Drawing lines from O to each vertex

Drawing lines from O to each vertex splits the triangle into three smaller triangles, as shown below.

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For each smaller triangle, the 'base' and 'height' are shown (the 'bases' are the sides of the original triangle, which are not horizontal). So, using Area $=\dfrac{1}{2}$ base$\times$height, the areas of the smaller triangles can be found:

Red triangle area $\dfrac{1}{2}\times13\times9$

Yellow triangle area $\dfrac{1}{2}\times20\times3.6$

Green triangle area $\dfrac{1}{2}\times21\times3$

So the total area of the triangle is

$\dfrac{13\times9+20\times3.6+21\times3}{2}=\dfrac{117+72+63}{2}=126$cm$^2$

Using trigonometry

The cosine rule can be used to find one of the angles in the triangle, and then the formula $\frac{1}{2}ab\sin{C}$ can be used to find the area.

For example, the angle between the 13 cm side and the 20 cm side can be found using the cosine rule:$$\begin{align}&21^2=13^2+20^2-2\times 13\times20\times\cos{\theta}\\\Rightarrow&441=569-520\cos{\theta}\\\Rightarrow&\cos{\theta}=\frac{569-441}{520}=\frac{128}{5200}=\frac{16}{65}\\\Rightarrow&\theta=75.75^{\text{o}}\end{align}$$

So the area of the triangle is $\frac{1}{2}\times13\times20\times\sin{75.75^{\text{o}}}=126$ cm$^2$.

Using Heron's Forumla

Heron's formula for the area of a triangle with sides $a$, $b$ and $c$ is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is half of the perimeter of the triangle.

For this triangle, $a$, $b$ and $c$ are $13$, $20$ and $21$, and the perimeter is $54$, so $s=27$.

So the area is $$\begin{split}\sqrt{27\times14\times7\times6}&=\sqrt{3^3\times2\times7\times2\times3}\\&=\sqrt{2\times2\times3^2\times3^2\times7\times7}\\&=\sqrt{(2\times3^2\times7)^2}\\&=2\times3^2\times7\\&=126\end{split}$$

So the area is $126$ cm$^2$.