Chapter 11.2, Problem 16WE

To determine

To Find: The area.

Answer to Problem 16WE

  $A=27\sqrt{3}sq.unit$

Explanation of Solution

Given: An equilateral triangle with height $9$ .

  

Using Pythagoras theorem

  $\begin{array}{l}{a}^2=h^2+{\left(\frac{a}{2}\right)}^2\\ h^2=\frac{3}{4}{a}^2\\ a^2=\frac{4}{3}{h}^2\\ a=\frac{2\sqrt{3}}{h}h\\ h=9\\ a=\frac{2\sqrt{3}}{h}(9)\\ a=6\sqrt{3}\end{array}$

Now the area is

  $\begin{array}{l}A=\frac{ah}{2}\\ A=\frac{\frac{2\sqrt{3}}{3}h\cdot h}{2}=\frac{\sqrt{3}}{3}{h}^2=\frac{\sqrt{3}}{3}(81)\\ A=27\sqrt{3}\end{array}$

Hence, $A=27\sqrt{3}sq.unit$