Question

Show that area of an equilateral triangle is √3/4x,^2 where side is x.

Answer 1

Let an equilateral triangle with the side of x units,

Now,

Draw a perpendicular height from any of the vertex.

The perpendicular will divide the triangle in two equal parts.

So,

Length of side in each triangle, on which perpendicular will be drawn, will be half of the total length.

Therefore,

base ( when applying Pythagoras Theorem ) = x / 2

By Pythagoras Theorem,

Length of perpendicular = √[ ( side )^2 - ( side / 2 )^2 ]

Length of perpendicular = √x^2 - x^2 / 4

Length perpendicular = √( 4x^2 - x^2 ) / 4

= √3x^2 / 4

= √3 x / 2

Now,

As we have drawn a perpendicular,

Area of right angled triangle = 1 / 2 × x × √3 x / 2

⇒ √3 / 4 x^2

Hence, proved.

Answer 2

Answer:

√3/4 * x^2

Step-by-step explanation:

Let ABC is a equilateral triangle of side AB =BC =CA = x

Altitude in equilateral triangle is always bisect baseline and make right angle triangle

so lets find length of altitude AD by pythagoras theorem,

In right angle triangle  ABD

AB^2 = BD^2 + AD^2

x^2 = (x/2)^2 + AD^2

=> AD =√ (x^2 -x^2/4) = √3x^2/4 = (√3x^2)/2

now area of triangle of ABC =  2 * Area of ABD

=> 2 * 1/2 * base * height => BD * AD = x/2 * (√3x^2)/2 => √3/4 * x^2