Chapter 10.3, Problem 25P

A painting is shaped like a golden rectangle. It’s length is 24 cm What is the painting’s width to the nearest tenth ol a cm?

To determine

To find: the painting’s width.

Answer to Problem 25P

The painting’s width is $14.83$ cm.

Explanation of Solution

Given information:

The length of the golden rectangle painting is $24\text{cm}$ .

Calculation: to find the width of the painting, first use the golden rectangle golden ratio formula and then simplify,

  $\begin{array}{l}\frac{a+b}{a}=\frac{a}{b}\\ a+b=24\\ b=24-a\text{ [subtract]}\\ \frac{24}{a}=\frac{a}{24-a}\text{ [substitute in the formula]}\\ a^2=576-24a\text{ [cross multiplication]}\\ a^2+24a-576=0\text{ [simplify]}\\ a=\frac{-24\pm \sqrt{{\left(24\right)}^2-4\left(1\right)\left(-576\right)}}{2\left(1\right)}\text{ [quadratic formula]}\\ a=\frac{-24\pm \sqrt{2880}}{2}\text{ [multiply and add]}\\ a=14.83\text{ [simplify]}\end{array}$

Thus, The painting’s width is $14.83$ cm.