Chapter 8.4, Problem 20WE

To determine

To find: the altitude of the equilateral triangle.

Answer to Problem 20WE

  $5\sqrt{3}\text{ units}$

Explanation of Solution

Given:

The length of each side of the equilateral triangle is 10.

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 leg and $\overline{AC}$ is the longer leg.

Now, by theorem 8-7, the hypotenuse is twice as long as the shorter leg, so

  $\begin{array}{c}AD=2CD\\ 10=2CD\\ \frac{10}{2}=CD\\ 5=CD\end{array}$

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

Thus,

  $\begin{array}{c}AC=\sqrt{3}CD\\ =\sqrt{3}\left(5\right)\\ =5\sqrt{3}\end{array}$

So, the length of the altitude of the given triangle is $5\sqrt{3}\text{ units}$ .