Chapter 9.7, Problem 21PSB

a

To determine

To calculate:The Perimeter of octagon.

Answer to Problem 21PSB

Perimeter of regular octagon is $48$ units.

Explanation of Solution

Given information:Here, we have given that side of central square is $6$ units long.

Formula used: Perimeter of regular octagon = $8$ (side)

Calculation: Side of central square = $6$ units.

Side of regular octagon = $6$ units (Parallel and equal to central square)

So, Perimeter of regular octagon = $8$ (side)

  $\begin{array}{l}\Rightarrow 8(6)\\ \Rightarrow 48\end{array}$

b

To determine

To calculate: The Span of octagon.

Answer to Problem 21PSB

Span of regular octagon is $(6+6\sqrt{2})$ units.

Explanation of Solution

Given information:Here, we have given that side of central square is $6$ units long.

Formula used:Interior angle of regular octagon = $\frac{(n-2)180°}{n}$

Calculation:Number of sides in octagon (n) = $8$

Interior angle of regular octagon = $\frac{(n-2)180°}{n}$

  $\begin{array}{l}\Rightarrow \frac{(8-2)180°}{8}\\ \Rightarrow \frac{(6)180°}{8}\\ \Rightarrow \frac{1080°}{8}\\ \Rightarrow 135°\end{array}$

Taking one right triangle,

The angles of triangle will be $90°,45°,45°$ ,

  $\therefore$ Hypotenuse will be $a\sqrt{2}$ and length of other sides be $a$ units each.

Here, hypotenuse = $6$

  $\begin{array}{l}\Rightarrow a\sqrt{2}=6\\ \Rightarrow a=\frac{6}{\sqrt{2}}\\ \Rightarrow a=\frac{6}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ \Rightarrow a=\frac{6\sqrt{2}}{2}\\ \Rightarrow a=3\sqrt{2}\end{array}$

  $\therefore$ Base and perpendicular = $3\sqrt{2}$

Now, span of octagon = $3\sqrt{2}+6+3\sqrt{2}$

  $\Rightarrow 6+6\sqrt{2}$