Heron's formula was first given by Heron of Alexandria. It is used to find the area of different types of triangles like equilateral, isosceles, and scalene triangles or quadrilaterals.
We will use some Pythagoras theorem, area of a triangle formula, and algebraic identities to derive Heron's formula. Let us take a triangle having lengths of sides, a, b, and c. Let the semi-perimeter of the triangle ABC is "s", the perimeter of the triangle ABC is "P" and the area of triangle ABC is "A". Let us assume the side length b is divided into two parts p and q as a perpendicular(h) falls from the vertex B on the side AC at point M. Consider the triangle below:
View attachment 1677
As we know, the area of a triangle
A= (1/2) b*h
where b is the base and h is the height of the triangle.
Let us begin to calculate the value of h.
Thus, as per the image, b = p + q
=> q = b - p ....(1)
On squaring both sides we get,
=> q^2 = b^2 + p^2 - 2bp ....(2)
Adding h^2on both sides we get,
q^2 + h^2 = b^2 + p^2 - 2bp + h^2 ....(3)
Applying Pythagoras Theorem in the triangle ABM we get,
h^2 + q^2 = a^2 ....(4)
Applying Pythagoras Theorem in the triangle ACM we get,
p^2 + h^2 = c^2 ....(5)
Substituting the value of (4) and (5) in (3) we get,
q^2 + h^2 = b^2 + p^2 - 2bp + h^2
a^2 = b^2 + c^2 - 2bp
p = (b^2 + c^2 - a^2)/(2b) ....(6)
From (5)
p^2 + h^2 = c^2
h^2 = c^2 - p^2 = (c + p) (c - p) ....(7) (As a^2 - b^2 = (a+b)(a-b))
Substituting (6) in (7) we get,
h^2 = (c + p) (c - p)
h^2 = (c + (b^2 + c^2 - a^2)/(2b)) (c - (b^2 + c^2 - a^2)/(2b))
h^2 = ((2bc + b^2 + c^2 - a^2)/(2b)) ((2bc - b^2 - c^2 + a^2)/2b)
h^2 = ((b + c)^2 - a^2)/(2b)) ((a^2 - (b - c)^2)(/2b))
h^2 = ((b + c + a)(b + c - a)(a + b - c)(a - b + c))/(4b^2) ....(8) (As a^2 - b^2 = (a+b)(a-b))
As perimeter of triangle is P = a + b + c and P = 2s. (Here s = semi-perimeter and s = P/2)
2s = a + b + c ....(9)
Substituting (9) in (8) we get,
h^2 = ((b + c + a)(b + c - a)(a + b - c)(a - b + c))/(4b^2)
h^2 = (2s* (2s - 2a) * (2s - 2b) * (2s - 2c))/(4b^2)
h^2 = (2s * 2(s - a) * 2(s - b) * 2(s - c))/(4b^2)
h^2 = (16s(s - a)(s - b)(s - c))/(4b^2)
h = sqrt(4s(s - a)(s - b)(s - c)/(4b^2))
h = (2sqrt((s(s - a)(s - b)(s - c)))/b ....(10)
Area of triangle ABC, A = (1/2) * base * height
A = (1/2)* b* h
A = (1/2) * b * (2sqrt((s(s - a)(s - b)(s - c)))/b
A = sqrt((s(s - a)(s - b)(s - c))
Thus, area of the triangle ABC is:
A = sqrt(s(s - a)(s - b)(s - c)) unit^2
