Question

Prove that :

The area of an equilateral triangle is $\frac{root3}{4} x^2$, where x is the side of the triangle.

Answer 1

$\textsf{\huge{Answer}}$

${\bold{To\:prove,}}$

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

$\textsf{\huge{Proof}}$

${\bold{Construction\::}}$

Draw a equilateral triangle say ABC with side as 'x' given in the question. Also draw AD perpendicular to BC.

In△ABD and△ACD,

AD = DA (common side)

AB = AC (side of an equilateral triangle)

BD = CD (⊥ bisector)

∴△ABD ≅△ACD

Now, we have to find ar(△ABD) by using pythagorean theorem. It states that, the square of hypotenuse is equal to the other two sides.

Now, we have to find the height,

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

$h^2 = x^2 - (\frac{x^2}{4})$

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

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

We know that,

Area of a △= $\frac{1}{2} \times base \times height.$

Area of △= $\frac{1}{2} \times x \times \frac{\sqrt{3}}{2} x$

Area of △= $\frac{\sqrt{3}}{4} x^2$

Hence, the area of an equilateral triangle is $\frac{\sqrt{3}}{4} x^2$

${\bold{Hence, the\:proof.}}$