Question

Prove that the area of an equilateral triangle is equal to √3÷4×a²  where a is the side of the triangle. 

Answer 1

Length of each side of equilateral triangle = a

Perimeter of equilateral triangle = 3 × a

Let s be the semi-perimeter.



By Heron’s formula,



Answer 2

AnswEr:

Given : An equilateral triangle ABC such that AB = BC = CA = a.

To prove : ar (∆ABC) -

$= \tt \frac{ \sqrt{3} }{4} \: \: {a}^{2}$

Construction : Draw AD perpendicular to BC.

Proof : In triangles ABD and ACD, we have

$\qquad \sf \: AB = AC \quad( \therefore \triangle \: ABC \: is \: equialteral) \\ \\ \qquad \sf \angle \: ABD = \angle \: ACD \quad \: (each \: equal \: to \: 90 \degree) \\ \\ \qquad \sf \: AD = AD \quad(common \: side)$

So, by RHS criterion of congruence, we have

$\qquad \sf \triangle \: ABD \: \cong \: \triangle \: ACD \\ \rightarrow \qquad \sf \: BD = CD \\ \\ \rm \: But - \: \: \: \sf \: BD + CD = a \\ \\ \implies \sf \: BD = CD = \frac{a}{2}$

Now, in Right triangles ABD, we have

$\qquad \rm {AB}^{2} = {AD}^{2} + {BD}^{2} \\ \\ \rm \rightarrow \qquad \: {a}^{2} = {AD}^{2} + ( \frac{a}{2}) {}^{2} \\ \\ \rightarrow \rm {AD}^{2} = {a}^{2} - { \frac{a}{4} }^{2} \implies \: {AD}^{2} = \frac{ {3a}^{2} }{4} \\ \\ \rm \rightarrow \: AD = \frac{ \sqrt{3}a }{2} \\ \\ \therefore \sf \: ar( \triangle \: ABC) = \frac{1}{2} (BC \times AD) \\ \\ \sf = \frac{1}{2} (a \times \frac{ \sqrt{3} }{2} a) = \frac{ \sqrt{3} }{4} {a}^{2}$