Chapter 12, Problem 11CT

To determine

To calculate: The new dimension of garden when a garden is currently $4\text{m}$ wide and $7\text{m}$ long. The area of the garden is to be doubled by increasing the width and the length by the same number of meters.

Answer to Problem 11CT

The new dimensions are width $6.132m$ and length $9.123m$ .

Explanation of Solution

Given information:

The garden is currently $4\text{m}$ wide and $7\text{m}$ long. The area of the garden is to be doubled by increasing the width and the length by the same number of meters.

Formula used:

The area of rectangle is $l\times b$ .

Where $l$ is length of rectangle and $b$ is width of rectangle.

If $a{z}^2+bz+c=0$ , $a\ne 0$ and $b^2-4ac\ge 0$ .

Then $z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The current wide $4\text{m}$ and length $7\text{m}$ of garden then area of garden;

The shape of a garden is rectangular so area of garden $l\times b=7\times 4=28{\text{cm}}^2$ .

The area of garden is double as $2\cdot 28=56{\text{cm}}^2$ .let assume sides are increase by $x$ then new width $=\left(x+4\right)$ and length $\left(x+7\right)$ .

Therefore, the area of garden is,

  $\left(x+4\right)\cdot \left(x+7\right)=56$

Furthermore,

  $\begin{array}{l}{x}^2+11x+28=56\\ x^2+11x-28=0\end{array}$

The equation is quadratic so using the quadratic formula.

The values of $a=1,b=11\text{and}c=-28$ put into the quadratic formula,

  $\begin{array}{l}x=\frac{\left(-11\right)\pm \sqrt{121-4\cdot 1\cdot \left(-28\right)}}{2\cdot 1}\\ x=\frac{-11\pm \sqrt{121+112}}{2}\\ x=\frac{-11\pm \sqrt{233}}{2}\end{array}$

Furthermore,

  $\begin{array}{l}x=\frac{-11\pm 15.26}{2}\\ x=2.132\text{or}x=-13.13\end{array}$

The dimensions are increase by $x=2.132$ .so new dimensions are width $=\left(2.132+4\right)=6.132m$ and length $=\left(2.123+7\right)=9.123m$ .

Thus, the new dimensions are width $6.132m$ and length $9.123m$ .