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In the figure, AB is parallel to CD and AB=CD=2 2 . If the

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In the figure, AB is parallel to CD and AB=CD=2 2 . If the  [#permalink]

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New post 15 Aug 2011, 05:30
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A
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C
D
E

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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+2pi
(C) 4+pi
(D) 4+2pi
(E) 4+4pi

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Re: Geometry not again!!  [#permalink]

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New post 15 Aug 2011, 05:46
Find length AD with pythagorean theorem.

(2root2)^2 + (2root2)^2 = AD^2

AD = 4

Area full circle = 4pi

Area full square = (2root2)*(2root2)= 8

Now area of the shaded region is the area of the full circle, minus 1/2 the difference between the area of the circle and the area of the inscribed square.


4pi - 1/2(4pi-8) = 4pi-2pi+4 = 2pi + 4 = D
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Re: Geometry not again!!  [#permalink]

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New post 15 Aug 2011, 06:04
eragotte wrote:
Find length AD with pythagorean theorem.

(2root2)^2 + (2root2)^2 = AD^2

AD = 4

Area full circle = 4pi

Area full square = (2root2)*(2root2)= 8

Now area of the shaded region is the area of the full circle, minus 1/2 the difference between the area of the circle and the area of the inscribed square.


4pi - 1/2(4pi-8) = 4pi-2pi+4 = 2pi + 4 = D


hmmmm but why did you minus 1/2??
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Re: Geometry not again!!  [#permalink]

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New post 15 Aug 2011, 06:10
DeeptiM wrote:
eragotte wrote:
Find length AD with pythagorean theorem.

(2root2)^2 + (2root2)^2 = AD^2

AD = 4

Area full circle = 4pi

Area full square = (2root2)*(2root2)= 8

Now area of the shaded region is the area of the full circle, minus 1/2 the difference between the area of the circle and the area of the inscribed square.


4pi - 1/2(4pi-8) = 4pi-2pi+4 = 2pi + 4 = D


hmmmm but why did you minus 1/2??




I am minusing 1/2 of the difference between the area of the circle and the area of the square. Think about it and you should see the the midpoint between the area of the circle and the area of the square is the area of the shaded region. Not really sure how to better explain it. Another way to do it is area of square + area of circle divided by two. Actually that is probably easier.

(4pi + 8)/2 = 2pi + 4
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Re: Geometry not again!!  [#permalink]

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New post 15 Aug 2011, 06:37
1
DeeptiM 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+2pi
(C) 4+pi
(D) 4+2pi
(E) 4+4pi

Attachment:
The attachment geometry_why_not.JPG is no longer available


Please see the attached picture.

\(ABCD\) is a square with \(Side=2\sqrt{2}\)

Area of square\(=(2\sqrt{2})^2=4*2=8\)

Let's say we join the diagonals of the square \(AD\) and \(BC\), intersecting at point O.

The four triangular regions(shown with greenish shades) within the square, viz \(AOB, BOD, DOC, COA\) will have equal areas each of 2 units i.e. 8/4=2

\(\triangle ACD\) is a right-angled triangle.
Using Pythagoras, AD=Diagonal=Diameter=4
Radius=r=4/2=2

Now, in order to find the area of the segment AC(orange area on left), we just need to find the area of the sector \(AOC\) (Dark Orange+ Darkest Green) and subtract the area of triangle \(AOC=2\) (Dark Green)

\(Area \hspace{2} of \hspace{2} sector=\frac{\theta}{360}*\pi*r^2\)

\(Area \hspace{2} of \hspace{2} sector=\frac{90}{360}*\pi*2^2=\pi\) {:Note: theta=AOC=90. Square's diagonals make \(90^{\circ}\) angle at intersection.}

Area of segment= Area of sector-Area of quarter of the square\(= \pi - 2\)

The shaded portion has two segments(Left Orange+Right Orange) and the square(sum of all green regions).

Total Area=2*Area of segment+Area of square\(=2(\pi -2)+8=2\pi-4+8=2\pi+4\)

Ans: "D"
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Re: Geometry not again!!  [#permalink]

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New post 15 Aug 2011, 09:59
From flukes diagram the green and the bleueish green forms one quarter of the area of the whole circle. This is because all the angles are equal.

so calculate the area of the complete circle and divide that value by 4 .

We are now left with 2 isosceles triangles for which we are to find the area. Very straight forward because we have the value of the base and hight.
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Re: Geometry not again!!  [#permalink]

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New post 15 Aug 2011, 12:01
ABCD is a square with side \(2\sqrt{2}\)
BC and AD will be diagonals which are perpendicular bisector at center 0 of circle.
Imagine 0 as center of circle.
Area of sector A0C = Area of sector B0D= Central angle/360 * pi r^2 = 90/360*pi*4=pi
Area of triangle 0CD = Area of triangle 0AB= 1/2 * 2 * 2 =2
Area of shaded region = Area of sector A0C + Area of sector B0D+Area of triangle 0CD + Area of triangle 0AB
= pi+pi+2+2=4+2pi

OA D.
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Re: Geometry not again!!  [#permalink]

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New post 15 Aug 2011, 18:38
Awsum guys, thnx a ton for the explanation..
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Re: Geometry not again!!  [#permalink]

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New post 17 Aug 2011, 02:52
Nice explanation fluke..
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Re: Geometry not again!!  [#permalink]

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New post 17 Aug 2011, 03:06
joining A-D and B-C we have 4 triangles.

area of top and bottom triangles = 2 * 1/2 * (base=2*2^(1/2)) * (altitude = 2^(1/2)

altitude = 2^(1/2) because distance is split equally at the center = [2* 2^(1/2)]/2

hence area = 4.

now area of left and right triangles = 2* [angle/360] * pi * r^2

r^2 = 4 hence r=2 also,
angles for the left triangle = 45-45-90 (center)

thus, 90/360 * pi * 4 * 2 = 2pi

thus 4+ 2pi. D
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Re: Geometry not again!!  [#permalink]

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New post 17 Aug 2011, 09:28
People have already done lot of hard work in explaining the solution, So agreeing to them I would just give my answer & that is D

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Re: In the figure, AB is parallel to CD and AB=CD=2 2 . If the &nbs [#permalink] 02 Nov 2018, 10:10
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