Question about area of circle word problem Picture of problem 
Please refer to the picture of the problem
I am trying to figure out: What percent of the total area is worth ten points?
So this is my attempt: the area of the 10 point region is pi * (4)^2 = 50.24
the total radius is 4+ 3+3+3 = 13
so the total area is 13^2 * pi = 530.66
So the percent is 50.23/530.66 = 10%
However, the answer is 30% which I do not get.
 A: You forgot to include the 10-point regions in the outer ring.  Their combined area is $\pi\dfrac{13^2-10^2}2\approx108$.  Add that to the area of the bull’s-eye to get the total area worth $10$ points.
A: there's three areas worth $10$ points.
The bullseye has area $\pi r_1^2 = \pi 4^2 = 16\pi$.
The area of the entire dart board has area $\pi r_2^2 = \pi (4+3*3)^2= 13^2\pi =169\pi$
The area of the three inner rings is $\pi r_3^2 = \pi (4+2*3)=10^2\pi =100\pi$.
So the area the $4$th ring only is the total area of the dart board minus the three inner rings, which is $169\pi -100\pi = 69\pi$.
The areas worth $10$ points are $\frac 14$  of that ring so area of those areas  worth $10$ are are each $\frac {69}4 \pi$.
So the total area worth $10$ point if we add them up is $16\pi + \frac {69}4\pi + \frac {69}4\pi = 50.5 \pi$
So the percentage of the dart board is $\frac {50.5\pi}{169\pi} =\frac {101}{338}\approx 30\%$
....
Now if I wanted to be a smart-aff I could note the radius of the dart board is $13$ inches.  The bullseye has radius of $4$ inches or $\frac 4{13}$ of the radius.
So the bullseye has $(\frac 4{13})^2$ of the area.
And the last ring has $\frac 3{13}$ of the radius.  So it has $(\frac 3{13})^2$ of the area.  And the parts worth ten points are half of that so,
The areas worth ten points are $(\frac 4{13})^2 + \frac{(\frac 3{13})^2}2 \approx 30\%$
