#### Answer

The new area is twice the original area .
In general, this result holds for any values of $\theta$ and $r$. (refer to the explanation below)

#### Work Step by Step

RECALL:
The area of a sector $(A)$ intercepted by a central angle $\theta$ on a circle whose radius is $r$ is given by the formula:
$A = \frac{1}{2}r^2\theta$, where $\theta$ is in radian measure.
From Number 109, the area of the sector is $\color{blue}{A \approx 8,060 \space yd^2}$.
If the angle is halved ($\frac{\pi}{9}$ and the radius length is doubled ($304 \text{yd}$), the area of the sector becomes:
$A=\frac{1}{2}r^2\theta
\\A=\frac{1}{2}(304^2)(\frac{\pi}{9})
\\\color{blue}{A \approx 16,130 \space yd^2}$
This is twice the area of the sector in Number 109.
This result hold in general for any values of $\theta$ and $r$ because if the radius is doubled and the angle is halved, then:
Area of Original Sector: $A = \frac{1}{2}r^2\theta$
Area of New Sector: $A=\frac{1}{2}(2r)^2\cdot \frac{\theta}{2}=\frac{1}{2}(4r^2)(\frac{\theta}{2})=r^2\theta$
Notice that the area of the new sector is twice the area of the original sector.