Chapter 7.3, Problem 37WE

a.

To determine

To find: breadth of original rectangle.

Answer to Problem 37WE

3.7082 units wide

Explanation of Solution

Given:

A large golden rectangle with length 6 units.

A square is cut off from the rectangle $x$ units.

  

Concept used:

A golden rectangle is a rectangle from which a square is cut off then the remaining rectangle will be similar to the original rectangle.

Quadratic formula : $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  

Breadth of the original rectangle $=x$ units.

Now,

  $\frac{6}{6-x}=\frac{6}{x}$

  $x^2=36-6x$

  $x^2+6x-36=0$

Now by using the quadratic formula to find the value of $x$ ,

  $\begin{array}{c}x=\frac{-6\pm \sqrt{36-4(1)(-36)}}{2}\\ =\frac{-6\pm \sqrt{180}}{2}\\ =\frac{-6\pm 6\sqrt{5}}{2}\\ =\frac{-6+6\sqrt{5}}{2},\frac{-6-6\sqrt{5}}{2}\\ =\frac{-6+6\sqrt{5}}{2}[\because x\text{can't be negative]}\end{array}$

  $\therefore x\approx 3.7082$

Conclusion:

The original rectangle is 3.7082 units wide.

b.

To determine

To write : the golden ratio

Answer to Problem 37WE

  $\frac{6}{3.7}=1.62$

Explanation of Solution

Given:

Length of the rectangle $=6$ units

Breadth of the rectangle $=3.7$ units

Calculation:

Golden ratio $=\frac{length}{breadth}$

  $\Rightarrow \frac{6}{3.7}$

Now by simplifying,

  $\Rightarrow 1.62$

Conclusion:

Therefore, the golden ratio is $\frac{6}{3.7}$ .