Chapter 4.10, Problem 3ST

To determine

the dimension of rectangular fish pond.

Answer to Problem 3ST

  $17\text{ft}\times 10\text{ft}$ .

Explanation of Solution

Given:

Pond is $7\text{ ft}$ longer than its width.

Wooden walk of $1\text{ft}$ is made around the park.

Combined area of pond and walk is $58{\text{ ft}}^2$ greater than area of pond alone.

Calculation:

Let width of the pond be $x\text{ft}$ then its length will be $\left(x+7\right)$ . $\begin{array}{l}\left(\text{Area of Pond+Area of Walk}\right)-\text{Area of Pond}=58\\ \left(x+9\right)\left(x+2\right)-x\left(x+7\right)=58\\ x^2+2x+9x+18-x^2-7x=58\\ 4x=40\\ x=10\text{ft}\end{array}$

Length of pond is.

  $\begin{array}{l}=x+7\\ =10+7\\ =17\text{ft}\end{array}$

Dimension of the pond is.

  $17\text{ft}\times 10\text{ft}$

Therefore, the answer is $17\text{ft}\times 10\text{ft}$ .