Question

In an equilateral triangle with side a prove that area = 3 root by 4 a ka square

Answer 1

with heron's formula
s=1/2×(a+b+c)
=1/2×(a+a+a)
=3a/2
area of triangle =
$\sqrt{s(s - a)(s - b)(s - c)}$
=
$\sqrt{3a \div 2 \times (3a \div 2 - a \times (3a \div 2 - a)(3a \div 2 - a)}$
=
$\sqrt{3a \div 2 \times (a \div 2 \times a \div 2 \times a \div 2}$
$\sqrt{3 } \div 4 \times {a}^{2}$

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 !