The area of a triangle is given by the expression:
[tex]A_{\text{triangle}}=\frac{1}{2}b\cdot h[/tex]Where b is the basis and h is the height.
Then, in the case of our problem:
[tex]\frac{1}{2}b\cdot h=36\sqrt[]{3}[/tex]There is something particular of the equilateral triangles, see the diagram below:
Notice that we can use trigonometric identities to find an expression that involves the length of a side and the height.
We can use the sine function
[tex]\begin{gathered} \sin (\theta)=\frac{O}{H} \\ \Rightarrow\sin (60degree)=\frac{h}{b} \end{gathered}[/tex]Furthermore, the exact value of sin(60°) is sqrt(3)/2
[tex]\sin (60degree)=\frac{\sqrt[]{3}}{2}[/tex]Therefore:
[tex]\begin{gathered} \frac{\sqrt[]{3}}{2}=\frac{h}{b} \\ \Rightarrow b=\frac{h}{\frac{\sqrt[]{3}}{2}}=\frac{2h}{\sqrt[]{3}} \end{gathered}[/tex]Finally, we can substitute this last result in the equation for the area of the triangle:
[tex]\begin{gathered} \frac{1}{2}b\cdot h=36\sqrt[]{3} \\ \Rightarrow\frac{1}{2}(\frac{2h}{\sqrt[]{3}})\cdot h=36\sqrt[]{3} \\ \Rightarrow h^2=\frac{2\cdot\sqrt[]{3}}{2}(36\sqrt[]{3}) \end{gathered}[/tex]We only need to simplify and solve for h, as shown below:
[tex]\begin{gathered} \Rightarrow h^2=36\cdot3 \\ \Rightarrow h=\sqrt[]{36\cdot3}=6\sqrt[]{3} \end{gathered}[/tex]The solution is then h=6*sqrt(3), option C
Explanation for sin(60°)=sqrt(3)/2
Actually, we can use an equilateral triangle which basis is equal to 1 to prove this:
The blue line is the height, and notice that it crosses the basis in the middle, so the orange segment is equal to 1/2. We can then use the Pythagoras Theorem to find the value of the blue segment:
[tex]\begin{gathered} x^2+(\frac{1}{2})^2=1^2=1 \\ \Rightarrow x=\sqrt[]{1-\frac{1}{4}}=\sqrt[]{\frac{3}{4}}=\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]Furthermore, we know that the identity sin(theta)=O/H, then:
[tex]\sin (60degree)=\frac{O}{H}=\frac{\frac{\sqrt[]{3}}{2}}{1}=\frac{\sqrt[]{3}}{2}[/tex]
