Chapter 8.4, Problem 46AYU

To determine

To find: The maximum grazing area of the cow.

Answer to Problem 46AYU

The total area that the cow grazes is $30857{\text{ft}}^2$ .

Explanation of Solution

Given:

The given diagram is shown in Figure

  

Figure

The length of the rope is $100\text{ft}$ .

Calculation:

Consider the cow grazes $\frac{3}{4}$ part of the whole area of the circle. Then the total area formed by the arc is,

  $TCU=\frac{3}{4}\pi r^2$

Then,

  $\begin{array}{c}TCU=\frac{3}{4}\pi {\left(100\right)}^2\\ =23561.93{\text{ft}}^{\text{2}}\end{array}$

Consider the barn is 10 ft by 10 ft square so at the corners the rope will remain 10 ft shorter or 90 ft only. As the cow comes from both the sides the arc of the paths overlap at point U as shown in Figure. The joining of the intersecting point to the corners of the barn forms the triangle PUR.

Consider the right angle triangle PQR by Pythagoras theorem is,

  $\begin{array}{c}P{R}^2=P{Q}^2+Q{R}^2\\ ={10}^2+{20}^2\\ =22.36\end{array}$

Consider the semi perimeter of the triangle is,

  $\begin{array}{c}s=\frac{22.36+90+80}{2}\\ =96.18\end{array}$

The area of the triangle by Heron’s formula is,

  $\begin{array}{c}A=\sqrt{96.18\left(96.18-90\right)\left(96.18-80\right)\left(96.18-22.36\right)}\\ =842.58{\text{ft}}^2\end{array}$

Consider the area of the gazing part inside is its difference of the area of the half barn inside the triangle that is,

  $842.58{\text{ft}}^2-\frac{20\times 10}{2}=742.58$

To determine the angle $B$ consider the law of cosine as,

` $\begin{array}{l}\cos B=\frac{{90}^2+{22.36}^2-{80}^2}{2\left(90\right)\left(22.36\right)}\\ \cos B=0.5466\\ B=56.8657°\end{array}$

Consider by angle sum property,

  $\begin{array}{l}45°+56.8657°+\angle TPS=180°\\ \angle TPS=78.1343°\end{array}$

To determine the angle $A$ consider the law of cosine as,

` $\begin{array}{l}\cos A=\frac{{80}^2+{22.36}^2-{90}^2}{2\left(80\right)\left(22.36\right)}\\ A=116.5665°\end{array}$

Consider by angle sum property,

  $\begin{array}{l}45°+116.5665°+\angle SRU=180°\\ \angle SRU=18.4335°\end{array}$

Then, the area of the sector TPS is,

  $\begin{array}{c}A=\frac{3.14159\left(90\right)\left(90\right)\left(78.1343°\right)}{360°}\\ =5522.98{\text{ft}}^2\end{array}$

Then, the area of the sector SRU is,

  $\begin{array}{c}A=\frac{3.14159\left(80\right)\left(80\right)\left(18.4335°\right)}{360°}\\ =102.52{\text{ft}}^2\end{array}$

Then, the total area that the cow grazes is,

  $\begin{array}{c}A=102.52{\text{ft}}^2+5522.98{\text{ft}}^2+742.58{\text{ft}}^2+23561.93{\text{ft}}^2\\ =30857{\text{ft}}^2\end{array}$