An Alternative Derivation of Heron’s Formula

In my previous blog titled “An Algebraic Proof of Heron’s Formula“, I algebraically derived the Heron’s formula concerning A, the area of a triangle:

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

where a, b, c are three sides of a triangle and s=\frac{a+b+c}{2}.

There is an alternative derivation which requires some basic trigonometry.

Here it is:

Screen Shot 2017-06-02 at 9.46.32 PM.png

Fig. 1

Let \theta = \angle ABC (see Fig. 1),  then

A=\frac{1}{2}a\;h = \frac{1}{2}a\;(c\sin{\theta})

i.e.,

A^2=\frac{1}{4}a^2 c^2 \sin^2{\theta} = \frac{1}{4} a^2 c^2 (1-\cos^2{\theta})\quad\quad\quad\quad\quad(1)

due to trigonometric identity \sin^2{\theta} + \cos^2{\theta} =1.

Moreover, by the law of cosines, b^2=a^2+c^2-2a c \cos{\theta} which implies

\cos{\theta} = \frac{a^2+c^2-b^2}{2 a c},

and (1) becomes

A^2=\frac{1}{4}a^2 c^2 (1-\frac{(a^2+c^2-b^2)^2}{4 a^2 c^2})

=\frac{1}{4}a^2 c^2 \frac{4a^2 c^2-(a^2+c^2-b^2)^2}{4a^2 c^2}

=\frac{1}{16}(2a c +a^2+c^2-b^2)(2a c-a^2-c^2+b^2)

=\frac{1}{16}((a+c)^2-b^2)(b^2-(a-c)^2)

=\frac{1}{16}(a+c+b)(a+c-b)(b+a-c)(b-a+c)

=\frac{1}{16}(a+b+c)(b+c-a)(a+c-b)(a+b-c)

=\frac{1}{16}\;2^4\;(\frac{a+b+c}{2})(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)

= s(s-a)(s-b)(s-c)

Hence,

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

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