Question

Prove that the area of an equilateral triangle is equal to âš3â„4 a 2 , where a is the side of the triangle.

Answer 1

Hope it helps you. See the attachment.

Answer 2

$\underline{\textbf{Hey mate Here is your answer }}$

$\underline{\textbf{To Prove }}$

$\boxed{\mathbf{Area\: Of\: an\: Equilateral \:Triangle \:= \:\frac{ \sqrt{3} }{4} {side}^{2} \:}}$

$\underline{\textbf{Proof ---}}$

Now let

$\textbf{∆ABC Be An Equilateral Triangle}$

$\textbf{And Measure Of Side = a }$

Now By $\textbf{Heron's Formula}$ That is

$\boxed{\mathbf{Area\:Of\; Triangle\:= \: \sqrt{s(s\: -\: a)(s\: - \:b)(s \:- \:c)} }}$

Therefore By This We Have

$\mathbf{ar(\triangle\:ABC\:)\:=\: \sqrt{s(s \:- \:a)(s\: -\: b)(s \:- \:c)} \:}$ ------ EQ ( 1 )

Where

S = Semi Perimeter And Sides = a , b , c

Therefore

$\mathbf{S \:= \:\frac{a+b+c}{2}}$

Since ∆ABC Is an Equilateral Triangle with a As side , So

$\mathbf{S \:= \:\frac{3a}{2}}$

Now Putting this value In EQ 1 We Have ​

$\mathbf{ar(\triangle\:ABC\:)\:=\: \sqrt{\frac{3a}{2}( \frac{3a}{2}\:- \:a)(\frac{3a}{2}\: -\: a)(\frac{3a}{2}\:- \:a)}}$

Now By Taking LCM And solving the Bracket We Have ​

$\mathbf{ar(\triangle\:ABC\:)\:=\: \sqrt{\frac{3a}{2}( \frac{a}{2}\:)(\frac{a}{2}\:)(\frac{a}{2}\:)}}$

That Is ​

$\mathbf{ar(\triangle\:ABC\:)\:=\: \sqrt{3 \times {( \frac{a}{2}) }^{2} {( \frac{a}{2}) }^{2} }}$

That Is Now Taking Squares Out We Have ​

$\mathbf{ar(\triangle\:ABC\:)\:=\: \frac{a}{2} \times \frac{a}{2} \: \sqrt{3 }}$

That Is

$\mathbf{ar(\triangle\:ABC\:)\:=\: \frac{ {a}^{2} }{4} \: \sqrt{3 }}$

Therefore We Have

$\boxed{\mathbf{ar(\triangle\: ABC\:)\: = \:\frac{ \sqrt{3} }{4} {a}^{2} }}$

$\underline{\mathbf{Hence\: Proved \: That }}$

$\boxed{\mathbf{ Area\: Of \: An\: Equilateral\:Triangle \:Is \:=\:\frac{ \sqrt{3} }{4} {a}^{2} }}$

$\large{\blue{Thanks...}}$

$\underline{\bold{\star\:\: BrainlyKing5\:\:\star}}$