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