$\"\"$

In equilateral triangle all sides are equal.

Consider $\"x\"$ to be one of the side of a triangle.

Find the height of the triangle using the figure shown below.

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From Pythagorean theorem,

$\"x^2=\\textrm{height}^2+(\\frac{x}{2})^2\"$

$\"\\textrm{height}^2=x^2-(\\frac{x}{2})^2\"$

$\"\\textrm{Height}=\\sqrt{x^2-(\\frac{x}{2})^2}\"$

$\"=\\sqrt{x^2-\\frac{x^2}{4}}\"$

$\"=\\sqrt{\\frac{4x^2-x^2}{4}}\"$

$\"=\\sqrt{\\frac{3x^2}{4}}\"$

$\"=\\frac{\\sqrt{3}}{2}x\"$

$\"\\textrm{Height}=\\frac{\\sqrt{3}}{2}x\"$ .

Area of a triangle is $\"A=\\frac{1}{2}(\\textrm{Base})(\\textrm{Height})\"$ .

$\"A=\\frac{1}{2}(x)(\\frac{\\sqrt{3}}{2}x)\"$

$\"=\\frac{\\sqrt{3}}{4}x^2\"$

$\"A(x)=\\frac{\\sqrt{3}}{4}x^2\"$ .

Domain:

The area of the function is $\"A(x)=\\frac{\\sqrt{3}}{4}x^2\"$ .

All possible values of $\"x\"$ is domain of the function.

Area of the triangle should be a greater than zero.

$\"\\frac{\\sqrt{3}}{4}x^2>0\"$

$\"x^2>0\"$

$\"x>0\"$

The domain of the function is $\"x>0\"$ .

$\"A(x)=\\frac{\\sqrt{3}}{4}x^2\"$ .

The domain of the function $\"A(x)=\\frac{\\sqrt{3}}{4}x^2\"$ is $\"x>0\"$ .