Chapter 11.5, Problem 21PSC

a.

To determine

To calculate:Thearea of the square.

Answer to Problem 21PSC

Thearea of the square is $291.42$ .

Explanation of Solution

Given information:

A square is formed by joining the alternate points of a regular octagon. A side of octagon is 10.

Formula used:

  $Areaofpolygon=\frac{1}{2}\times a\times P$

where a = Apothem,

P = Perimeter

Apothem of the octagon $=5+5\sqrt{2}$

Diagonal of a square= $2\times$ Apothem of the octagon

  $Areaofsquare=\frac{1}{2}\times (d_1\times d_2)$

  $d=Diagonalofsquare.$

Calculation:

  

In the shaded triangle, i.e. ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$

  $\begin{array}{l}10=x\sqrt{2}\\ x=\frac{10}{\sqrt{2}}\\ x=5\sqrt{2}\end{array}$

Since the apothem is $5+x$ and $x=5\sqrt{2}$ , it is $5+5\sqrt{2}$ .

  $\begin{array}{l}Areaofpolygon=\frac{1}{2}\times a\times P\\ A=\frac{1}{2}[(5+5\sqrt{2})\times 80]\\ A=\frac{1}{2}(400+400\sqrt{2})\\ A=200+200\sqrt{2}\end{array}$

  

Apothem of the octagon $=5+5\sqrt{2}$

Diagonal of a square= $2\times$ Apothem of the octagon

Diagonal of a square= $2\times (5+5\sqrt{2})=10+10\sqrt{2}$

The diagonals are congruent and are perpendicular bisector of each other.

  $\begin{array}{l}{A}_{Square}=\frac{1}{2}\times (d_1\times d_2)\\ A_{Square}=\frac{1}{2}\times {(}^{10}\\ A_{Square}=\frac{1}{2}\times (100+200\sqrt{2}+200)\\ A_{Square}=150+100\sqrt{2}\\ A_{Square}=291.42\end{array}$

b.

To determine

To calculate: The area of the shaded region.

Answer to Problem 21PSC

The area of the shaded region is $191.42$ .

Explanation of Solution

Given information:

A square is formed by joining the alternate points of a regular octagon. A side of octagon is 10.

Formula used:

  $Areaofpolygon=\frac{1}{2}\times a\times P$

where a = Apothem,

P = Perimeter

  $Areaofregularoctagon=200+200\sqrt{2}$

  $A_{ShadedRegion}=A_{Octagon}-A_{Square}$

Calculation:

  

In the shaded triangle, i.e. ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$

  $\begin{array}{l}10=x\sqrt{2}\\ x=\frac{10}{\sqrt{2}}\\ x=5\sqrt{2}\end{array}$

Since the apothem is $5+x$ and $x=5\sqrt{2}$ , it is $5+5\sqrt{2}$ .

  $\begin{array}{l}Areaofpolygon=\frac{1}{2}\times a\times P\\ A=\frac{1}{2}[(5+5\sqrt{2})\times 80]\\ A=\frac{1}{2}(400+400\sqrt{2})\\ A=200+200\sqrt{2}\end{array}$

  

Apothem of the octagon $=5+5\sqrt{2}$

Diagonal of a square= $2\times$ Apothem of the octagon

Diagonal of a square= $2\times (5+5\sqrt{2})=10+10\sqrt{2}$

The diagonals are congruent and are perpendicular bisector of each other.

  $\begin{array}{l}{A}_{Square}=\frac{1}{2}\times (d_1\times d_2)\\ A_{Square}=\frac{1}{2}\times {(}^{10}\\ A_{Square}=\frac{1}{2}\times (100+200\sqrt{2}+200)\\ A_{Square}=150+100\sqrt{2}\end{array}$

  $A_{Octagon}=200+200\sqrt{2}$

  $\begin{array}{l}{A}_{ShadedRegion}=A_{Octagon}-A_{Square}\\ A_{ShadedRegion}=(200+200\sqrt{2})-(150+100\sqrt{2})\\ A_{ShadedRegion}=50+100\sqrt{2}\\ A_{ShadedRegion}=191.42\end{array}$