB = CD =2*2^(1/2). If the [#permalink]

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New post 14 Dec 2017, 15:25
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Bunuel wrote:
Image
In the figure, AB is parallel to CD and \(AB=CD=2 \sqrt{2}\). If the distance between AB and CD is \(2\sqrt{2}\), what is the area of the shaded region?

(A) \(2+\pi\)

(B) \(2+2\pi\)

(C) \(4+\pi\)

(D) \(4+2\pi\)

(E) \(4+4\pi\)


[Reveal] Spoiler:
Attachment:
The attachment Geo.JPG is no longer available

Attachment:
Geoedit.JPG
Geoedit.JPG [ 26.87 KiB | Viewed 451 times ]

The algebra here is beastly, not the geometry.
I solved the first time in a more traditional way. My answer had subtraction.
I realized that the only way to get a plus sign was to include subtracting another subtraction.

So: Area of shaded = (area of circle) - (area of two unshaded regions)

Find the circle's area

Connect A to C, and B to D.
That is a square with four equal sides whose length = \(2\sqrt{2}\)
Given: AB = CD = \(2\sqrt{2}\)
Given: distance between AB and CD [which are parallel] is \(2\sqrt{2}\), so side AC = BD = \(2\sqrt{2}\)

The diagonal of the square:
\(s\sqrt{2}=(2\sqrt{2}*\sqrt{2})=4\)

Diagonal = diameter of circle, \(d\)
\(4 = d = 2r\)
\(r = 2\)

Area of circle: \(\pi r^2 = 4\pi\)

Find the area of the UNshaded regions
If the shaded area were just the square:
Each unshaded region would have an area calculated by (circle area - square area)/4
Area of the drawn square ABCD = \(s^2 = (2\sqrt{2})^2 = 8\)
\(\frac{4\pi - 8}{4} = (\pi - 2)\) = area of each unshaded region

OR
The square's diagonals are perpendicular bisectors, such that
Sectors' central angles are 90
Each sector's total area \(= \pi\)
-- \(\frac{90}{360} = \frac{1}{4} =\frac{?}{4\pi} =\)
\(\pi =\) sector area

The area of the triangle in each sector (where sides = radii = 2) is \(\frac{2 * 2}{2} = 2\)
So an UNshaded region in a sector would have area calculated by (Sector area - triangle area), namely, \((\pi - 2)\)

Find the shaded area

There are two sectors that contain unshaded regions. The area of each UNshaded region \(= (\pi - 2)\).
There are two sector areas, fully shaded, whose shaded area is simply \(\frac{1}{4}\) of the circle's area: \(\pi\).
Ignore the two fully shaded sectors (because if you do not, as far as I can tell, your answers will look nothing like the choices*).

To find the shaded area, subtract the area of the unshaded regions from the whole circle's area.
In other words, use (circle area) - (area of unshaded region in 2 of the sectors)

(Total circle area) - (area of TWO unshaded regions) = shaded area
\(4\pi - (\pi - 2) - (\pi - 2) =\)
\(4\pi - \pi + 2 -\pi + 2 =\)
\(2\pi + 4\)


Answer
[Reveal] Spoiler:
D


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In the figure, AB is parallel to CD and AB = CD =2*2^(1/2). If the   [#permalink] 14 Dec 2017, 15:25
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In the figure, AB is parallel to CD and AB = CD =2*2^(1/2). If the

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