Chapter 12.2, Problem 1C

To determine

To evaluate: Construct an equilateral triangle then find the value of $x+y+z$ .

Answer to Problem 1C

The equilateral triangle is shown in Figure (1) and the value of $x+y+z=h$ .

Explanation of Solution

Given information: An equilateral triangle such that x , y and z are the perpendicular distances from any point inside the triangle to each side.

Calculation:

The triangle drawn below is an equilateral triangle.

  

Figure (1)

Let $P$ be any point inside the triangle and x , y and z are the perpendicular distances from this point to each side.

Draw a line from $P$ to each of $A$ , $B$ , and $C$ forming three triangles $PAB$ , $PBC$ , and $PCA$ .

The formula to find the area of triangle is:

  $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$

So,

  $\begin{array}{l}\text{Area}\text{of}\Delta PAC=\frac{a.z}{2}\\ \text{Area}\text{of}\Delta PAB=\frac{a.x}{2}\\ \text{Area}\text{of}\Delta PCB=\frac{a.y}{2}\end{array}$

The area of $\Delta ABC$ can be calculated as:

  $\begin{array}{c}\Delta ABC=\Delta PAC+\Delta PAB+\Delta PCB\\ \frac{a.h}{2}=\frac{a.z}{2}+\frac{a.x}{2}+\frac{a.y}{2}\\ x+y+z=h\end{array}$

Observation is that the sum of the distances from any interior point to the sides of the equilateral $\Delta ABC$ is equal to the length of triangle’s altitude.

Therefore, the equilateral triangle is shown in Figure (1) and the value of $x+y+z=h$ .