The shaded portion is a minor sector with diameter = 28m
therefore,
Radius ,r = diameter/2
Radius, r=28m/2=14m
Next, we have to calculate the angle of the sector
The angle of the sector is connected to the 140° on a straight line
Let the angle of the sector be
![\phi](https://img.qammunity.org/2023/formulas/mathematics/college/m5bsf12weu2hzjy1txmi8mzd1rf3kl1rdh.png)
Therefore,
![\begin{gathered} \phi+140^0=180^0 \\ \phi=180^0-140^0 \\ \phi=40^0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ov1xxfk60mpz3m3qdpd1f3ej3bfn8s8abt.png)
Then we can calculate the area of a sector using the formula
![\begin{gathered} \text{Angle of sector=}(\phi)/(360)*\pi r^2 \\ \text{Where,} \\ r=14m \\ \phi=40^0 \\ \pi=3.14(\text{ but the answer has to be in terms of }\pi) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ubwhnfprc5o6sy5tqur5vra90nvht9mo7i.png)
On substitution, we will have the area of the sector as
![\begin{gathered} \text{Area of shaded sector}=(40)/(360)\pi*14^2 \\ \text{Area of shaded sector=}21.78\pi m^2 \\ \text{Area of shaded sector=21.78}\pi\text{m}^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tml9y65l38y5jhtbbgrktxkdctz5l56hqp.png)
Therefore,
The area of the shaded sector in terms of π = 21.78πm²