Chapter 11.2, Problem 33WE

(a)

To determine

To Show: The area is $A=\frac{s^2}{4}\sqrt{3}$ .

Explanation of Solution

Given: An equilateral triangle has side of length of $s$

  

In triangle $ABC$ , draw the altitude

Now

Here two ${30}^{°}-{60}^{°}-{90}^{°}$ triangles are formed.

In triangle $ADC$

  $\begin{array}{l}AC=2DC\\ AD=\sqrt{3}DC\\ Thus\\ AD=\sqrt{3}\left(\frac{AC}{2}\right)\\ AD=\sqrt{3}\left(\frac{s}{2}\right)\end{array}$

Now in triangle $ABC$

  $\begin{array}{l}b=BC=s\\ h=AD=\frac{s}{2}\sqrt{3}\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(s)\left(\frac{s}{2}\sqrt{3}\right)\\ A=\frac{s^2}{4}\sqrt{3}\end{array}$

Hence proved.

(b)

To determine

To Find: The area when $s=7$ .

Answer to Problem 33WE

  $A=\frac{49\sqrt{3}}{4}$

Explanation of Solution

Given: An equilateral triangle has side of length of $s$

  

Area of triangle $ABC$

  $A=\frac{s^2}{4}\sqrt{3}$

Substitute $s=7$

We get

  $\begin{array}{l}A=\frac{s^2}{4}\sqrt{3}\\ A=\frac{7^2}{4}\sqrt{3}\\ A=\frac{49\sqrt{3}}{4}\end{array}$

Hence, $A=\frac{49\sqrt{3}}{4}$