Heron’s Formula

Heron’s formula to find area of triangle:

If length of three sides of triangle are a, b, c then,

The semiperimetre $\text{s = }\frac{\text{a + b + c}}{\text{2}}$

And the area of $\Delta \text{ABC = }\sqrt{\text{s}\left(\text{s - a}\right)\left(\text{s - b}\right)\left(\text{s - c}\right)}$

Example:

Find area of triangular plot as in figure.

Here sides of triangle are $\text{5, 7, 8}$ units

$\therefore \text{a = 5}$ units, $\text{b = 7}$ units, $\text{c = 8}$ units

$\therefore s\text{ = }\frac{\text{a + b + c}}{\text{2}}$

$\text{= }\frac{\text{5 + 7 + 8}}{\text{2}}$

$\text{= }\frac{\text{20}}{\text{2}}$

$\text{= 10}$

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

$\text{= }\sqrt{\text{10}\left(\text{10 - 5}\right)\left(\text{10 - 7}\right)\left(\text{10 - 8}\right)}$

$\text{= }\sqrt{\text{10 × 5 × 3 × 2}}$

$\text{= }\sqrt{\text{5 × 2 × 5 × 3 }\times \text{ 2}}$

$\text{= 10}\sqrt{3}$ sq units

Application of Heron’s formula:

Area of irregular polygon:

We can apply heron’s formula to find area of irregular polygon by dividing it into number of triangles.

Example: Find area of polygon $\text{PQRST}$.

Area of polygon $\text{= A}\left(\Delta \text{PQT}\right)\text{ + A}\left(\Delta \text{TQS}\right)\text{ + A}\left(\Delta \text{SQT}\right)$

For $\Delta \text{PQT}$, $\text{a = 4, b = 3, c = 5}$

$\therefore s\text{ = }\frac{\text{a + b + c}}{\text{2}}$

$\text{= }\frac{\text{4 + 3 + 5}}{\text{2}}$

$\text{= }\frac{\text{12}}{\text{2}}$

$\text{= 6}$ units

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

$\text{= }\sqrt{\text{6}\left(\text{6 - 4}\right)\left(\text{6 - 3}\right)\left(\text{6 - 5}\right)}$

$\text{= }\sqrt{\text{6 × 2 × 3 × 1}}$

$\text{= 6}$ sq.Units

For $\Delta \text{TQS}$, $\text{a = 4, b = 5, c = 7}$

$\therefore s\text{ = }\frac{\text{a + b + c}}{\text{2}}$

$\text{= }\frac{\text{4 + 5 + 7}}{\text{2}}$

$\text{= }\frac{\text{16}}{\text{2}}$

$\text{= 8}$ units

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

$\text{= }\sqrt{\text{8}\left(\text{8 - 4}\right)\left(\text{8 - 5}\right)\left(\text{8 - 7}\right)}$

$\text{= }\sqrt{\text{8 × 4 × 3 × 1}}$

$\text{= 4}\sqrt{\text{6}}$ sq. units

For $\Delta \text{SQT}$, $\text{a = 5, b = 7, c = 6}$

$\therefore s\text{ = }\frac{\text{a + b + c}}{\text{2}}$

$\text{= }\frac{\text{5 + 7 + 6}}{\text{2}}$

$\text{= }\frac{\text{18}}{\text{2}}$

$\text{= 9}$ units

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

$\text{= }\sqrt{\text{9}\left(\text{9 - 5}\right)\left(\text{9 - 7}\right)\left(\text{9 - 6}\right)}$

$\text{= }\sqrt{\text{9 × 4 × 2 × 3}}$

$\text{= 6}\sqrt{\text{6}}$ sq. units

Area of polygon = $\text{A}\left(\Delta \text{PQT}\right)\text{ + A}\left(\Delta \text{TQS}\right)\text{ + A}\left(\Delta \text{SQT}\right)$

$\text{= 6 + 4}\sqrt{\text{6}}\text{ + 6}\sqrt{\text{6}}$

$\text{= 6 + 10}\sqrt{\text{6}}$ sq. units.