Triangle

# Area of a Triangle

In this section, we shall discuss about the different formulae to find the area of a triangle.

Let $ABC$ be a triangle. Let us denote the angle $BAC$, $CBA$, $ACB$ of the triangle $ABC$ by $A$, $B$, $C$ respectively; and the lengths of the sides $BC$, $CA$, $AB$ by $a$, $b$, $c$ respectively. Let $s$ be the semi perimeter of the triangle $ABC$. $\therefore s=\frac{a+b+c}{2}$

From the sine law, we know that the area of any triangle is half the product of any two sides and the sine of the angle between them. $\Delta=\frac{1}{2}bc\sin A=\frac{1}{2}ca\sin B=\frac{1}{2}ab\sin C$

Also, we have the half angle formulae; $\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$ $\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$ $\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

Now, let us combine above formulae to find out different formulae for the area of triangle $ABC$.

$\Delta=\frac{1}{2}ab\sin C$ $=\frac{1}{2}ab\cdot 2\sin\frac{C}{2}\cos\frac{C}{2}$ $=ab\sqrt{\frac{(s-a)(s-b)}{ab}}\sqrt{\frac{s(s-c)}{ab}}$ $\therefore\Delta=\sqrt{s(s-a)(s-b)(s-c)}\;\;\text{ ___(1)}$ This gives the area of triangle $ABC$. This formula is also known as Heron’s formula.

Now, putting the value of $s=\frac{a+b+c}{2}$ in equation $\text{(1)}$, we get, $\begin{array}{l}\;\;\;\;\;\Delta ABC\\=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}\\ =\frac{1}{4}\sqrt{2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4}\end{array}$

If $R$ be the circum-radius of the triangle $ABC$, then from sine law, we have $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$

Now, $\Delta=\frac{1}{2}bc\sin A$ $=\frac{1}{2}bc\cdot\frac{a}{2R}$ $\therefore\Delta=\frac{abc}{4R}$

We can also prove the following relations relating the area of triangle $ABC$. $1.\;\;\;\tan\frac{A}{2}=\frac{(s-b)(s-c)}{\Delta}\;\;\;\text{etc.}$

$2.\;\;\;\tan A=\frac{\sin A}{\cos A}=\frac{a}{R}\cdot\frac{bc}{b^2+c^2-a^2}$ $=\frac{\Delta}{b^2+c^2-a^2}$

Similarly, we can prove, $\tan B=\frac{\Delta}{c^2+a^2-b^2}$ $\tan C=\frac{\Delta}{a^2+b^2-c^2}$

$3.\;\;\;\tan\frac{A}{2}=\frac{(s-b)(s-c)}{\Delta}\;\;\;\text{etc.}$ $4.\;\;\;\cot\frac{A}{2}=\frac{s(s-a)}{\Delta}\;\;\;\text{etc.}$