# Heron’s formula

Do you remember how can you calculate the area of a triangle? Of course, there are several ways but the most common way is to use a formula

$$A = \frac{a \cdot v_{a}}{2},$$

where $a$ is the length of the arbitrary side of a triangle, while $v_{a}$ is the length of the height of a triangle.

But, what if you don’t know the length of the height of a triangle, and you only know the lengths of the sides of a triangle? In these cases you can use Heron’s formula. In addition, it is named after Hero of Alexandria, who was a Greek mathematician in $10 – 70$ AD.

## Heron’s formula

The area of a triangle whose side lengths are $a, b$ and $c$ is:

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

where $s$ is a semi – perimeter of a triangle, i.e. $s = \frac{a + b + c}{2}$.

## Examples

Example 1:  Calculate the area of a triangle whose side lengths are $29 \ cm$, $25 \ cm$ and $6 \ cm$.

Solution:

$a = 29 \ cm$

$b = 25 \ cm$

$\underline{c = 6 \ cm}$

$A = ?$

Since we know side lengths, we can use Heron’s formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$.

Furthermore, we need to calculate the semi – perimeter of a triangle:

$$s = \frac{a + b + c}{2} = \frac{29 + 25 + 6}{2} = \frac{60}{2} = 30 \ cm.$$

As a result, the area of a triangle is

$$A = \sqrt{30 \cdot (30 – 29) \cdot (30 – 25) \cdot (30 – 6)} = \sqrt{3600} = 60 \ cm^2.$$

In conclusion, the area of a triangle is $A = 60 \ cm^2.$

Example 2:  Calculate the length of the longest height of a triangle if side lengths are $15 \ cm$, $112 \ cm$ and $113 \ cm$.

Solution:

$a = 15 \ cm$

$b = 112 \ cm$

$\underline{c = 113 \ cm}$

$v = ?$

The longest height of a triangle is the height corresponding to the shortest side of a triangle.

We can calculate the height by using the formula: $A = \frac{a \cdot v_{a}}{2}$.

We know the length of the side $a$. Furthermore, we can easily get the area of a triangle by using Heron’s formula, since we also know the lengths of the other two sides. As a result, we have

$$s = \frac{a + b + c}{2} = \frac{15 + 112 + 113}{2} = 120 \ cm$$

$$A = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{120 \cdot (120 – 15) \cdot (120 – 112) \cdot (120 – 113)} = 840 \ cm^2.$$

Therefore,

$$840 = \frac{15 \cdot v_{a}}{2} \rightarrow v_{a} = 112 \ cm.$$

In conclusion, the length of the longest height of a triangle is $112 \ cm$.

Example 3:  Calculate the area of a trapezium whose parallel sides have lengths $22 \ cm$ and $16 \ cm$ respectively, while the other two sides have lengths $25 \ cm$ and $29 \ cm$.

Solution:

$a = 22 \ cm$

$b = 29 \ cm$

$c = 16 \ cm$

$\underline{d = 25 \ cm}$

$A = ?$

A formula for the area of a trapezium is: $A = \frac{a + c}{2} \cdot v$.

Furthermore, we know the lengths of the parallel sides so we need to calculate the length of the height of a trapezium.

Also, notice that we can draw a parallel with $BC$ through point $D$. Therefore, $\vert DE\vert = b = 29 \ cm$.

Since $EBCD$ is a parallelogram, we know that $\vert EB \vert = c = 16 \ cm$. Therefore, $\vert AE \vert = 22 – 16 = 6 \ cm$.

In other words, now we know the lengths of all sides of a triangle $AED$ so we can calculate its area $A’$:

$s = \frac{a + b + c}{2} = \frac{6 + 29 + 25}{2} = 30 \ cm$

$A’ = \sqrt{30(30 – 6)(30 – 29)(30 – 25)} = \sqrt{3600} = 60$ $cm^2$

$60 = \frac{6 \cdot v}{2} \rightarrow v = 20 \ cm$

Finally,

$A = \frac{a + c}{2} \cdot v = \frac{22 + 16}{2} \cdot 20 = 380$ $cm^2$.

In conclusion, the area of a trapezium is $A = 380$ $cm^2$.

Example 4:  Calculate the area of a parallelogram if the length of one of his side is $51 \ cm$ and lengths of diagonals are $40 \ cm$ and $74 \ cm$.

Solution:

$a = 51 \ cm$

$e = 40 \ cm$

$\underline{f = 74 \ cm}$

$A = ?$

If we look at the triangles $ABS$ and $CDS$, we can notice that they are congruent by $SSS$ theorem because of the side of length $a$ and the fact that each diagonal bisects the other (revise congruent triangle postulates).

In other words, this means that areas of triangles $ABS$ and $CDS$ are equal.

Furthermore, triangles $BCS$ and $DAS$ are also congruent by $SSS$ theorem (sides of lengths $b, \frac{e}{2}, \frac{f}{2}$). Therefore, areas of those triangles are equal.

Now let’s observe a triangle $CDS$. Let $v$ be the height of a triangle on side $SC$.

Since there is an obtuse angle in the vertex $S$, we can see that the foot of the perpendicular will be on the $\overline{AS}$. This perpendicular (height) $v$ is also a height in a triangle $DAS$.

Furthermore, the fact that $\vert AS \vert = \vert SC \vert = \frac{e}{2}$ tells us that areas of triangles $DAS$ and $CDS$ are equal.

In other words, we simply need to calculate the area of one triangle and then multiply that number by $4$  to find out the area of a parallelogram.

First we will calculate the area $A’$ of a triangle $ABS$. We know

$a = 51 \ cm$

$\frac{e}{2} = \frac{40}{2} = 20 \ cm$

$\frac{f}{2} = \frac{74}{2} = 37 \ cm$

Also,

$s = \frac{a + \frac{e}{2}+ \frac{f}{2}}{2} = \frac{51 + 20 + 37}{2} = 54 \ cm$

$A’ = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{54(54 – 51)(54 – 20)(54 – 37)} = 306$ $cm^2$

Finally, the area of a parallelogram $ABCD$ is

$A = 4 \cdot 306$ $cm^2$ = $1224$ $cm^2$.