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

Question

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

eragotte · Answer

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

DeeptiM · Answer

hmmmm but why did you minus 1/2??

eragotte · Answer

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

fluke · Answer

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

riangle 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{	heta}{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"

manishgeorge · Answer

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.

jamifahad · Answer

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.

DeeptiM · Answer

Awsum guys, thnx a ton for the explanation..

mission2009 · Answer

Nice explanation fluke..

amit2k9 · Answer

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

hence area = 4.

now area of left and right triangles = 2*   * 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

anordinaryguy · Answer

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

bumpbot · Answer

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

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