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

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

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:
[[(75°)/(360°)]* π*r^(2 ) =>[5/24]* π*(4√3)^2=>10π

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

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.

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

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