In the figure above, point O is the center of the circle

Question

photo.JPG In the figure above, point O is the center of the circle and line segment BD is tangent to the to the circle at point C. If BC = 4, OB = 8, and OC = CD, then what is the area of the region whose perimeter is radius OA, arc ACE, and radius OE? A. 8π B. 10π C. 12π D. 16π E. 20π

DoItRight · Answer

My question is how did you know where to put point A and point E? Was the image included in the problem or did you draw it?

Angle OCD is 90, and given OC=CD, we know angle COD and CDO =45 degrees. 
OB=8, CB=4, and properities of the special 30, 60, 90 triangle. x, 2x,  x\sqrt{3} , we find out the radius is  4\sqrt{3} . 
We also know angle COB, which is opposite of x in the triangle = 30 degrees
Radius=  4\sqrt{3}  Area=48pie
Angle AOE =45+30=75 degree

\frac{75}{360} *48pie=10pie

alex1233 · Answer

It was part of the question I did not draw it

EMPOWERgmatRichC · Answer

Hi All,

The GMAT is based heavily on patterns, so building up your 'pattern-matching' skills is a valuable part of your training. When complex-looking questions appear, they are almost certainly going to be based on a series of overlapping patterns (since it's difficult to make a question complex by accident), so you should be on the lookout for "little" patterns, then think about how they 'connect' to one another.

Here, the first rule that you need to know is that lines that are TANGENT to a circle always form 90 degree angles. This means that triangles OBC and ODC are both RIGHT triangles.

With triangle OBC, we're given two of the sides: one of the legs is 4 and the hypotenuse is 8. You should be thinking....."what type of right triangle has a hypotenuse that is exactly DOUBLE one of its legs.....?" The pattern is that it's a 30/60/90 right triangle.

Next, with triangle ODC, we're told that the two legs of that right triangle are equal. What type of right triangle has two legs that are EQUAL....? The pattern is that it's an ISOSCELES right triangle, so we're dealing with a 45/45/90 right triangle.

From here, it's just a few more steps to figure out the central angle of the circle and the sector area of that piece of circle.

As you continue to study, remember that you're not expected to do every step of a question 'all at once.' Break prompts into small pieces, look for patterns and do the work on the pad.

GMAT assassins aren't born, they're made,
Rich

Senthil7 · Answer

OB = 8 = 2*4
CB = 4
Using pythagoras therom we can deduce OC = 4√3
As we can see this triangle follows 30-60-90 Right angled triangle hence angle COB = 30 degrees.
We know from the question OC = DC hence 
Computing OD will be 4√6 which means triangle ODC is a 45-45-90 right triangle hence angle DOC = 45degrees.
Finally, OE=OC=OA = 4√3 as all are the radii of the circle.
The area covering OE,OC,OA arcs EC and CA is to be calculated using the proportion of central angle as follows:
 * π*r^(2 ) => * π*(4√3)^2=>10π

Divyadisha · Answer

OB= 8, BC = 4

Since DB is tangent on point C, it makes 90 degrees angle at C

So OC = \sqrt{OB^2 - CB^2}= 4\sqrt{3}

Sides are in ratio 1:2:\sqrt{3}, Hence angle COB= 30 degrees

OC=DC; that means ODC is an isosceles triangle with angle DOC= 45 degrees

Angle DOB= 30+45= 75 degrees
Radius = OC= 4\sqrt{3}

Area= 75/360 *π \sqrt{3}^2= 10π

B is the answer

exc4libur · Answer

Hey  EMPOWERgmatRichC

If I didn't recognize that the sides of the triangle were in proportion 30:60:90=x:x√3:2x, then how could we proceed?

Thanks.

EMPOWERgmatRichC · Answer

Hi exc4libur,

Even if you did not recognize the 30/60/90 triangle, you still have 2 sides of a right triangle - and any time a GMAT question gives you that type of information, then it's almost a certainty that you will have to figure out the 3rd side of the triangle (which you can still do with the Pythagorean Theorem: X^2 + Y^2 = Z^2).

GMAT assassins aren't born, they're made,
Rich

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

In the figure above, point O is the center of the circle

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In the figure above, point O is the center of the circle

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards