Question

Find the length of an equilateral triangle having an area of 16" title="\sqrt{3}" alt="\sqrt{3}" align="absmiddle" class="latex-formula"> $cm^{2}$

Answer 1

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, $a$.

Let $h$ be the height of the triangle:

$(\frac{a}{2})^2+h^2=a^2$

Let's solve for h in terms of $a$.

$\frac{a^2}{4}+h^2=a^2$

Subtract $\frac{a^2}{4}$ on both sides:

$h^2=a^2-\frac{a^2}{4}$

$h^2=\frac{4}{4}a^2-\frac{1}{4}a^2$

$h^2=\frac{4-1}{4}a^2$

$h^2=\frac{3}{4}a^2$

Now square root both sides:

$h=\frac{\sqrt{3}}{2}a$

So the area of the triangle is $\frac{1}{2} \cdot a \cdot \frac{\sqrt{3}}{2}a$.

Let's simplify that a bit: $\frac{\sqrt{3}}{4}a^2$.

We are also given a numerical value for the area, $16\sqrt{3}$.

So this will give us the equation $\frac{\sqrt{3}}{4}a^2=16\sqrt{3}$ so that we can solve for $a$.

Multiply both sides by $\frac{4}{\sqrt{3}}$:

$a^2=16 \sqrt{3} \cdot \frac{4}{\sqrt{3}}$

Simplify the right hand side:

$a^2=16 \cdot 4$

$a^2=64$

Take the square root of both sides:

$a=\sqrt{64}$

$a=8$

Answer 2

Answer:

Thus , The length of each side of the given equilateral triangle is 8