Chapter 11, Problem 24RP

To determine

To find: The area of a regular hexagon whose span is 36.

Answer to Problem 24RP

Area of the regular hexagon with span 36 is $648\sqrt{3}.$

Explanation of Solution

Given Information:

Span of a regular hexagon is $\left(a\right)=36$

Formula used:

  $\tan \theta =\frac{\text{Opposite}\text{side}}{\text{Adjacent}\text{side}}$

Area of a regular hexagon $\left(A\right)=\frac{3\sqrt{3}}{2}{s}^2$

Calculation:

Span of a regular hexagon is $\left(a\right)=36$

Join the diagonals of the regular hexagon and then join midpoints of BC and EF

  

We get $\Delta FOH$

Here $\angle FHO={90}^{\circ },\angle OFH={60}^{\circ }\text{and}\angle FOH={30}^{\circ }$

We have $OH=18$

Using trigonometry, we have in $\Delta FOH$

  $\begin{array}{l}\tan {30}^{\circ }=\frac{FH}{OH}\\ \frac{1}{\sqrt{3}}=\frac{FH}{18}\\ FH=6\sqrt{3}\\ \therefore EF=2FH=12\sqrt{3}\end{array}$

So, each side of a regular hexagon $\left(s\right)=12\sqrt{3}$

Area of a regular hexagon $\left(A\right)=\frac{3\sqrt{3}}{2}{s}^2$

Where $s$ is the side of the regular hexagon

  $\begin{array}{l}\therefore A=\frac{3\sqrt{3}}{2}{\left(12\sqrt{3}\right)}^2\\ \therefore A=648\sqrt{3}\end{array}$

Hence, area of the regular hexagon quadrilateral is $648\sqrt{3}.$