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:

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 areaConnect 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 regionsIf 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 areaThere 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

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

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"