Answer:
8 cm
Step-by-step explanation:
An equilateral triangle has 3 sides all being congruent to each other.
If I draw a line segment from one vertex to the opposite side at it's midpoint, I would have halved the triangle into two right triangles.
Let's each side of this equilateral triangle have measurement, [tex]a[/tex].
Let [tex]h[/tex] be the height of the triangle:
[tex](\frac{a}{2})^2+h^2=a^2[/tex]
Let's solve for h in terms of [tex]a[/tex].
[tex]\frac{a^2}{4}+h^2=a^2[/tex]
Subtract [tex]\frac{a^2}{4}[/tex] on both sides:
[tex]h^2=a^2-\frac{a^2}{4}[/tex]
[tex]h^2=\frac{4}{4}a^2-\frac{1}{4}a^2[/tex]
[tex]h^2=\frac{4-1}{4}a^2[/tex]
[tex]h^2=\frac{3}{4}a^2[/tex]
Now square root both sides:
[tex]h=\frac{\sqrt{3}}{2}a[/tex]
So the area of the triangle is [tex]\frac{1}{2} \cdot a \cdot \frac{\sqrt{3}}{2}a[/tex].
Let's simplify that a bit: [tex]\frac{\sqrt{3}}{4}a^2[/tex].
We are also given a numerical value for the area, [tex]16\sqrt{3}[/tex].
So this will give us the equation [tex]\frac{\sqrt{3}}{4}a^2=16\sqrt{3}[/tex] so that we can solve for [tex]a[/tex].
Multiply both sides by [tex]\frac{4}{\sqrt{3}}[/tex]:
[tex]a^2=16 \sqrt{3} \cdot \frac{4}{\sqrt{3}}[/tex]
Simplify the right hand side:
[tex]a^2=16 \cdot 4[/tex]
[tex]a^2=64[/tex]
Take the square root of both sides:
[tex]a=\sqrt{64}[/tex]
[tex]a=8[/tex]
