Question

Prove that area of equilateral triangle is root 3 by 4 a square

Answer 1

let AB = a , AC = b and BC = c.

since, the triangle is an Equilateral triangle.

$\therefore$All its sides are equal.

=> AB = BC = AC

$\bf{\underline{Construction}}$ : Draw an altitude AD which is perpendicular to BC such that it divides BC into 2 equal parts ; BD and CD.

now, ∆ABC is divided into 2 parts : right ∆ABD and right ∆ACD.

In ∆ABD,

it is a right angled triangle.

Hypotenuse (AB) = a

base (BD) = $\frac{1}{2}$ × BC = $\frac{1}{2}$ × a = $\frac{a}{2}$

and , perpendicular (AD) = ?

By Pythagoras theorem,

(Hypotenuse)² = (base)² + (perpendicular)²

=> (AB)² = (BD)² + (AD)²

=> (a)² = ($\frac{a}{2}$)² + (AD)²

=> (AD)² = (a)² - ($\frac{a}{2}$)²

use the identity , a² - b² = (a+b)(a-b) :-

=> (AD)² = ( a + $\frac{a}{2}$ )( a - $\frac{a}{2}$ )

=> (AD)² = ($\frac{2a+a}{2}$)($\frac{2a-a}{2}$)

=> (AD)² = ($\frac{3a}{2}$)($\frac{a}{2}$)

Multiply the terms on Right hand side,

=> (AD)² = $\frac{3{a}^{2}}{4}$

=> AD = $\sqrt{\frac{3{a}^{2}}{4}}$

=> AD = $\frac{\sqrt{3} }{2}a$

then, we know that :-

Area of a triangle ,

= $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AD

= $\frac{1}{2}$ × a × $\frac{\sqrt{3}}{2}a$

$\bf \huge = \frac{ \sqrt{3} }{4} {a}^{2}$

hence , proved that area of equilateral triangle = $\frac{\sqrt{3}}{4}$a²

Answer 2

Given

Side of square = a

To prove

That area of an equilateral ∆ = √3/4 × a²


Here, we will apply Heron's Formula ;)

According to Heron's Formula, area of a ∆

=
$\sqrt{s(s - a)(s - b)(s - c)}$


Where s is the semiperimeter, and a,b and c are sides.
s = (a + b + c)/2

For an equilateral triangle, all sides are equal

=> a = b = c

=> s = (a + b + c)/2

=> s = (a + a + a)/2

=> s = 3a/2

Now substitute the value

$\sqrt{s(s \times a)(s - b)(s - c)}$

$\sqrt{ \frac{3a}{2}( \frac{3a}{2} - a)( \frac{3a}{2} - a)( \frac{3a}{2} - a)}$


Now,
$\sqrt{ \frac{3a}{2}( \frac{3a}{2} - \frac{2a}{2} )( \frac{3a}{2} - \frac{2a}{2} )( \frac{3a}{2} - \frac{2a}{2} )}$

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


Hence Proved ;)