Question

Taking 'a' as a side of an equilateral triangle, prove that area of an equilateral triangle is √3/4 a2

Answer 1

Given, side of an equilateral is a.

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We know that,

★ All the sides of an Equilateral triangle are equal.

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Therefore,

★ AB = BC = CA = a

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Now, Finding the altitude of ∆ABC,

Firstly draw a perpendicular 'h' from point A to base BC.

Therefore,

AD ⊥ BC

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(2,2)\qbezier(0,0)(0,0)(4,0)\qbezier(2,2)(4,0)(4,0)\put(2.7,1.1){ \boldsymbol / }\put(1.1,1.1){ \boldsymbol \backslash }\put(1.8,2.1){ \bf A }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\qbezier(3.6,0.4)(3.4,0.2)(3.6,0)\qbezier(0.4,0.4)(0.6,0.3)(0.4,0)\qbezier(2,2)(2,0)(2,0)\put(1.9,-0.4){ \bf D }\put(2.3,0){\line(0,1){0.3}}\put(2, 0.3){\line(1,0){0.3}}\put(0.8,1.5){ \bf a }\put(3,1.4){ \bf a }\put(1, - 0.5){ \bf \frac{a}{2} }\put(3, - 0.5){ \bf \frac{a}{2} }\end{picture}$

⠀⠀

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☯ Using Pythagoras theorem,

In ∆ADC

$\implies\sf h^2 = AC^2 - DC^2$

$\implies\sf h^2 = a^2 - \bigg( \dfrac{a}{2} \bigg)^2\qquad\qquad\bigg\lgroup\bf DC = \dfrac{1}{2} \times BC \bigg\rgroup$

$\implies\sf h^2 = a^2 - \dfrac{a^2}{4}$

$\implies\sf h^2 = \dfrac{4a^2 - a^2}{4}$

$\implies\sf h^2 = \dfrac{3a^2}{4}$

$\implies\sf \sqrt{h^2} = \sqrt{ \dfrac{3a^2}{4}}$

$\implies{\boxed{\frak{\purple{h = \dfrac{ \sqrt{3}}{2}a}}}}\;\bigstar$

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☯ Now, As we know that,

$\star\;{\boxed{\sf{\pink{Area_{\;(triangle)} = \dfrac{1}{2} \times b \times h}}}}$

Putting values,

$\implies\sf \dfrac{1}{2} \times a \times \dfrac{ \sqrt{3}a}{2}$

$\implies{\boxed{\frak{\purple{ \dfrac{ \sqrt{3}}{2}a^2}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;Area\;of\;an\; equilateral\; \triangle\;is\; \bf{ \dfrac{ \sqrt{3}}{2}a^2}.}}}$