A Detailed Derivation of the Heron’s Formula

Heron’s Formula can be used to find the area of a triangle given the lengths of the three sides. A triangle with side lengths a, b, and c, its area A can be calculated using the Heron’s formula

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

where

s = \displaystyle \frac{a+b+c}{2}

is the semiperimeter (half the perimeter) of the triangle.

In this post, I will provide a detailed derivation of this formula. 

Some Preliminaries

The area of a triangle is half the product of its base and its altitude. In the figure below, \overline{AM} is the altitude of triangle ABC. If the length of the altitude is not given, and an angle measure is given, we can use Trigonometry to calculate the altitude.

Heron's Formula

In the figure above, the altitude \overline{AM} forms right triangle AMC. We know that

\sin \theta = (length of opposite side)/(length of hypotenuse) 

so,

\sin C = \displaystyle\frac{AM }{AC}.

Simplifying, we have AM = AC \sin C which is equivalent to

AM = b \sin C

Since the base of triangle is \overline{BM} = a and its altitude is b \sin C, its area A is given by the formula

A = ab \sin C. (1)

 

The Derivation

Now we apply the preceding formula, the Cosine Law  and the Pythagorean identity \sin \theta + \cos \theta = 1 to derive the Heron’s formula.

Using the Pythagorean identity, and manipulating algebraically

\sin^2 C + \cos^2 C = 1

\sin^2 C = 1 - \cos^2 C

\sin^2 C = (1 + \cos C)(1- \cos C)  (2)

By the Cosine Law, in a triangle ABC with side lengths a, b, and c

c^2 = a^2 + b^2 -2ab \cos C.

Calculating for \cos C, we have

\cos C = \displaystyle\frac{a^2 + b^2 - c^2}{2ab}.

Substituting the preceding equation to (2), we have

\sin^2 C = \left (1 + \displaystyle\frac{a^2 + b^2 - c^2}{2ab} \right ) \left ( 1 - \displaystyle \frac{a^2 + b^2 - c^2} {2ab} \right ).

= \left (\displaystyle\frac{2ab + a^2 + b^2 - c^2}{2ab} \right ) \left ( \displaystyle \frac{2ab - a^2 - b^2 + c^2} {2ab} \right ).

=\displaystyle \frac{ [(a + b)^2 - c^2][c^2 - (a-b)^2 ]}{4a^2b^2}.

\sin^2 C = \displaystyle \frac {( a + b + c)(a + b - c)( c + a - b) (c - a + b)}{4a^2b^2}.

Getting the square root of both sides, we have

\sin C = \displaystyle \frac{ \sqrt{(a + b +c)(a + b - c)(c + a - b) (c -a + b)}}{2ab} (3).

Using (1) and (3), we calculate the area of the triangle ,

A = ab \sin C

A = \frac{1}{2} ab \displaystyle \frac{ \sqrt{(a + b +c)(a + b - c)(c + a - b) (c -a + b)}}{2ab}

A = \displaystyle \frac{1}{4} \sqrt{(a + b + c)(a + b - c)(c + a - b)(c - a + b)} (4)

Now, if we let s be the semiperimeter (half the perimeter) of triangle ABC, then

a + b + c = 2s.

Now, a + b + c - 2a = b + c - a = 2s -2a = 2(s - a).

Also, a - b + c = 2(s - b) and a + b - c = 2(s - c).

Substituting the expressions with s to (4), we have

A = \frac{1}{4} \sqrt{2s [2(s-a)][2(s-b)][2(s-c)]}

= \frac{1}{4} \sqrt{16s(s-a)(s-b)(s-a)}

which is equivalent to

A = \frac{4}{4} \sqrt{s(s-a)(s-b)(s-a)}.

Simplifying, we have A = \sqrt{s(s-a)(s-b)(s-c)}, the Heron’s formula.

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2 Comments

  1. Very nice proof. I’ve seen some really ugly ones to verify the Heron’s formula. This one relies on the formula for the area of a triangle, the law of cosines, and the Pythagorean Identity. Straightforward and easily digestible. Thank you

    Reply

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