Chapter 8, Problem 35RE

To determine

To find: The approximate area of the lake.

Answer to Problem 35RE

The approximate area of the lake is $5611{\text{ft}}^2$ .

Explanation of Solution

The value of $x$ is calculated as,

  $\begin{array}{c}{x}^2={50}^2+{70}^2-2\left(70\right)50\cos 100°\\ x=92.8\end{array}$

Calculate the semi perimeter.

  $\begin{array}{c}s=\frac{a+b+c}{2}\\ =\frac{50+70+92.8}{2}\\ =106.4\end{array}$

The area $A_1$ is calculated as,

  $\begin{array}{c}{A}_1=\sqrt{106.4\left(106.4-70\right)\left(106.4-50\right)\left(106.4-92.8\right)}\\ =1723.57{\text{ft}}^2\end{array}$

The value of $y$ is calculated as,

  $\begin{array}{c}{y}^2={50}^2+{92.8}^2-2\left(92.8\right)50\cos 32°\\ y=56.9\end{array}$

Calculate the semi perimeter.

  $\begin{array}{c}{s}_2=\frac{a+b+c}{2}\\ =\frac{56.9+50+92.8}{2}\\ =99.85\\ \approx 100\end{array}$

The area $A_2$ is calculated as,

  $\begin{array}{c}{A}_2=\sqrt{100\left(100-56.9\right)\left(100-50\right)\left(100-92.8\right)}\\ =1245.6{\text{ft}}^2\end{array}$

The third semi perimeter is,

  $\begin{array}{c}{s}_3=\frac{56.9+125+100}{2}\\ =140\end{array}$

The area $A_3$ is calculated as,

  $\begin{array}{c}{A}_3=\sqrt{140\left(140-56.9\right)\left(140-125\right)\left(140-100\right)}\\ =2642{\text{ft}}^2\end{array}$

Therefore, the area of the lake is,

  $\begin{array}{c}A=A_1+A_2+A_3\\ =1723.57+1245.6+2642\\ =5611{\text{ft}}^2\end{array}$

Therefore, the approximate area of the lake is $5611{\text{ft}}^2$ .