In the figure given above, it is the area shaded in yellow.<\/span><\/p>\n <\/span><\/p>\n In an equilateral triangle, the measure of all the three sides is equal to each other.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n In\u00a0 \u25b3ABC<\/span><\/p>\n Hence it is an equilateral triangle.<\/span><\/p>\n <\/p>\n Where, a = length of a side<\/span><\/p>\n Derivation of the formula<\/strong><\/p>\n There are many ways to derive the above formula and here we are going to try the following 3 ways.<\/span><\/span><\/p>\n <\/p>\n <\/b><\/p>\n \u25b3<\/span>ABC is an equilateral triangle.<\/span><\/p>\n AB = BC = CA = a<\/span><\/p>\n AD is a perpendicular line drawn from A on line BC and this line AD bisects BC.<\/span><\/p>\n \\mathrm{BD}=\\mathrm{DC}=\\frac{a}{2}<\/span><\/span><\/p>\n Hence <\/span>\u25b3<\/span>ADC is a right-angled triangle with <\/span>\u2220<\/span>ADC = 90\u00b0<\/span><\/p>\n According to Pythagoras theorem, in\u00a0 <\/span>\u25b3<\/span>ADC<\/span><\/p>\n \\mathrm{AD}^{2} +\\mathrm{DC}^{2}=\\mathrm{CA}^{2} <\/span><\/span><\/p>\n <\/p>\n <\/span><\/p>\n In triangle ABC,\u00a0<\/span><\/p>\n AD =height\u00a0<\/span><\/p>\n BC = base<\/span><\/p>\n Area of triangle A B C=\\frac{1}{2} \\times base \\times height<\/span><\/span><\/p>\n=\\frac{1}{2} \\times \\mathrm{BC} \\times \\mathrm{AD} <\/span>\n <\/span><\/p>\n Hence, it is proved that the area of the equilateral triangle ABC is \\frac{\\sqrt{3}}{4} a^{2}<\/span><\/span>\u00a0<\/span>, where a is the length of each side.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n According to heron\u2019s formula, the area of a triangle having sides a, b and c is given by the formula<\/span><\/p>\n <\/span><\/p>\n \\text { Area }=\\sqrt{s(s-a)(s-b)(s-c)}<\/span><\/span><\/p>\n Where s = semi perimeter of the triangle Where s = semi perimeter of the triangle <\/span><\/span><\/p>\n <\/span><\/p>\n In an equilateral triangle,<\/span><\/p>\n <\/span><\/p>\n \\text { 2. } s=\\frac{a+b+c}{2}=\\frac{a+a+a}{2}=\\frac{3 a}{2}<\/span><\/span><\/p>\n <\/span><\/p>\n Area of an equilateral triangle\u00a0\u00a0<\/span><\/p>\n =\\sqrt{\\frac{3 a^{4}}{16}}=\\frac{\\sqrt{3}}{4} \\mathrm{a}^{2}<\/span><\/span><\/p>\n <\/p>\n <\/span><\/p>\n Hence, it is proved that the area of the equilateral triangle ABC is \\frac{\\sqrt{3}}{4} a^{2}<\/span><\/span>, where a is the length of each side.<\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n \u25b3<\/span>ABC is an equilateral triangle.<\/span><\/p>\n AB = BC = CA = a<\/span><\/p>\n AD is a perpendicular line drawn from A on line BC and this line AD bisects BC.<\/span><\/p>\n \\mathrm{BD}=\\mathrm{DC}=\\frac{a}{2}<\/span><\/span><\/p>\n <\/span><\/p>\n Also, we know that<\/span><\/p>\n \u2220<\/span>ABC =<\/span>\u2220<\/span>BCA = <\/span>\u2220<\/span>CAB = 60\u00b0<\/span><\/p>\n In \u25b3ABD,\u00a0<\/span><\/p>\n So, we can write\u00a0<\/span><\/p>\n AD = AB sin 60\u00b0<\/span><\/p>\n <\/span><\/p>\n \\Rightarrow \\mathrm{AD}=\\mathrm{a} \\times \\frac{\\sqrt{3}}{2} <\/span><\/span><\/p>\n <\/p>\n In triangle ABC,\u00a0<\/span><\/p>\n AD =height\u00a0<\/span><\/p>\n BC = base.<\/span><\/p>\n Area of triangle ABC <\/span><\/p>\n Hence, it is proved that the area of the equilateral triangle ABC is \\frac{\\sqrt{3}}{4} a^{2}<\/span><\/span>, where a is the length of each side.<\/span><\/p>\n <\/span><\/p>\n It is an equilateral triangle <\/span> Side = a = 8 cm<\/span><\/p>\n \\text { Area }= \\frac{\\sqrt{3}}{4} \\mathrm{a}^{2} <\/span><\/span><\/p>\nWhat is an equilateral triangle?<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-equilateral-triangle-01.png\")
<\/span><\/p>\n\n
The formula for finding the area<\/b><\/h2>\n\\text { Area }=\\frac{\\sqrt{3}}{4} \\mathrm{a}^{2}<\/span>\n
\n
Derivation using Pythagoras theorem<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-equilateral-triangle-02.png\")
<\/b><\/p>\n
\\Rightarrow \\mathrm{AD}^{2}=\\mathrm{CA}^{2}-\\mathrm{DC}^{2} <\/span><\/span><\/p>\n
\\Rightarrow \\mathrm{AD}^{2}=\\mathrm{a}^{2}-\\left(\\frac{a}{2}\\right)^{2} <\/span><\/span><\/p>\n
\\Rightarrow \\mathrm{AD}^{2}=\\frac{3}{4} \\mathrm{a}^{2} <\/span><\/span><\/p>\n
\\Rightarrow \\mathrm{AD}=\\left(\\frac{3}{4} \\mathrm{a}^{2}\\right)^{1 \/ 2} <\/span><\/span><\/p>\n
\\Rightarrow\u00a0 \\mathrm{AD}=\\frac{\\sqrt{3}}{2} \\mathrm{a}<\/span><\/span><\/p>\n
=\\frac{1}{2} \\times \\mathrm{a} \\times\\left(\\frac{\\sqrt{3}}{2} \\mathrm{a}\\right) <\/span><\/span><\/p>\n
=\\frac{\\sqrt{3}}{4} \\mathrm{a}^{2}<\/span>
<\/span><\/span><\/p>\nDerivation using heron\u2019s formula<\/b><\/h2>\n
\n
<\/span>(all sides are equal)<\/span><\/li>\n<\/ol>\n
=\\sqrt{s(s-a)(s-b)(s-c)} <\/span><\/span><\/p>\n
=\\sqrt{\\frac{3 a}{2}\\left(\\frac{3 a}{2}-a\\right)\\left(\\frac{3 a}{2}-a\\right)\\left(\\frac{3 a}{2}-a\\right)} <\/span><\/span><\/p>\n
=\\sqrt{\\frac{3 a}{2} \\times \\frac{a}{2} \\times \\frac{a}{2} \\times \\frac{a}{2}} <\/span>
<\/span><\/p>\nDerivation using trigonometry<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-equilateral-triangle-03.png\")
<\/b><\/p>\n\n
\n
\\Rightarrow \\mathrm{AD}=\\frac{\\sqrt{3}}{2} \\mathrm{a}<\/span><\/span><\/p>\n
=\\frac{1}{2} \\times \\text { base } \\times \\text { height } <\/span><\/span><\/p>\n
=\\frac{1}{2} \\times \\mathrm{BC} \\times \\mathrm{AD} <\/span><\/span><\/p>\n
=\\frac{1}{2} \\times \\mathrm{a} \\times \\frac{\\sqrt{3}}{2} \\mathrm{a} <\/span><\/span><\/p>\n
=\\frac{\\sqrt{3}}{4} \\mathrm{a}^{2}<\/span>
<\/span><\/p>\nSolved Examples<\/b><\/h2>\n
\n
<\/span>(all sides are equal)<\/span><\/p>\n
=\\frac{\\sqrt{3}}{4} 8^{2} <\/span><\/span><\/p>\n
=\\frac{\\sqrt{3}}{4} \\times 64 <\/span><\/span><\/p>\n