Question

if a the side equilateral triangle then prove A=root3/4 a²

Answer 1

Length of the side = a

Find semiperimeter:

$\text {Perimeter = } a + a + a = 3a$

$\text {Semiperimeter = } 3a \div 2 = \dfrac{3a}{2}$

Find area:

$\text {Area = } \sqrt{\dfrac{3a}{2}\bigg(\dfrac{3a}{2}- a\bigg) \bigg(\dfrac{3a}{2}- a\bigg) \bigg(\dfrac{3a}{2}- a\bigg) }$

$\text {Area = } \sqrt{\dfrac{3a}{2}\bigg(\dfrac{3a}{2}- a\bigg)^3}$

$\text {Area = } \sqrt{\dfrac{3a}{2}\bigg(\dfrac{3a-2a}{2}\bigg)^3}$

$\text {Area = } \sqrt{\dfrac{3a}{2}\bigg(\dfrac{a}{2}\bigg)^3}$

$\text {Area = } \sqrt{\dfrac{3a}{2}\bigg(\dfrac{a^3}{2^3}\bigg)}$

$\text {Area = } \sqrt{\dfrac{3a^4}{16}$

$\text {Area = } \dfrac{\sqrt{3a^4} }{\sqrt{16} }$

$\text {Area = }\dfrac{a^2\sqrt{3} }{4 }$

$\text {Area = }\dfrac{\sqrt{3} }{4 } a^2$

Answer 2

Solution :

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Derivation of Area of an equilateral triangle ;

Let ABC be an equilateral triangle with sides 'a'. Now, draw AD perpendicular to BC.

Here, we have ΔABD = ΔADC.

We will find area of ΔABD using pythagorean theorem, according to which, the square of hypotenuse is equal to the sum of the squares of the other two sides.

Here, we have ;

$\sf a {}^{2} = h {}^{2} + (\frac{a}{2} ) {}^{2} \\ \\ \sf h {}^{2} = a {}^{2} - \frac{a {}^{2} }{4} \\ \\ \sf h {}^{2} = \frac{3a {}^{2} }{4} \\ \\ \sf h = \frac{ \sqrt{3} }{2} a$
Now, we get the height ;

$\sf area \: of \: \Delta = \frac{1}{2} \times base \times height \\ \\ \sf area \: of \: \Delta = \frac{1}{2} \times a \times \frac{ \sqrt{3} }{2} a \\ \\ \sf area \: of \: \Delta = \frac{ \sqrt{3} }{4} a {}^{2}$

Hence, area of equilateral triangle is

$\sf area \: of \: \Delta = \frac{ \sqrt{3} }{4} a {}^{2}$

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Thanks for the question !