Area of a Triangle
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Area of a triangle when height and base is known

Right-angled Triangle



h b
This can be viewed as the half of rectangle formed by sides h and b:

h b h b
$$ \begin{aligned} area_{rect} & = b \cdot h\\ area_{triangle} & = {1 \over 2} \cdot b \cdot h\\ \end{aligned} $$

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Acute Triangle

Now let's look at an acute triangle:

b h b1 b2
Here we can see that the line $h$ dissects the triangle into two smaller right-angular triangles and we know that for each of them we can calculate the area and then sum them together. Let's take the bases to be $b1$ and $b2$ where $b = b1 + b2$ then:
$$ \begin{aligned} area_{triangle} & = {1 \over 2} \cdot b1 \cdot h + {1 \over 2} \cdot b2 \cdot h\\ & = {1 \over 2} \cdot (b1 + b2) \cdot h\\ & = {1 \over 2} \cdot b \cdot h\\ \end{aligned} $$

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Obtuse Triangle

For the obtuse triangle we see the following:

b h b1 b2
The area is now the area of the larger triangle with base $b1$ minus the area of the smaller triangle with base $b2$, here $b = b1 - b2$:
$$ \begin{aligned} area_{triangle} & = {1 \over 2} \cdot b1 \cdot h - {1 \over 2} \cdot b2 \cdot h\\ & = {1 \over 2} \cdot (b1 - b2) \cdot h\\ & = {1 \over 2} \cdot b \cdot h\\ \end{aligned} $$

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Acute Triangle

Now let's look at an acute triangle:

b h b1 b2
Here we can see that the line $h$ dissects the triangle into two smaller right-angular triangles and we know that for each of them we can calculate the area and then sum them together. Let's take the bases to be $b1$ and $b2$ where $b = b1 + b2$ then:
$$ \begin{aligned} area_{triangle} & = {1 \over 2} \cdot b1 \cdot h + {1 \over 2} \cdot b2 \cdot h\\ & = {1 \over 2} \cdot (b1 + b2) \cdot h\\ & = {1 \over 2} \cdot b \cdot h\\ \end{aligned} $$

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Area by using Limits



b h f(x) dx
The idea is to slice $h$ up into $n$ slices and estimate the area of the slice and then add all the slices together. The bigger we make $n$ the nearer we get to the are of the triangle. $f(x)$ is the length of the slice at the x'th slice.
$$ \begin{aligned} dx & = {h \over n}\\ f(x) & = x \cdot {b \over n}\\ area_{triangle} & = \lim_{n \to \infty} \sum_{i = 0}^{n} f(i) \cdot dx\\ & = \lim_{n \to \infty} \sum_{i = 0}^{n} i \cdot {b \over n} \cdot {h \over n}\\ & = {b \cdot h} \cdot \lim_{n \to \infty} {1 \over n^2} \cdot \sum_{i = 0}^{n} i\\ & = {b \cdot h} \cdot \lim_{n \to \infty} {1 \over n^2} \cdot { { n^2 - n} \over 2}\\ & = {1 \over 2} \cdot {b \cdot h} \cdot \lim_{n \to \infty} 1 - {1 \over n}\\ & = {1 \over 2} \cdot {b \cdot h}\\ \end{aligned} $$

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Area when only the lengths of the sides are known

Let's assume we have some arbitrary triangle and we would like to determine what is the area of a triangle is if we only know the length of it's sides. We can view it for example as follows:

a b c
We approach this from straight geometric knowledge or trigonometric knowledge.

Geometric approach

Let's start with the fact that we can determine the height of a triangle and from that and the length of the opposite side we can determine the area. Let's view above as follows:

a b h c
From "Height of a Triangle we can determine the value of h, and we know the formula to determine the area of a triangle.
$$ \begin{aligned} area & = {1 \over 2} \cdot h \cdot c\\ & = {1 \over 2} \cdot c \cdot \sqrt{a^2 - \Big({{a^2 - b^2 + c^2} \over {2 \cdot c}}\Big)^2}\\ & = {1 \over 4} \cdot \sqrt{4 \cdot a^2 \cdot c^2 - ({a^2 - b^2 + c^2})^2}\\ \end{aligned} $$

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Trigonometric Approach

Let's assume for now the angle between $a$ and $b$ is $\Theta$ then our image looks as follows:

a b c Θ
$$ \begin{aligned} b^2 & = a^2 + c^2 - 2 \cdot a \cdot c \cdot \cos(\Theta)\\ \cos(\Theta) & = {{a^2 + c^2 - b^2} \over {2 \cdot a \cdot c}}\\ \sin(\Theta) & = \sqrt{1 - \cos^2(\Theta)}\\ & = \sqrt{1 - \Big({{a^2 + c^2 - b^2} \over {2 \cdot a \cdot c}}\Big)^2}\\ & = {{\sqrt{4 \cdot a^2 \cdot c^2 - (a^2 + c^2 - b^2)^2}} \over {2 \cdot a \cdot c}}\\ area_{triangle} & = {1 \over 2} \cdot a \cdot c \cdot \sin(\Theta)\\ & = {1 \over 2} \cdot a \cdot c \cdot {{\sqrt{4 \cdot a^2 \cdot c^2 - (a^2 + c^2 - b^2)^2}} \over {2 \cdot a \cdot c}}\\ & = {1 \over 4} \cdot \sqrt{4 \cdot a^2 \cdot c^2 - ({a^2 - b^2 + c^2})^2}\\ \end{aligned} $$

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