Question

Using Heron’s formula, find the area of an equilateral triangle of side $ 'a' $ units.

Answer 1

Hint: We are asked to use Heron’s formula and we are given an equilateral triangle. An equilateral triangle is one whose all sides have the same measurement. In our case length of each side is units. We will use this fact in Heron’s formula to obtain our solution
1) Area of a triangle with length of sides as $ a,b,c $ is given by the formula:
 $ A = \sqrt {s(s - a)(s - b)(s - c)} $ , where \[s\] is semi perimeter of the triangle
The above formula is called the Heron’s formula.
2) A triangle with length of sides as $ a,b,c $ has semi perimeter as:
\[s = \dfrac{{a + b + c}}{2}\]

Complete step-by-step answer:

Here we are given that the triangle is an equilateral triangle
Hence, all the sides will be units.
Heron's formula is given by:
 $ A = \sqrt {s(s - a)(s - b)(s - c)} $
Where $ a,b,c $ are three sides of the triangle.
But as we are give that the triangle is an equilateral triangle hence,
 All the values of all sides will be equal
That is $ a = b = c = a $
Now \[s\] is semi perimeter of the triangle
Hence, we will first find out the value of \[s\]
Semi-perimeter of the triangle having side units is:
\[s = \dfrac{{a + a + a}}{2} = \dfrac{{3a}}{2}\]
Now let’s put the value in the final formula and find the area of given triangle
 $ \Rightarrow A = \sqrt {s(s - a)(s - b)(s - c)} $
Here in place of \[s\] we will put \[\dfrac{{3a}}{2}\]and $ a = b = c = a $
 $ \Rightarrow A = \sqrt {\dfrac{{3a}}{2}(\dfrac{{3a}}{2} - a)(\dfrac{{3a}}{2} - a)(\dfrac{{3a}}{2} - a)} $
 $ = \sqrt {\dfrac{{3a}}{2}(\dfrac{{3a - 2a}}{2})(\dfrac{{3a - 2a}}{2})(\dfrac{{3a - 2a}}{2})} $
 $ = \sqrt {\dfrac{{3a}}{2} \times \dfrac{a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2}} $
 $ = \sqrt {\dfrac{{3{a^4}}}{{{2^4}}}} = \dfrac{{\sqrt 3 {a^2}}}{4} $
Hence, the area is $ \dfrac{{\sqrt 3 {a^2}}}{4} $

Note: While solving for the operation under square root you should be very careful unless the whole value may change. There is no need to put negative sign after solving the rooted terms because area is always a positive quantity.