Chapter 1.6, Problem 91E

To determine

Find the original length of the reed.

Answer to Problem 91E

  $x=\frac{1\pm \sqrt{61}}{2}cubits$

Explanation of Solution

Calculation:

Here, let us suppose:

  $\begin{array}{l}x=Lengthofreed\\ w=Widthoffield\\ l=Lengthoffield\\ A=Areaoffield\end{array}$

Now, we have:

  $1ninda=12cubits$

We will convert it to square cubits, the area of field becomes:

  $\begin{array}{l}A=375nind{a}^2\\ A=\frac{375nind{a}^2}{1}.\frac{144cubit{s}^2}{1nind{a}^2}\\ A=54000cubit{s}^2\end{array}$

We have to convert the word problem into equations and then solve it.

Now, we will translate the first measurement into an equation for the length of the field yields:

  $l=60\left(x-1\right)$

Now, the second portion of the problem gives us equation for the width of the field in terms of the measurement of the reed:

  $w=60x$

Now, we are given the total area of the field so recalling the formula for area of a rectangle:

  $A=w.l$

Here, $A$ is the area, $w$ is the width, and $l$ is the length.

  $A=\left(60\left(x-1\right)\right)\left(60x\right)$

Here, we are given the value for the area of the field, and we have already converted it to proper units, thus we get:

  $\begin{array}{l}54000=\left(60\left(x-1\right)\right)\left(60x\right)\\ 54000=\left(60x-60\right)\left(60x\right)\\ 54000=3600x^2-3600x\\ 3600x^2-3600x-54000=0\end{array}$

Now, we will solve it by using quadratic formula:

  $\begin{array}{l}x=\frac{-\left(-3600\right)\pm \sqrt{{\left(-3600\right)}^2-4\left(3600\right)\left(-54000\right)}}{2\left(3600\right)}\\ x=\frac{3600\pm \sqrt{12960000+777600000}}{7200}\\ x=\frac{3600\pm \sqrt{790560000}}{7200}\\ x=\frac{3600\pm 3600\sqrt{61}}{7200}\\ x=\frac{1\pm \sqrt{61}}{2}\end{array}$

Hence, the required answer is $x=\frac{1\pm \sqrt{61}}{2}cubits$ .