## To all you math whizzes, this might be easy, but for me? I’m pretty darn proud I figured it out*

February 20, 2009 by Well Heeled Blog

Circle *O (see below)* has center *O, *diameter *AB *and a radius of 6. Line *CD* is parallel to the diameter.

What is the perimeter of the shaded region?

The perimeter will be the length of the *arc CAE* and the* line segments CB* and* EB*.

Calculate *arc CAE*

Step 1: Angle X = 30 degrees, because they are *alternate interior angles*

Step 2: Angle CBE = 2X = 2*30 degrees = 60 degrees

Step 3: Angle COE = 120 degrees because COE is a *corresponding angle* of the *inscribed angle *CBE. Corresponding angles = 2x inscribed angle’s measure

Step 4: arc = (120 degrees / 360 degrees) * circumference = 1/3 * 2(radius)*pi* = 1/3 * (2)(6)*pi* = 4*pi*

Calculate* lines CB *and *EB*

Step 1: Triangle ACB is a 30-60-90 triangle because one it’s in a semicircle. Angle C = 90 degrees.

Step 2: 30-60-90 triangle = x – rad3(x) – 2x

Step 3: *line segment CB* is opposite 60 degrees, so the length is radical3(x) = radical3(6)

Step 4: *line segment EB* is the same as CB, so EB = radical3(6)

Add them together:

perimeter of shaded region = 4*pi* + radical3(6) + radical3(6)

perimeter of shaded region = 4*pi* + 12 radical3

*after the timed practice test & looked at my notes, but still. I RULE (for one, brief, shining moment).

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on February 20, 2009 at 9:24 pm |nysdelightI have forgotten how cool math was!! If you ask me know I’m clueless??? Great job! I love those shinning moments as well!

on February 20, 2009 at 9:28 pm |ChantalleIck. This is going to give me nightmares!

on February 21, 2009 at 11:17 am |frugalCPAVery nice.

on February 21, 2009 at 4:14 pm |JonathanI got tired after figuring out 4pi. What class is this for?

on February 22, 2009 at 11:35 pm |Paragon2PiecesLooks like studying for the GMAT is going well π

on February 22, 2009 at 11:57 pm |CDWH,

Here is an easier way.

1-Draw two lines between OC and OE. The OAC and OAE are 60 each. 120 at the total.

2-Total perimeter = 2Pr => 2P6 * 120/360 = 4P

3-Draw a line between AC. The ACB is 90, since it can see the diameter. (30,60,90 triangle) If AB = 12, then AC = 6 rad3. Since there are to identical ones, 12 rad3.

4- At the total 4P + 12rad3

Try to time yourself against computer based test. Time is very crucial at GMAT quantitative.

good luck

on February 24, 2009 at 10:18 pm |Serendipityyou lost me at perimeter. but I’m proud of you for getting it! π