Chapter 8.4, Problem 6ST1

To determine

To find: the altitude of the equilateral triangle.

Answer to Problem 6ST1

  $6\sqrt{3}\text{ units}$

Explanation of Solution

Given:

The length of each side of the equilateral triangle is 12.

Concept Used:

Theorem 4-4: RHS rule states that the right triangles are congruent if any of side and hypotenuse of right triangles are congruent.

Theorem 8-7: In a $30°-60°-90°$ triangle, the longer Side is $\sqrt{3}$ times as long as the shorter Side and the hypotenuse of right triangle are two times as long as the shorter side of triangle.

First draw an equilateral triangle as mentioned in the question:

Here $△ABD$ is an equilateral triangle, so

$AB=BD=AD$ ….. (1)

$m\angle A=m\angle B=m\angle C=60$ ….. (2)

Consider the right triangles $△ACB$ and $△ACD$

$AB=AD$ [By equation (1)]

$AC=AC$ [Common side]

So, by HL-theorem $△ACB\cong △ACD$

Thus,

  $m\angle BAC=m\angle DAC$

Now consider,

  $\begin{array}{c}m\angle A=60\\ m\angle BAC+m\angle DAC=60\\ m\angle DAC+m\angle DAC=60\end{array}$

  $\begin{array}{c}2m\angle DAC=60\\ m\angle DAC=\frac{60}{2}\\ m\angle DAC=30\end{array}$

Apply theorem 3-11 on the triangle $△ABC$ ,

  $\begin{array}{c}m\angle A+m\angle B+m\angle C=180\\ m\angle A+90+60=180\\ m\angle A+150=180\end{array}$

  $\begin{array}{c}m\angle A=180-150\\ m\angle A=30\end{array}$

Thus, the triangle $△ACD$ is a $30°-60°-90°$ . Here $\overline{AD}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{CD}$ is the shorter side and $\overline{AC}$ is the longer side.

Now, by theorem 8-7, the hypotenuse is two times as long as the shorter side, so

  $\begin{array}{c}AD=2CD\\ 12=2CD\\ \frac{12}{2}=CD\\ 6=CD\end{array}$

Also, by theorem 8-7, the longer side is $\sqrt{3}$ times as long as the shorter side.

Thus,

  $\begin{array}{c}AC=\sqrt{3}CD\\ =\sqrt{3}\left(6\right)\\ =6\sqrt{3}\end{array}$

So, the length of the altitude of the given triangle is $6\sqrt{3}\text{ units}$ .