**Heron’s formula**

- In geometry, Heron’s formula, named after Hero of Alexandria gives us the area of any triangle whose three sides lengths are already known.
- So, we can find area of triangle from their three sides also.
- If a, b, c are the lengths of the three sides of the triangle then area of that triangle is given by,

Area of triangle = √s(s-a)(s-b)(s-c)

Where, s = (a + b + c)/ 2

- This is the formula which is known as Heron’s Formula.

__Derivation for Heron’s Formula:__

__Derivation for Heron’s Formula:__

- Let us consider the ΔABC with three sides length a, b and c as shown in following figure.
- We drawn the perpendicular AP on base BC and l(BC) = b, l(AP) = h.
- Let x be the length of BP, l(BP) = x and hence l(PC) = (b – x).
- In general, we know that area of triangle ABC having base b and height h is given by,

Area of triangle = ½ * base * height

Area of triangle = ½*b*h

In ΔABP, by Pythagoras theorem

Thus, a^{2} = x^{2} + h^{2}

Hence, h^{2} = a^{2} – x^{2}

In ΔAPC, by Pythagoras theorem

Thus, c^{2} = h^{2} + (b – x)^{2}

Hence, h^{2} = c^{2} – (b^{2} -2bx + x^{2}) = c^{2} – b^{2} + 2bx – x^{2}

We equate the value of h^{2} from both triangles,

a^{2} – x^{2}= c^{2} – b^{2} + 2bx – x^{2}

a^{2} = c^{2} – b^{2} + 2bx

a^{2} – c^{2} + b^{2} = 2bx

thus, x = (a^{2} – c^{2} + b^{2})/ 2b

By putting the value of x in equation h^{2} = a^{2} – x^{2},

We get, h^{2 }= a^{2} – [ (a^{2} – c^{2} + b^{2})/ 2b]^{2}

Thus, h^{2} = [4a^{2}b^{2} – (a^{2} – c^{2} + b^{2})^{2}]/ 4b^{2}

Hence, h^{2} = 1/4b^{2}*[(2ab)^{2} – (a^{2} – c^{2} + b^{2})^{2}]

By using the formula here, (A^{2} – B^{2}) = (A – B) *(A + B)

We get, h^{2} = 1/4b^{2}*[(2ab + a^{2} – c^{2} + b^{2}) (2ab – a^{2} + c^{2} – b^{2})

Thus, h^{2} = 1/4b^{2}*[ (a + b)^{2} – c^{2}] [c^{2} – (a – b)^{2}]

Hence, h^{2} = 1/4b^{2}*[(a + b +c) (a + b – c)] [(c + a – b) (c – a + b)]

But, s = (a + b + c)/ 2

Thus, h^{2} = 1/4b^{2}*[2s*(2s – 2c) *(2s – 2b) *(2s – 2a)]

Hence, h^{2} = 1/4b^{2}*4[4*(s – a) *(s – b) *(s – c)]

h^{2} = 1/b^{2}*[4*(s – a) *(s – b) *(s – c)]

h = 2/b*√[(s – a) *(s – b) *(s – c)]

Now, we know that area of triangle ABC = ½*base*height

= ½*b*h

=1/2*b*2/b*√ [(s – a) *(s – b) *(s – c)]

= √ [(s – a) *(s – b) *(s – c)]

Thus, area of triangle if the lengths of three sides of triangle are known is given by,

Area of triangle = √ [(s – a) *(s – b) *(s – c)]

This formula is known as **Heron’s Formula**.

Hence proved.