A cow is tethered to one corner of a

square barn, 10 feet by 10 feet, with a rope 100 feet long. What is the maximum grazing area for the cow?

We are given a diagram,

Area A₂ consists of two equal sectors with radius 90ft and central angle is given by $\delta$

Now we find $\delta$

We get

role="math" localid="1646790759664" $$\begin{gathered} \delta +40.5^o={90}^o \\ \delta =49.5^o \end{gathered}$$

Now we find the area A₂by doubling the area of one of the sectors

role="math" localid="1646790934563" $$\begin{gathered} A_2=2\left(\frac{\delta }{360}\right)r^2\pi \\ {A}_2=2\frac{49.5^o}{{360}^o}{\left({90}^o\right)}^2\pi \\ {A}_2=2\left(\frac{11}{80}\right)8100\pi \\ {A}_2=6997.9f{t}^2 \end{gathered}$$

we have,

Area A₃is divided into two equal triangles dashed by a line

We find area of one triangle and then double it

First find the measure of angle $\alpha$ which completes a straight line with 45^o angle

We have,

$$\begin{gathered} \alpha +{45}^o={180}^o \\ \alpha ={135}^o \end{gathered}$$

Now we find the values of $\beta ,\gamma$

$$\begin{gathered} \frac{\sin \gamma }{10}=\frac{\sin 135}{90} \\ \sin \gamma =\frac{\sin 135}{90}\left(10\right) \\ \sin \gamma =0.0785 \\ \gamma =4.5^o \end{gathered}$$

Now

$$\begin{gathered} \alpha +\beta +\gamma ={180}^o \\ \beta ={180}^o-{135}^o-4.5^o \\ \beta =40.5^o \end{gathered}$$

And finally the area can be given as

$$\begin{gathered} A_3=2\left(\frac{1}{2}\right)\left(10\right)\left(90\right)(\sin 40.5) \\ {A}_3=584.5f{t}^2 \end{gathered}$$

Area A₁Consists of $\frac{3}{4}th$of the circle

Hence

$$\begin{gathered} A_1=\frac{3}{4}\left(\pi r^2\right) \\ {A}_1=\frac{3}{4}\left(\pi {100}^2\right) \\ {A}_1==23561.945f{t}^2 \end{gathered}$$

Now find the total area

$$\begin{gathered} A=A_1+A_2+A_3 \\ A=23561.945+6997.9+584.5 \\ A=31144.345f{t}^2 \end{gathered}$$