Chapter 8.4, Problem 40AYU

To determine

To find: The area of the lake.

Answer to Problem 40AYU

The area of the shaded region is $143.299{\text{ft}}^2$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the value of the angle,

  $\begin{array}{c}\angle DEC=180°-100°\\ =80°\end{array}$

Consider the law of cosine as,

  ${\left(CD\right)}^2=D{E}^2+E{C}^2-2DE\left(EC\right)\cos \angle DEC$

Then,

  $\begin{array}{l}{\left(CD\right)}^2={45}^2+{20}^2-2\left(45\right)\left(20\right)\cos 0°\\ {\left(CD\right)}^2=2112.43328\\ CD=45.96\text{ft}\end{array}$

Consider the semi-perimeter of the above triangle is,

  $\begin{array}{c}s=\frac{CD+DE+EC}{2}\\ =\frac{45.96+45+20}{2}\\ =55.48\end{array}$

Consider the area of the triangle $\Delta CDE$ is,

  $$

  $\begin{array}{c}A=\sqrt{s\left(s-CD\right)\left(s-DE\right)\left(s-EC\right)}\\ =\sqrt{55.48\left(55.48-45.96\right)\left(55.48-45\right)\left(55.48-20\right)}\\ =443.15856{\text{ft}}^2\end{array}$

Consider the value of side BC is,

  $\begin{array}{l}{\left(BC\right)}^2={\left(AB\right)}^2+{\left(CA\right)}^2-2\left(AB\right)\left(CA\right)\cos \Delta \left(\angle BAC\right)\\ {\left(BC\right)}^2={\left(35\right)}^2+{\left(80\right)}^2-2\left(35\right)\left(80\right)\cos \left(15°\right)\\ {\left(BC\right)}^2=2215.81537\\ BC=47.07\text{ft}\end{array}$

Consider the semi perimeter of the triangle ABC is,

  $\begin{array}{c}s=\frac{35+47.07+80}{2}\\ =81.035\end{array}$

Consider the area of the triangle $\Delta ABC$ by the law of cosine is,

  $\begin{array}{c}\Delta ABC=\sqrt{81.035\left(81.035-35\right)\left(81.035-47.07\right)\left(81.035-80\right)}\\ =362.13156{\text{ft}}^{\text{2}}\end{array}$

Consider the semi perimeter of the triangle $\Delta BCD$ is,

  $\begin{array}{c}s=\frac{40.07+45.96+40}{2}\\ =63.015\end{array}$

Consider the area of the triangle $\Delta BCD$ by the law of cosine is,

  $\begin{array}{c}\Delta ABC=\sqrt{63.015\left(63.015-47.07\right)\left(63.015-45.96\right)\left(63.015-40\right)}\\ =628.00859{\text{ft}}^{\text{2}}\end{array}$

The area of the lake is sum of the area of all the triangle,

  $\begin{array}{c}A=443.15856+362.13156+628.00859\\ =143.299{\text{ft}}^2\end{array}$