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Area of Equilateral Triangle Formula

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Last updated date: 09th Apr 2024
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An Introduction to the Area of an Equilateral Triangle



The triangle is a very important shape in geometry. There are different types of triangles. The most simple triangle of geometry is the equilateral triangle. An equilateral triangle is one in which all three sides are equal in length. Consequently, each of its three interior angles measures a third of 180°, which is 60° each. In the article, we will learn how to obtain the formula to find the area of an equilateral triangle as well as solved examples that will help us to grasp the concept. 


Derivation of Formula of Area of an Equilateral Triangle

One of the most interesting and useful properties of an equilateral triangle is that its altitude, angle bisector and median from any of its vertices are coincident (they are the same line segment). This can be proved using the concept of congruence of triangles. Consider the following equilateral triangle ABC whose side length is ‘a’ units. 


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Since AD is the median to BC, BD = DC = BC/2 = a/2 units. Since AD is also perpendicular to BC, we can use the Pythagorean theorem to determine its length (which is the height of the triangle corresponding to the side BC).


\[A{C^2} = A{D^2} + D{C^2} \Rightarrow A{D^2} = A{C^2} - D{C^2} = {a^2} - {\left( {\frac{a}{2}} \right)^2} = {a^2} - \frac{{{a^4}}}{4} = \frac{{3{a^2}}}{4}\]


\[\Rightarrow AD = h = \frac{{\sqrt 3 a}}{2}\]


area\[\left( {\Delta ABC} \right) = \frac{1}{2}bh = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times a \times h = \frac{1}{2} \times a \times \frac{{\sqrt 3 a}}{2} = \frac{{\sqrt 3 }}{4}{a^2}\]

 

Note that we can derive this formula using Heron’s formula as well.


\[s = \frac{{\left( {a + b + c} \right)}}{2} = \frac{{3a}}{2}\]


area\[\left( {\Delta ABC} \right) = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - b} \right)} \]


area\[\left( {\Delta ABC} \right) = \sqrt {\left( {\frac{{3a}}{2}} \right)\left( {\frac{{3a}}{2} - a} \right)\left( {\frac{{3a}}{2} - a} \right)\left( {\frac{{3a}}{2} - a} \right)} \]


area(ABC) \[\sqrt {\left( {\frac{{3a}}{2}} \right)\left( {\frac{a}{2}} \right)\left( {\frac{a}{2}} \right)\left( {\frac{a}{2}} \right)} \]


area(ABC) \[ = \sqrt {\frac{{3{a^4}}}{{16}}} \]


area(ABC) \[ = \frac{{\sqrt 3 }}{4}{a^2}\]


Let’s look at an example to see how to use this formula.


Solved Examples

Question: Find the area of an equilateral triangle whose altitude is 33cm.


Solution:

\[h = 3\sqrt 3 \,cm\]


\[h = \frac{{\sqrt 3 a}}{2}\, \Rightarrow \frac{{\sqrt 3 a}}{2} = 3\sqrt 3 \Rightarrow a = 6\,cm\]


area\[\left( {\Delta ABC} \right) = \frac{{\sqrt 3 }}{4}{a^2} = \frac{{\sqrt 3 }}{4}{6^2} = 9\sqrt 3 \,c{m^2}\]


Why don’t you try to solve a problem for practice?


Question: Find the area of an equilateral triangle whose perimeter is 12 cm.


Options:

(a)  \[2\sqrt 3 \,c{m^2}\]

(b) \[4\sqrt 3 \,c{m^2}\]

(c) \[6\sqrt 3 \,c{m^2}\]

(d) \[8\sqrt 3 \,c{m^2}\]


Answer: (b)


Solution:

perimeter \[ = 3a = 12\,\,cm\,\,\, \Rightarrow a = 4\,\,cm\]


area \[ = \frac{{\sqrt 3 }}{4}{a^2} = \frac{{\sqrt 3 }}{4}{4^2} = 4\sqrt 3 \,\,c{m^2}\]


Conclusion

This was all about the methods of finding the area of an equilateral triangle which is helpful in various domains. The solved examples provided in the article will help students to understand the use of the area of equilateral triangle formula.

FAQs on Area of Equilateral Triangle Formula

1. What is the relation between the area and perimeter of an equilateral triangle?

We know that the formula for the perimeter of a triangle with length of sides equal to ‘a’ is given as following:


perimeter =3a


Therefore, \[a=\frac{perimeter}{3}\]


The formula for the area of an equilateral triangle with length of sides ‘a’ is given as following:


\[area= \frac{\sqrt{3a^{2}}}{4}=\frac{\sqrt{3}}{4}\left ( \frac{perimeter}{3}^{2} \right )=\frac{perimeter^{2}}{12\sqrt{3}}\]

2. What are the formulas for area and perimeter of an equilateral triangle in terms of its altitude, median and angle bisectors?

We already know that the altitude, median and angle bisectors of an equilateral triangle are the same line segment.


As we saw earlier, the altitude of an equilateral triangle with side length ‘a’ is given as:


\[\text{altitude}=\text{median} =\text{angle bisector} = h=\frac{\sqrt{3a}}{2}\]


\[\Rightarrow a=\frac{2h}{\sqrt{3}}\]


Now, its perimeter and area can be given as:


\[\text{perimeter}=3a=3\left (\frac{2h}{\sqrt{3}}  \right )=2\sqrt{3h}\]


\[\text{area}\frac{\sqrt{3a^{2}}}{4}=\frac{\sqrt{3}}{4}\left ( \frac{2h}{\sqrt{3}} \right )^{2}=\frac{h^{2}}{\sqrt{3}}\]