Question

If x is the length of the median of an equilateral triangle. then its area is

• A. $x^2$
• B. $\frac{\sqrt{3}}{2}{x}^2$
• C. $\frac{\sqrt{3}}{3}{x}^2$
• D. $\frac{1}{2}{x}^2$

Answer 1

The correct option is C $\frac{\sqrt{3}}{3}{x}^2$

In an equilateral triangle the median $\left(x\right)$ is the height of the triangle.
Let each of the sides be $a$.
Then the median bisects the base each of length $\frac{a}{2}$.
Applying Pythagoras theorem, we have
$x^2+{\left(\frac{a}{2}\right)}^2=a^2$
$x^2=a^2-{\left(\frac{a}{2}\right)}^2=\frac{{3a}^2}{4}$
$a$ $=\frac{2x}{\sqrt{3}}$.
Now area of the triangle $=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times a\times x$.
$=\frac{1}{2}\times \frac{2x}{\sqrt{3}}\times x$
$=\frac{x^2}{\sqrt{3}}$
$=\frac{\sqrt{3}{x}^2}{3}$