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"

Attachments

Square_Inscribed_In_Circle.JPG [ 17.9 KiB | Viewed 4544 times ]

_________________

~fluke

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