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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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13 Dec 2017, 23:33
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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 [ 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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14 Dec 2017, 15:25
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Bunuel wrote:

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 [ 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

*I stand ready to be corrected; about this issue it is likely.
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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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