Answer:
![144\pi\ un^2.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kyhz043mifwb3m6s1hd5hak67xij8stnv2.png)
Explanation:
In the attached diagram, circle R is shown. Line segments QR and SR are radii and QR = SR = 18 units.
The measure of the central angle QRS is
![(8\pi)/(9)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/twbyi3vn7pu7t04opkaeaa2of182o54zbx.png)
1. Find the area of the whole circle:
![A_(circle)=\pi r^2=\pi \cdot 18^2=324\pi \ un^2.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1iwj3zzyl1vhprym8fnh03ccavcr8mb9h7.png)
2. Note that the whole circle is determined by the full rotation angle with measure
radians. So,
![\begin{array}{cc}\text{Angle}&\text{Area}\\ \\2\pi &324\pi \\ \\(8\pi )/(9)&A_(sector)\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9kyr7utzi55t2asj6nu5aezv5nsxwhq675.png)
So, write a proportion:
![(2\pi)/((8\pi)/(9))=(324\pi)/(A_(sector))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9t0vi09u1lro2sq7y29yxzz24dvzmbqbdq.png)
Cross multiply
![2\pi \cdot A_(sector)=324\pi \cdot (8\pi )/(9)\\ \\A_(sector)=162\pi \cdot (8)/(9)=144\pi\ un^2.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t8b7jcr64621i37p0snlv7o8od2upqlv76.png)