Let, ABC be a triangle, in which, side BC = \(a\) unit, CA = \(b\) unit and AB = \(c\) unit. Perpendicular AD is drawn from the vertex A to the base BC, i.e, \(AD \perp BC\). In the base side Bc, if we suppose DC by \(x\) unit then BD will be \((a - x)\) unit. Let the height of the triangle ABC be \(AD = h\) unit.

A B C D c b h a – x x a

Now, the perimeter of the triangle ABC \((P) = a + b + c\) and its semi-perimeter \((s)\) becomes

\[ (s) = \frac{P}{2} = \frac{a + b + c}{2} \]

Here, in the right angled triangle ADB,

\[ \begin{aligned} & AD^2 + BD^2 = AB^2 \\ \text{or, } & h^2 + (a - x)^2 = c^2 \\ \text{or, } & h^2 = c^2 - (a - x)^2 \text{ ................. (i)} \end{aligned} \]

Again in the right angled triangle ADC

\[ \begin{aligned} & AD^2 + DC^2 = AC^2 \\ \text{or, } & h^2 + x^2 = b^2 \end{aligned} \]
\[ \text{or, } h^2 = b^2 - x^2 \text{ ............... (ii)} \]

From equation (i) and (ii)

\[ \begin{aligned} c^2 - (a - x)^2 &= b^2 - x^2 \\ \text{or, } c^2 &= b^2 - x^2 + (a - x)^2 \\ \text{or, } c^2 &= b^2 - x^2 + a^2 - 2ax + x^2 \\ \text{or, } c^2 &= b^2 + a^2 - 2ax \\ \text{or, } 2ax &= b^2 + a^2 - c^2 \\ \text{or, } x &= \frac{b^2 + a^2 - c^2}{2a} \text{ ........... (iii)} \end{aligned} \]

Substituting the value of \(x\) in equation (ii),

\[ \begin{aligned} h^2 &= b^2 - \left(\frac{b^2 + a^2 - c^2}{2a}\right)^2 \\ \text{or, } h^2 &= b^2 - \frac{(b^2 + a^2 - c^2)^2}{4a^2} \\ \text{or, } h^2 &= \frac{4a^2b^2 - (a^2 + b^2 - c^2)^2}{4a^2} \\ \text{or, } h^2 &= \frac{(2ab)^2 - (a^2 + b^2 - c^2)^2}{4a^2} \\ \text{or, } h^2 &= \frac{(2ab + a^2 + b^2 - c^2)(2ab - a^2 - b^2 + c^2)}{4a^2} \\ \text{or, } h^2 &= \frac{[(a + b)^2 - c^2][c^2 - (a - b)^2]}{4a^2} \\ \text{or, } h^2 &= \frac{(a + b + c)(a + b - c)(c + a - b)(c - a + b)}{4a^2} \text{ ......... (iv)} \end{aligned} \]

From above, \( s = \frac{a + b + c}{2} \)

\[ \begin{aligned} \text{or, } & a + b + c = 2s \text{ ......... (v)} \\ \text{or, } & a + b = 2s - c \end{aligned} \]

Subtracting \(c\) from both sides

\[ \begin{aligned} \text{or, } & a + b - c = 2s - c - c \\ \text{or, } & a + b - c = 2s - 2c = 2(s - c) \\ \text{or, } & a + b - c = 2(s - c) \text{ ......... (vi)} \end{aligned} \]

Similarly, \( a + c - b = 2s - 2b = 2(s - b) \) ......... (vii)

\( b + c - a = 2s - 2a = 2(s - a) \) ......... (viii)

from equations (iv), (v), (vi), (vii) and (viii),

\[ \begin{aligned} h^2 &= \frac{2s \times 2(s - c) \times 2(s - b) \times 2(s - a)}{4a^2} \\ \text{or, } h^2 &= \frac{16s(s - a)(s - b)(s - c)}{4a^2} \\ \text{or, } h &= \frac{2\sqrt{s(s - a)(s - b)(s - c)}}{a} \end{aligned} \]

we know that,

\[ \begin{aligned} \text{Area of triangle ABC } &= \frac{1}{2} \times \text{BC} \times \text{AD} = \frac{1}{2} \times a \times h \\ &= \frac{1}{2} \times a \times \frac{2\sqrt{s(s - a)(s - b)(s - c)}}{a} \\ &= \sqrt{s(s - a)(s - b)(s - c)} \\ \therefore\ \text{Area of triangle ABC } &= \sqrt{s(s - a)(s - b)(s - c)} \text{ square units} \end{aligned} \]

Formula to find the area of scalene triangle,

Area of scalene triangle = \( \sqrt{s(s - a)(s - b)(s - c)} \), where s is the semi perimeter of the triangle. It is called Heron's formula