Area of equilateral triangle – formula, derivation and examples

Area of equilateral triangle

It is the area enclosed by three equal sides of a triangle.

In the figure given above, it is the area shaded in yellow.

What is an equilateral triangle?

In an equilateral triangle, the measure of all the three sides is equal to each other.

In  △ABC

1. AB = BC = AC = a
2. ∠ABC =∠BCA = ∠CAB = 60°

Hence it is an equilateral triangle.

The formula for finding the area

Where, a = length of a side

Derivation of the formula

There are many ways to derive the above formula and here we are going to try the following 3 ways.

1. Using Pythagoras theorem
2. Using herons’ formula
3. Using trigonometry

Derivation using Pythagoras theorem

△ABC is an equilateral triangle.

AB = BC = CA = a

AD is a perpendicular line drawn from A on line BC and this line AD bisects BC.

\mathrm{BD}=\mathrm{DC}=\frac{a}{2}

Hence △ADC is a right-angled triangle with ∠ADC = 90°

According to Pythagoras theorem, in  △ADC

\mathrm{AD}^{2} +\mathrm{DC}^{2}=\mathrm{CA}^{2}

\Rightarrow \mathrm{AD}^{2}=\mathrm{CA}^{2}-\mathrm{DC}^{2}

\Rightarrow \mathrm{AD}^{2}=\mathrm{a}^{2}-\left(\frac{a}{2}\right)^{2}

\Rightarrow \mathrm{AD}^{2}=\frac{3}{4} \mathrm{a}^{2}

\Rightarrow \mathrm{AD}=\left(\frac{3}{4} \mathrm{a}^{2}\right)^{1 / 2}

\Rightarrow  \mathrm{AD}=\frac{\sqrt{3}}{2} \mathrm{a}

In triangle ABC, 

AD =height 

BC = base

Area of triangle A B C=\frac{1}{2} \times base \times height

=\frac{1}{2} \times \mathrm{a} \times\left(\frac{\sqrt{3}}{2} \mathrm{a}\right)

=\frac{\sqrt{3}}{4} \mathrm{a}^{2}

Hence, it is proved that the area of the equilateral triangle ABC is \frac{\sqrt{3}}{4} a^{2} , where a is the length of each side.

Derivation using heron’s formula

According to heron’s formula, the area of a triangle having sides a, b and c is given by the formula

\text { Area }=\sqrt{s(s-a)(s-b)(s-c)}

Where s = semi perimeter of the triangle Where s = semi perimeter of the triangle

In an equilateral triangle,

1. a = b = c 
   (all sides are equal)

\text { 2. } s=\frac{a+b+c}{2}=\frac{a+a+a}{2}=\frac{3 a}{2}

Area of an equilateral triangle  

=\sqrt{s(s-a)(s-b)(s-c)}

=\sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)}

=\sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}

=\sqrt{\frac{3 a^{4}}{16}}=\frac{\sqrt{3}}{4} \mathrm{a}^{2}

Hence, it is proved that the area of the equilateral triangle ABC is \frac{\sqrt{3}}{4} a^{2}, where a is the length of each side.

Derivation using trigonometry

△ABC is an equilateral triangle.

AB = BC = CA = a

AD is a perpendicular line drawn from A on line BC and this line AD bisects BC.

\mathrm{BD}=\mathrm{DC}=\frac{a}{2}

Also, we know that

∠ABC =∠BCA = ∠CAB = 60°

In △ABD, 

1. ∠ABD = 60°
2. ∠ADB = 90°

So, we can write 

AD = AB sin 60°

• AD = a sin 60

\Rightarrow \mathrm{AD}=\mathrm{a} \times \frac{\sqrt{3}}{2}

\Rightarrow \mathrm{AD}=\frac{\sqrt{3}}{2} \mathrm{a}

In triangle ABC, 

AD =height 

BC = base.

Area of triangle ABC
=\frac{1}{2} \times \text { base } \times \text { height }

=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AD}

=\frac{1}{2} \times \mathrm{a} \times \frac{\sqrt{3}}{2} \mathrm{a}

=\frac{\sqrt{3}}{4} \mathrm{a}^{2}

Hence, it is proved that the area of the equilateral triangle ABC is \frac{\sqrt{3}}{4} a^{2}, where a is the length of each side.

Solved Examples

1. A triangle has three equal sides and each side measures 8 cm. Find the area of this triangle.

It is an equilateral triangle
(all sides are equal)

Side = a = 8 cm

\text { Area }= \frac{\sqrt{3}}{4} \mathrm{a}^{2}

=\frac{\sqrt{3}}{4} 8^{2}

=\frac{\sqrt{3}}{4} \times 64

=16 \sqrt{3} \mathrm{~cm}^{2}

2. The area of an equilateral triangle is 19 \sqrt{3} \mathrm{~cm}^{2}. Find the perimeter of this triangle.

\text { Area }=\frac{\sqrt{3}}{4} \mathrm{a}^{2}

\Rightarrow 19 \sqrt{3}=\frac{\sqrt{3}}{4} \mathrm{a}^{2}

\Rightarrow \mathrm{a}^{2}=19 \sqrt{3} \times \frac{4}{\sqrt{3}}

\Rightarrow \mathrm{a}^{2}=19 \times 4=76

\Rightarrow \mathrm{a}=76^{1 / 2}

\Rightarrow \mathrm{a}=2 \sqrt{19} \mathrm{~cm}

Perimeter = 3a 

Hence the perimeter of the triangle is=6 \sqrt{19} \mathrm{~cm}