Chapter 5.6, Problem 42E

Solution

To find: The lengths of the diagonal $HB$, $HC$, and $HD$ in the octagon.

The length of the diagonals $HB$ is $37\text{ft}$, $HC$ is $48.3\text{ft}$, and $HD$ is $52.3\text{ft}$ in the octagon.

Given information:

The length of the side of the octagon is $20\text{ft}$ .

Formula used:

For a regular polygon the interior angle with $n$ sides is given as $\frac{180°\left(n-2\right)}{n}$ .

Law of cosines for the give octagon:

  $H{B}^2=H{A}^2+B{A}^2-2\times HA\times BA\cos HAB$

  $H{C}^2=H{B}^2+B{C}^2-2\times HB\times BC\cos HBC$

  $H{D}^2=H{C}^2+C{D}^2-2\times HC\times CD\cos HCD$

Calculation:

Consider triangle $HBA$ .

For a regular octagon the interior angle with $8$ sides.

Substitute $8$ for $n$ in $\frac{180°\left(n-2\right)}{n}$ to calculate $\angle HAB$

  $\begin{array}{c}\angle HAB=\frac{180°\left(8-2\right)}{8}\\ =135°\end{array}$

Substitute $20$ for $HA$, $20$ for $BA$, and $135°$ for $\angle HAB$ in the formula to calculate the length of the diagonal $HB$ .

  $\begin{array}{c}H{B}^2=H{A}^2+B{A}^2-2\times HA\times BA\cos HAB\\ HB=\sqrt{H{A}^2+B{A}^2-2\times HA\times BA\cos HAB}\\ =\sqrt{{20}^2+{20}^2-2\times 20\times 20\times \cos 135}\\ \approx 37\text{ft}\end{array}$

Consider triangle $HBA$ .

  $HA=BA$ since it is an isosceles triangle. Therefore, $\angle BHA$ is given as $\angle HAB=\angle BHA=\frac{180-135}{2}=22.5°$ .

Consider triangle $HCB$

  $\angle HBC=135-22.5=112.5°$

Substitute $37$ for $HB$, $20$ for $BC$, and $112.5°$ for $\angle HBC$ in $H{C}^2=H{B}^2+B{C}^2-2\times HB\times BC\cos HBC$ to calculate the length of the diagonal $HC$ .

  $\begin{array}{c}H{C}^2=H{B}^2+B{C}^2-2\times HB\times BC\cos HBC\\ HC=\sqrt{H{B}^2+B{C}^2-2\times HB\times BC\cos HBC}\\ =\sqrt{{37}^2+{20}^2-2\times 37\times 20\times \cos 112.5}\\ \approx 48.3\text{ft}\end{array}$

Consider triangle $HDC$ .

  $\angle HAB=\angle ABC=\angle HCB=45°$

  $\angle HCD=135-45=90°$

Substitute $48.3$ for $HC$, $20$ for $CD$, and $90°$ for $\angle HCD$ in $H{D}^2=H{C}^2+C{D}^2-2\times HC\times CD\cos HCD$ to calculate the length of the diagonal $HD$ .

  $\begin{array}{c}H{D}^2=H{C}^2+C{D}^2-2\times HC\times CD\cos HCD\\ HD=\sqrt{H{C}^2+C{D}^2-2\times HC\times CD\cos HCD}\\ =\sqrt{{48.3}^2+{20}^2-2\times 48.3\times 20\times \cos 90}\\ \approx 52.3\text{ft}\end{array}$

Therefore, the length of the diagonals $HB$ is $37\text{ft}$, $HC$ is $48.3\text{ft}$, and $HD$ is $52.3\text{ft}$ in the octagon.