Find a fraction of a part that is shaded 
In the figure , O is the centre of the two circles . The circles are divided into sectors of equal sizes. Given that the area of the shaded portion A is twice of the area of the shaded portion B 
What is the fraction of the figure that is shaded ? 
I'm not sure how to solve this problem..
Thanks in advance !
 A: Common sense.  The circle consists of 8 arches the size of of A.  And 8 wedges the size of the 2 that make of B.  
We are told an Arch is twice as big as 2 wedges so an arch is the size of 4 wedges.  
So the entire circle of 8 arches and 8 wedges is the size of 32 + 8 = 40 wedges.  
1 arch and 2 wedges = 6 wedges are shaded.
So 6/40 = 3/20 of the circle is shaded.
A: Without loss of generality let the inner radius be $1$ and the outer radius be $r$. The area of $B$ is $\frac{\pi}{4}$ and the area of $A$ is $\frac{\pi (r^2-1)}{8}$. The area of $A$ is twice that of $B$ so:
$$2\times\frac{\pi}{4}=\frac{\pi (r^2-1)}{8}$$
$$4=r^2-1$$
$$r^2=5$$
$$r=\sqrt{5}$$
So the area of the entire shape is $5\pi$ and the area of the shaded parts is $\frac{\pi}{4}+\frac{\pi}{2}=\frac{3\pi}{4}$. So the shaded fraction is:
$$\frac{\frac{3\pi}{4}}{5\pi}=\frac{3}{20}$$
A: Okay, I'm not sure how much algebra you've had, so I'll try to keep algebraic equations out of it. 
First we need to know the total area in terms of A and B
We can see that there are four middle sections the size of B. Also, there are 8 outer sections the size of A. So the circle is made up of 8 A's and 4 B's.
Since A's are the same as 2 B's, we can change that to say there are 8 * 2 = 16 B's on the outside and 4 B's on the inside. So the circle is made up of 16 + 4 = 20 B's. That's our denominator.
Now the shaded section is A + B, but since we know A is just 2 B's, we can see that it's equal to 3 B's. That's our numerator.
Put those together and you have your fraction.
