Radius of circle center O is 3 times the radius of circle center C.

Question

https://gmatclub.com/forum/download/file.php?id=30223 Radius of circle center O is 3 times the radius of circle center C. Angle ACB = Angle POQ. If the shaded area of circle C is 2 then what is the area of the shaded part of circle O ? A. 6 B. 12 C. 18 D. 36 E. 3/2 2016-05-13_1307.png

Kurtosis · Answer

Radius of circle with center O = 3*(Radius of circle with center C)
Area of circle with center O = 9*(Area of circle with center C)

Since the central angle is the same, Area of shaded region in Circle with center O = 9 * (Area of shaded region in circle with center C) = 9 * 2 = 18

Answer: C

Awaited · Answer

Area of the sector = (θ/360*) * π r2

So for C : (θ/360*) * π r2 = 2

r2 = (2/pi)*6 = 12/pi

Therefore r = 2 root 3 /Pi

Now for O : r = 3*2 root 3 /Pi = 6root3/pi

Area for O = 1/6 * pi(6root3/pi)2 = 1/6* 36 * 3 = 18

Hence Answer is C

generis · Answer

1. New area = old area * (scale factor) ^2

2. Call the scale factor K.*

3. Radius of Circle O is 3 times the radius of Circle C.  \frac{3r}{r} = 3. Scale factor K = 3

4. The scale factor for the sector areas is K ^2 , or 3 ^2 = 9

5. Circle C's sector area is 2. Circle O sector area?
 
New area = old area * K ^2

2 * 9 = 18

Answer C

I tested values of sector angles = 60, Circle C radius = 6, Circle O radius therefore = 18 to be positive that 9 was the multiplier/scale factor. 
With those values you get area of sector CAB = 6 \pi  and area of sector OPQ = 54 \pi . Correct.

*For a change from one length to another length, multiply by K.
For a change from one area to another  area, multiply by K ^2   (because area is length * length)
For a change from one volume to another volume, multiply by K ^3  (because volume is length * length * length)

leeum · Answer

Another way to do this one:

Solve total area for both circles

Tiny circle = 4pi
Big circle = 36pi

Now lets just assume that the shaded region is an arbitrary amount of space of the overall circle (and since we know the interior angle is the same, we can assume that the % taken up by each region is the same % for both circles).

If the region is 10% of the circle, then it equals

Tiny circle shaded = .1 * 4pi = .4pi
Big circle shaded = .1 * 36pi = 3.6pi

This tells us two things, 1) the shaded region can't be bigger than 36pi since that is the overall area (so throw away answer D). 2) The regions are different by a factor of 9 (.4 * 9 = 3.6), so, the answer should also be able to divide by 9 evenly. THe only one that can do this is 18, or C.

Answer C

src86 · Answer

R=3.r
x= angle
Area of small circle(A1) = (x/360)*pi.r^2
Area of big circle (A2) = (x/360)*pi.R^2

A1/A2=1/9

A2 = 9.A1 = 18(C)

bumpbot · Answer

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Radius of circle center O is 3 times the radius of circle center C.

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Radius of circle center O is 3 times the radius of circle center C.

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