91Ó°ÊÓ

The barn is rectangular by $10×20feet$

Consider the figure:

The maximum area = Area of circle OBDO + Area of triangle OAC+ Area of sector ACB + Area of sector OCD -Area of rectangle OEFA

The rope is $100feet,sor=100$

Area of the region role="math">OBD)=34Ï€(100)2=23,571.429

$$\begin{gathered} \tan ∠AOF=\frac{20}{10} \\ ∠AOF={\tan }^{-1}\left(2\right) \\ =63.435° \end{gathered}$$

$$\begin{gathered} AC=100-OA \\ =100-10 \\ =90 \end{gathered}$$

$∠AOC=∠AOF=63.435°$

By applying Law of sines,

$$\begin{gathered} \frac{\sin ∠COA}{AC}=\frac{\sin ∠OAC}{OC}=\frac{\sin ∠ACO}{OA} \\ \frac{\sin 63.435°}{90}=\frac{\sin ∠OAC}{OC}=\frac{\sin ∠ACO}{10} \end{gathered}$$

Combine first and last,

$$\begin{gathered} \sin ∠ACO=\frac{63.435°}{9} \\ ∠ACO={\sin }^{-1}(0.0993808) \\ ∠ACO=5.703° \end{gathered}$$

By Interior angle property of triangles,

$$\begin{gathered} ∠AOC+∠OAC+∠OCA=180° \\ ∠OAC=180°-63.435°-5.703° \\ ∠OAC=110.862° \end{gathered}$$

Area of $∆OAC=\frac{1}{2}\left(OA\right)\left(AC\right)\sin ∠OAC$
$$\begin{gathered} =\frac{1}{2}\left(10\right)\left(90\right)\sin 110.862° \\ =420.4984 \end{gathered}$$

$$\begin{gathered} ∠OAC+∠CAB=180° \\ ∠CAB=180°-110.862° \\ =69.138° \end{gathered}$$

Area of the sector $ACB=\frac{angle}{360°}×{\mathrm{Ï€°ù}}^2$
localid="1647440938685">=69.138°360°×π(90)2=4889.044

$$\begin{gathered} ∠COD=90°-∠AOF \\ =90°-63.435° \\ =26.565° \end{gathered}$$

Area of the sector $=\frac{26.565°}{360°}×\mathrm{π}{\left(100\right)}^2$
$=2319.1667$

Area of rectangle$=20×10$

$=200$

Area grazed by the cow$=23571.429+420.4984+4889.044+2319.667-200$

$=31,000.14$