Chapter 9.6, Problem 45PS

Solution

To find: The area of quadrilateral shape land plot.

The area of plot is $\text{19}\text{.77}\text{acre}$ .

Given data:

The shape of plot is shown in Fig.1.

  

Method/Formula used:

The Law of Cosine states that in a triangle ABC , $\cos \angle B=\frac{c^2+a^2-b^2}{2ac}$

And the area of triangle ABC is $\frac{1}{2}ca\sin \angle B$ .

Calculations:

The area of plot ABCD is

  $\text{Area}\text{}ABCD=\text{Area}\text{}\Delta DAB+\text{Area}\text{}\Delta DCB$ ... (1)

In triangle DAB ,

  $\cos \angle A=\frac{D{A}^2+A{B}^2-D{B}^2}{2DA\cdot AB}$

Substitute, 743 for DA , 1210 for AB and 1480 for DB ,

  $\begin{array}{c}\cos \angle A=\frac{{743}^2+{1210}^2-{1480}^2}{2\cdot 743\cdot 1210}\\ \cos \angle A=-0.097\\ \angle A={\cos }^{-1}\left(-0.097\right)\\ =95.57°\end{array}$

The area of triangle $\text{}\Delta DAB$ is

  $\text{}\Delta DAB=\frac{1}{2}DA\cdot AB\sin \angle A$

Substitute, 743 for DA , 1210 for AB and $95.57°$ for $\angle A$ ,

  $\begin{array}{c}\text{}\Delta DAB=\frac{1}{2}743\cdot 1210\cdot \sin 95.57°\\ =\frac{1}{2}743\cdot 1210\cdot \left(0.9953\right)\leftarrow \sin 95.57°=0.9953\\ =447392.55\end{array}$

In triangle DCB ,

  $\cos \angle C=\frac{D{C}^2+C{B}^2-D{B}^2}{2DC\cdot CB}$

Substitute, 1000 for DA , 858 for AB and 1480 for DB ,

  $\begin{array}{c}\cos \angle C=\frac{{1000}^2+{858}^2-{1480}^2}{2\cdot 1000\cdot 858}\\ \cos \angle C=-0.265\\ \angle C={\cos }^{-1}\left(-0.265\right)\\ =105.35°\end{array}$

The area of triangle $\text{}\Delta DCB$ is

  $\text{}\Delta DCB=\frac{1}{2}DC\cdot CB\sin \angle C$

Substitute, 1000 for DC , 858 for CB and $105.35°$ for $\angle C$ ,

  $\begin{array}{c}\text{}\Delta DCB=\frac{1}{2}1000\cdot 858\cdot \sin 105.35°\\ =\frac{1}{2}1000\cdot 858\cdot \left(0.9643\right)\leftarrow \sin 105.35°=0.9643\\ =413696.19\end{array}$

Substitute $447392.55$ for $\text{Area}\text{}\Delta DAB$ and $413696.19$ for $\text{Area}\text{}\Delta DCB$ in result (1),

  $\begin{array}{c}\text{Area}\text{}ABCD=447392.55+413696.19\\ =861088.74\end{array}$

Thus, the area of plot is $\text{861088}\text{.74}\text{f}{\text{t}}^{\text{2}}$ . As one acre equals $43560$ square ft, therefore, the area of plot in in acre is

  $\frac{\text{861088}\text{.74}\text{f}{\text{t}}^{\text{2}}}{\text{43560f}{\text{t}}^{\text{2}}}\text{=19}\text{.77}\text{acre}$ .

The area of plot is $\text{19}\text{.77}\text{acre}$ .

The area of the plot can be determined from the Fig. 3, if the length of diagonal AC is known.