Feed the goat, help the farmer

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Lets say that the circle with center $A$ (and radius $10m$ ) is the field and the goat is tied at point $B$ .

So, the goat can move around in an area equivalent to the area of the circle with center $B$ . Lets also assume that the length of the rope is $r$ . Hence, the radius of this circle also is $r$ .

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What we'll do is break the required area into two parts (segments) $A_1$ and $A_2$ , find their area, and add them up and finally equate the resulting expression with $\frac{\pi 10^2}{2}$ which is $50 \pi$

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$A_1+A_2=A$ which is the required area.

$\rightarrow$ Consider the following--

$\bullet PQ=2s$

$\bullet OB=p$ , so $OA=(10-p)$ as $AB=10$ (Radius of the field)

Now, applying the Pythagoras theorem on $\triangle BOP$ and $\triangle AOP$ respectively, we get

$\rightarrow r^2=s^2+p^2$ and

$\rightarrow 100=(10-p)^2+s^2$ .

Solving for $s$ and $p$ , we find that

$\rightarrow s=\frac{r}{20} \sqrt{400-r^2}$

and

$\rightarrow p=\frac{r^2}{20}$

Now, lets consider $\triangle BPQ$ and the arc subtended by it.

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Area of the segment( $A_1$ )=(Area of the sector)-(area of the triangle).

Lets call $\angle OBP$ as $\alpha$ .

So, $\cos \alpha=\dfrac{p}{r}=\dfrac{\frac{r^2}{20}}{r}=\dfrac{r}{20}$ .

$\rightarrow \alpha=\cos^{-1} \left(\dfrac{r}{20}\right)$

Hence the area of the sector $=\dfrac{2\alpha}{2\pi}\pi r^2$

$=r^2 \cos^{-1} \left(\dfrac{r}{20}\right)$

Now, area of $\triangle PQB=\frac{1}{2} \times$ base $\times$ height

= $\frac{1}{2} \times 2s \times p$

$=\dfrac{r^3}{400} \sqrt{400-r^2}$

Subtracting the area of the triangle from that of the sector yields the area of the segment $A_1=r^2 \cos^{-1} \left(\dfrac{r}{20}\right)-\dfrac{r^3}{400} \sqrt{400-r^2}$

or simply

$A_1=r^2 \cos^{-1} \left(\dfrac{r}{20}\right)-sp$

Similarly, calculating the area of the other segment $A_2$ yields

$A_2=100 \cos^{-1} \left(\dfrac{200-r^2}{200}\right)-\frac{r}{2} \sqrt{400-r^2}+sp$

So, $A_1+A_2=A=100 \cos^{-1} \left(\dfrac{200-r^2}{200}\right)-\frac{r}{2} \sqrt{400-r^2}+r^2 \cos^{-1} \left(\dfrac{r}{20}\right)$

Now we equate $A$ with half the area of the field, which comes out to be $50 \pi$

So, the final equation stands as follows

$100 \cos^{-1} \left(\dfrac{200-r^2}{200}\right)-\frac{r}{2} \sqrt{400-r^2}+r^2 \cos^{-1} \left(\dfrac{r}{20}\right)=50\pi$

Solving for $r$ gives $r=\boxed{11.58m}$

Let $C_1$ be the circle with radius $10$ .

$C_2$ be the circle with unknown radius $r$ .

The area that exactly $\frac{1}{2}$ of the field is $\frac{1}{2}10^2π = 50π$

The two segments bounded by the two circles $C_1$ & $C_2$ is equal to $50π$

Let $A$ & $B$ be the angle subtended by the arc formed by $C_1$ & $C_2$ .

Segment at $C_1$ is equal to $10^2π(\frac{A}{360}) - \frac{10^2}{2}\sin A$

Segment at $C_2$ is equal to $r^2π(\frac{B}{360}) - \frac{r^2}{2}\sin B$

$\boxed{[10^2π(\frac{A}{360}) - \frac{10^2}{2}\sin A] + [r^2π(\frac{B}{360}) - \frac{r^2}{2}\sin B] = 50π}$ ---> $eq. 1$

$\sin \frac{A}{4} = \frac{r}{(2)(10)}$

$\boxed{r = 20\sin \frac {A}{4}}$ ---> $eq. 2$

$\frac{B}{2} = 90 - \frac{A}{4}$

$\boxed{B = \frac{360-A}{2}}$ ---> $eq. 3$

Substituting $eq. 2$ & $eq. 3$ in $eq. 1$ will result to:

$\frac{5}{18}(Aπ-180\sin A) + \frac{(20\sin \frac{A}{4})^2}{720}[(360 - A)π - 360\sin \frac{A}{2}] = 50π$

Note: All values of angles in the above formula are in degree unit.

Solve for angle $A$

$A = 141.6233553 deg$

solve for radius $r$ by substituting $A = 141.6233553 deg$ to $eq. 2$

$\boxed{r = 11.58728473 m}$

$The~goat~is~tied~at~Q.\\ From~ the ~above~ we~ see\\ " f(R)= 100*Cos^{-1}(1-\dfrac{R^2} {100}+\dfrac{R^2} 2* Cos^{-1}(\dfrac{R} {20} - 10R*Sin( Cos^{-1}(1-\dfrac{R^2}{100})+Cos^{-1}(1-\dfrac{R}{20} ) )- 50*\pi " \\ Since~ R~ has~ to~ be~ greater~ than~ 10, ~the~ values~ tried~ were ~R=11,~ R=12,~ R=11.5 ,~ R=11.6,~ and~ then~ refined ~it ~to ~R=11.587285..\\ So~ stopped~ at~ f(11.587285)=5.9581 E-6 .~~ Instead~ of~ 0~ this~ is ~the~ error.~~By~ the~ way~ the~ answer~ should ~have ~been ~\Large~ 11.59.$