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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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Joined: 02 Sep 2009
Posts: 53063
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, 22:33
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55% (hard)

Question Stats:

58% (02:53) correct 42% (01:54) wrong based on 53 sessions

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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) $$4+4\pi$$

Attachment:

Geo.JPG [ 6.7 KiB | Viewed 967 times ]

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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, 14:25
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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) $$4+4\pi$$

Attachment:

Geoedit.JPG [ 26.87 KiB | Viewed 143 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
$$4\pi - (\pi - 2) - (\pi - 2) =$$
$$4\pi - \pi + 2 -\pi + 2 =$$
$$2\pi + 4$$

D
Attachments

Geoedit.JPG [ 26.87 KiB | Viewed 670 times ]

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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, 14:25
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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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