Chapter 12, Problem 12.1P

To determine

The area enclosed within polygon ABDFGA.

Answer to Problem 12.1P

The area enclosed within polygon ABDFGA is $A_{ABFDGA}=280842.66{\text{ft}}^2$

Explanation of Solution

Given:

A closed polygon ABDFGA

Concept Used:

The formula for area of a triangle is

  $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$     ....... (1)

The formula for semi-perimeter of triangle is

  $s=\frac{a+b+c}{2}$     ....... (2)

Here, $A$ is area of triangle and $s$ is semi-perimeter of triangle and $a,b,c$ are the three sides of the triangle.

The area of polygon ABDFGA can be calculated by considering smaller triangles which together form the polygon.

  $A_{ABFDGA}=A_{DFG}+A_{BDG}+A_{ABG}$     ....... (3)

Calculation:

Consider the triangle DFG

Substitute $219.1\text{ft}$ for $a$ , $567.6\text{ft}$ for $b$ , $588.7\text{ft}$ for $c$ in formula (2)

  $\begin{array}{c}s=\frac{219.1+567.6+588.7}{2}\\ =\frac{1375.4}{2}\\ =687.7\text{ft}\end{array}$

Substitute $219.1\text{ft}$ for $a$ , $567.6\text{ft}$ for $b$ , $588.7\text{ft}$ for $c$ , $687.7\text{ft}$ for $s$ in formula (1)

  $\begin{array}{c}{A}_{DFG}=\sqrt{687.7\left(687.7-219.1\right)\left(687.7-567.6\right)\left(687.7-588.7\right)}\\ =\sqrt{687.7\times 468.6\times 120.1\times 99}\\ =\sqrt{3831594230}\\ =61899.87{\text{ft}}^2\end{array}$

Consider the triangle BDG

Substitute $298.3\text{ft}$ for $a$ , $588.7\text{ft}$ for $b$ , $551.8\text{ft}$ for $c$ in formula (2)

  $\begin{array}{c}s=\frac{298.3+588.7+551.8}{2}\\ =\frac{1438.8}{2}\\ =719.40\text{ft}\end{array}$

Substitute $298.3\text{ft}$ for $a$ , $588.7\text{ft}$ for $b$ , $551.8\text{ft}$ for $c$ , $719.40\text{ft}$ for $s$ in formula (1)

  $\begin{array}{c}{A}_{BDG}=\sqrt{719.40\left(719.40-298.3\right)\left(719.40-588.7\right)\left(719.40-551.8\right)}\\ =\sqrt{719.40\times 421.1\times 130.7\times 167.6}\\ =\sqrt{6635983183}\\ =81461.54{\text{ft}}^2\end{array}$

Consider the triangle ABG

Substitute $536.6\text{ft}$ for $a$ , $551.8\text{f}$ for $b$ , $610.5\text{ft}$ for $c$ in formula (2)

  $\begin{array}{c}s=\frac{536.6+551.8+610.5}{2}\\ =\frac{1698.9}{2}\\ =849.45\text{ft}\end{array}$

Substitute $536.6\text{ft}$ for $a$ , $551.8\text{f}$ for $b$ , $610.5\text{ft}$ for $c$

  $849.45\text{ft}$ , for $s$ in formula (1)

  $\begin{array}{c}{A}_{ABG}=\sqrt{849.45\left(849.45-536.6\right)\left(849.45-551.8\right)\left(849.45-610.5\right)}\\ =\sqrt{849.45\times 312.85\times 297.65\times 238.95}\\ =\sqrt{18901091350}\\ =137481.24{\text{ft}}^2\end{array}$

Substitute $137481.24{\text{ft}}^2$ for $A_{ABG}$ , $81461.54{\text{ft}}^2$ for $A_{BDG}$ , $61899.87{\text{ft}}^2$ for $A_{DFG}$ in the equation (3),

  $\begin{array}{c}{A}_{ABFDGA}=A_{DFG}+A_{BDG}+A_{ABG}\\ =61899.87+81461.54+137481.24\\ =280842.66{\text{ft}}^2\end{array}$

Conclusion:

The area enclosed within polygon ABDFGA is $A_{ABFDGA}=280842.66{\text{ft}}^2$