Question

prove that area of equilateral triangle is root 3 by 4 into side square

Answer 1

Answer:

Proof:

Step 1: Since all the 3 sides of the triangle are same,

AB = BC = CA = a

Step 2: Find the altitude of the △ABC.  

Draw a perpendicular from point A to base BC, AD ⊥ BC

By using Pythagoras theorem

In △ ADC

h2 = AC2 - DC2

= a2 - (a2)2 [Because, DC = a2 ]

= a2 - a24

h = 3√a2

Step 3: We know that, Area of a triangle = 12 * Base * Height

= 12 * a * 3√a2

= 3√4a2

The area of a equilateral triangle = 3√4a2.

Step-by-step explanation:

Answer 2

Solution :

_______________________

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

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