Consider that the area of a triangle with base (b) and height (h) is given by,
![\text{Area of triangle}=(1)/(2)* b* h](https://img.qammunity.org/2023/formulas/mathematics/college/4ygd92xa7l1ununrx89gshhm83gdktdyu6.png)
According to the given problem,
![\begin{gathered} b=56\text{ m} \\ h=56\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6wp5g1v2pexzkphdzycc1w2vwnaes4jst9.png)
So the area of the triangle becomes,
![\begin{gathered} A_T=(1)/(2)*56*56 \\ A_T=1568 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9szo08hjhonvwh4ae3s3uvxkw1pwyt1xdy.png)
Consider that the area of the circle with diameter (d) is given by,
![A_C=(\pi)/(4)d^2](https://img.qammunity.org/2023/formulas/mathematics/college/wo0vesm5xkm7drk3r01ag8n6g7pz9la28b.png)
According to the given problem,
![d=20\text{ m}](https://img.qammunity.org/2023/formulas/mathematics/college/77s0zkei1qhu5affcls0x0d4hrq3xzfrwh.png)
Then the area of the circle becomes,
![\begin{gathered} A_C=(\pi)/(4)(20)^2 \\ A_C=100\pi \\ A_C\approx314.16 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cra00rfbt955wh5ki3fww5sembv0r5ld4o.png)
Now, the area of the shaded region (A) is calculated as,
![\begin{gathered} A=A_T-A_C \\ A=1568-314.16 \\ A=1253.84 \\ A\approx1254 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jiqv8som0lv737inwki09wwegtvczsd5xj.png)
Thus, the area of the shaded region is 1254 sq. meters approximately.
Therefore, the 3rd option is correct choice.