Question

using heroine formula find the area of an equilateral triangle of side a unit ​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\sf\underline{\underline{\:\:\:\:Given:-\:\:\:\:}}$

• Side of equilatera∆ = a

$\sf\underline{\blue{\:\:\:\:To\: Find:-\:\:\:\:}}$

The area of triangle

Now

$\sf\underline{\pink{\:\:\:\: Solution:-\:\:\:\:}}$

$\boxed{\sf{Area\:of\: \triangle=\sqrt{s(s-a)(s-b)(s-c)}}}$

• a,b, and c are the sides of ∆

$\sf \implies s=\dfrac{a+b+c}{2}\\ \\ \sf\:\:a,b,\:and\:c= a\\ \\ \sf\implies s=\dfrac{a+a+a}{2}\\ \\ \sf\implies s= \dfrac{3a}{2}$

$\sf \bullet (s-a)=\dfrac{3a}{2}-a\\ \\ \sf (s-a)=\dfrac{3a-2a}{2}\\ \\ \sf \bullet (s-a)=\dfrac{a}{2}$

• side =a

$\sf \bullet (s-b)=\dfrac{3a}{2}-a\\ \\ \sf (s-b)=\dfrac{3a-2a}{2}\\ \\ \sf \bullet (s-b)=\dfrac{a}{2}$

• side =a

$\sf \bullet (s-c)=\dfrac{3a}{2}-a\\ \\ \sf (s-c)=\dfrac{3a-2a}{2}\\ \\ \sf \bullet (s-c)=\dfrac{a}{2}$

Now the area

$\sf\implies Area=\sqrt{s(s-a)(s-b)(s-c)}\\ \\ \sf\implies Area=\sqrt{\dfrac{3a}{2}\times \dfrac{a}{2}\times \dfrac{a}{2}\times \dfrac{a}{2}}\\ \\ \sf\implies Area=\sqrt{\dfrac{3a{}^{4}}{16}}\\ \\ \sf\implies Area=\dfrac{\sqrt{3}}{4}a{}^{2}$

$\boxed{\sf{Area\:of\: equilateral\triangle=\dfrac{\sqrt{3}}{4}a{}^{2}}}$