Question

Prove Heron's formula

Answer 1

Given:

Heron's formula

Sol:

Heron's formula:

for finding the area of a triangle in terms of the lengths of its sides. In symbols, if a, b, and c are the lengths of the sides

In Heron's formula is:

$S=\frac{a+b+c}{2}$

Heron's formula to find the area of a triangle is:

So, the area is:

$\text{ Area }=\sqrt{s(s-a)(s-b)(s-c)}$

Area of triangle is :

$\begin{gathered} =\frac{1}{2}\times\text{ base }\times\text{ height} \\ \\ =\frac{1}{2}\times b\times h \end{gathered}$

Use cosine law then:

$\begin{gathered} A=\frac{1}{2}ab\sin\gamma \\ \\ =\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} \\ \\ =\frac{1}{4}\sqrt{(2ab-(a^2+b^2-c^2)(2ab+(a^2+b^2-c^2)} \\ \\ \end{gathered}$

Solve then:

$\begin{gathered} =\frac{1}{4}\sqrt{(c^2-(a-b)^2)((a+b)^2-c^2)} \\ \\ =\sqrt{\frac{(c+(a-b))(c-(a-b))((a+b)-c)((a+b))+c)^{16}}{16}} \\ \\ =\sqrt{\frac{(b+c-a)}{2}\frac{(a+c-b)}{2}\frac{(a+b-c)}{2}\frac{(a+b+c}{2}} \\ \\ =\sqrt{s(s-a)(s-b)(s-c)} \end{gathered}$

In cosine law.

Let us prove the result using the law of cosines:

Let a, b, c be the sides of the triangle and

$\alpha,\beta,\gamma$

are opposite angle to the side.

We know that low of cosine is:

$\begin{gathered} \cos\gamma=\frac{a^2+b^2-c^2}{2ab} \\ \\ \text{ And} \\ \\ \sin\gamma=\sqrt{1-\cos^2\gamma} \\ \\ =\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)}}{2ab} \end{gathered}$

Here base of triangle = a

Height is:

$H=\sin\gamma$