Chapter P, Problem 16RE

Solution

Show that the figure determined by the given points is equilateral triangle.

No, the given figure is not equilateral triangle

Given information:

Given points are $\left(0,1\right),\left(4,1\right),\text{ and }\left(2,1-\sqrt{3}\right)$

Formula used:

1. A triangle is equilateral triangle length of all sides of the triangle are equals.
2. The distance between two points, $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$, is given by:

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

Calculation:

To proof that the triangle is equilateral triangle prove that the length of all sides of the triangle are equals.

First, calculate the length of the sides of the triangle in given figure use the formula for the distance between two points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ that is

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

The length of the side with end points $\left(0,1\right),\text{ and }\left(4,1\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(4-0\right)}^2+{\left(1-1\right)}^2}=\sqrt{4^2+0^2}\\ =\sqrt{4^2}\\ =4\end{array}$

Similarly, the length of the side with end points $\left(0,1\right),\text{ and }\left(2,1-\sqrt{3}\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(2-0\right)}^2+{\left(1-\sqrt{3}-1\right)}^2}=\sqrt{2^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{4+3}\\ =\sqrt{7}\end{array}$

And, the length of the side with end points $\left(4,1\right),\text{ and }\left(2,1-\sqrt{3}\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(2-4\right)}^2+{\left(1-\sqrt{3}-1\right)}^2}=\sqrt{{\left(-2\right)}^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{4+3}\\ =\sqrt{7}\end{array}$

Thus, the length of the sides of the triangle in given figure are 4, $\sqrt{7}$ , and $\sqrt{7}$ .

Since, all the lengths are not equal, i.e. the given figure is not equilateral triangle.