Question about area of circle word problem

Question

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.

Answer 1

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.

Answer 2

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 $(\frac 4{13})^2 + \frac{(\frac 3{13})^2}2 \approx 30\%$