Question

In a equilateral triangle with side 'a' .Prove that the area of the triangle is ((3^1/2)/4)*a^3

Answer 1

In equilateral triangle ABC,

AB= BC = CA.

Draw a line B to touch AC, mark it as D.This is the height of the triangle.

We know that, Area of triangle = 1/2 * base * height.

Now, Let's find height BD which we don't know.

Consider the sides AB= BC=CA=s

Now, We can see that AD+DC= AC,

DC=AC/2 [ As BD is bisecting the angle B and the two triangles ∆ABD,∆ACD are congruent ]

Hence, DC=AC/2=s/2.

We know that, as Angle BDC=90°,
By pythagoras theorem ;

BC²=DC²+BD²

s²=(s/2)²+BD²

BD²=s²-(s²/4)=(4s²-s²)/4

BD=√(3s²/4)=√3s/2

Now area=1/2* base * height

=1/2*s* √3s/2

=√3/4s²

If the given side is a ,then area = √3/4a²

HENCE PROVED!!!^^