Question

Luis has a square pasture with a surface area of 1800 m² and is not fenced. In the center of the pasture there is a tree from which he ties his horse with a rope that reaches exactly to the corners of the pasture and allows the horse to surround the terrain.What area can the horse tread outside of the pasture?

Answer 1

`A=1027.43m^2`

Step-by-step explanation

First, let us make a sketch of the situation:

From the diagram:

=> The square with side length L represents the square pasture with an area of 1800 m²

=> The straight line, D, represents the distance between the corners of the square (the diagonal of the square)

=> The circle represents the area that the horse can tread in total.

=> The inner circle represents the tree at the center.

To solve this, we first have to find the radius of the circle, in order to find the area of the circle (the area the horse can tread in total).

The area of the square is 1800 m², this means that the length of the side of the square can be found using the formula for the area of a square:

`\begin{gathered} A=L^2 \\ L=\text{length of the side of the square} \\ \Rightarrow1800=L^2 \\ L=\sqrt[]{1800} \\ L=30\sqrt[]{2}m \end{gathered}`

Now, we can find the length of the diagonal of the square, D, using the Pythagoras theorem:

`\begin{gathered} D^2=L^2+L^2 \\ \Rightarrow D^2=2L^2 \\ D^2=2(30\sqrt[]{2})^2 \\ D^2=2\cdot1800=3600 \\ D=\sqrt[]{3600} \\ D=60m \end{gathered}`

D represents the diameter of the circle, but to find its area, we need to find its radius, R, using :

`\begin{gathered} D=2\cdot R \\ R=(D)/(2) \end{gathered}`

Therefore:

`R=(60)/(2)=30m`

The area of the circle is:

`\begin{gathered} A=\pi R^2 \\ A=\pi\cdot30^2 \\ A=2827.43m^2 \end{gathered}`

Finally, to find the area that the horse can tread outside the pasture, we have to subtract the area of the square pasture from the circle that the horse can tread in total, that is:

`A_{\text{treadoutside}}=A_{\text{treadtotal}}_{}-A_{\text{pasture}}`

Therefore, the answer is:

`\begin{gathered} A=2827.43-1800 \\ A=1027.43m^2 \end{gathered}`