Question

A rectangular fish pond has length 2m more than twice it's width. solve for the length and the width if the area of the pond is 63 meters square.

Answer 1

Given:

The area of the pond, A=63 m^2.

Let l be the length and w be the width of the rectangular pond.

It is given that the pond has length 2m more than twice it's width.

Hence, the expression for the length of the pond is,

`l=2+2w\text{ ---(1)}`

Now, the area of the rectangular pond can be expressed as,

`\begin{gathered} A=lw \\ =(2+2w)w \\ =2w+2w^2 \end{gathered}`

Since A=63, we get

`\begin{gathered} 63=2w+2w^2\text{ } \\ 2w^2+2w-63=0\text{ ---(2)} \end{gathered}`

The above equation is in the form of a quadratic equation given by,

`ax^2+bx+c=0\text{ ---(3)}`

Comparing equations (2) and (3), we get a=2, b=2, c=-63 and x=w.

Solve equation (2) for w using discriminant method.

`\begin{gathered} w=\frac{-b\pm\sqrt[\square]{b^2-4ac}}{2a} \\ =\frac{-2\pm\sqrt[]{2^2-4*2*(-63)}}{2*2} \\ =\frac{-2\pm2\sqrt[]{127}}{4} \end{gathered}`

Since width w cannot be negative ,

`\begin{gathered} w=\frac{-2+2\sqrt[]{127}}{4} \\ =5.13 \end{gathered}`

Put w=5.13 in equation (1).

`\begin{gathered} l=2+2*5.13 \\ l=12.26 \end{gathered}`

Therefore, the length of pond is approximately 12.26 m and its width is approximately 5.13 m.