In an equilateral triangle with side \'a\' prove that its area = root

1.	In an equilateral triangle with side \'a\' prove that its area = root 3/4a square
Answer» According to the question, we have to prove that in an equilateral triangle with side a, area =\xa0{tex}\\frac { \\sqrt { 3 } } { 4 } a ^ { 2 }{/tex}Let\xa0{tex}\\triangle{/tex}ABC be an equilateral triangle with side a.Then, AB = AC = BC = a.Draw AD{tex}\\perp{/tex}BC.In\xa0{tex}\\triangle{/tex}ADB and\xa0{tex}\\triangle{/tex}ADC, we haveAB = AC ({tex}\\Delta ABC{/tex}\xa0equilateral triangle){tex}\\angle{/tex}ADB=\xa0{tex}\\angle{/tex}ADC = 90° (AD is altitude)\xa0AD\xa0=\xa0AD\xa0(Common){tex}\\therefore \\Delta A D B \\cong \\triangle A D C{/tex}\xa0[ RHS Congurency rule]\xa0{tex}\\therefore{/tex}\xa0BD = DC =\xa0{tex}\\frac { a } { 2 }{/tex}From right\xa0{tex}\\triangle{/tex}ADB, we haveAB2\xa0= AD2\xa0+ BD2\xa0(By Pythagoras theorem){tex}\\Rightarrow{/tex}\xa0AD =\xa0{tex}\\sqrt { A B ^ { 2 } - BD ^ { 2 } }{/tex}{tex}\\Rightarrow A D = \\sqrt { a ^ { 2 } - \\left( \\frac { a } { 2 } \\right) ^ { 2 } }{/tex}{tex}\\Rightarrow A D = \\sqrt { a ^ { 2 } - \\frac { a ^ { 2 } } { 4 } }{/tex}{tex}\\Rightarrow A D = \\sqrt { \\frac { 3 a ^ { 2 } } { 4 } }{/tex}So, the altitude is, AD =\xa0{tex}\\frac { \\sqrt { 3 } a } { 2 }{/tex}Area of {tex}\\triangle{/tex}ABC =\xa0{tex}\\frac { 1 } { 2 } \\times B C \\times A D{/tex}{tex}= \\frac { 1 } { 2 } \\times a \\times \\frac { \\sqrt { 3 } a } { 2 }{/tex}{tex}= \\frac { \\sqrt { 3 } a ^ { 2 } } { 4 }{/tex}Hence proved.

In an equilateral triangle with side \'a\' prove that its area = root 3/4a square