Right triangle

Quick Solution:

The ubiquitous $3\text{-}4\text{-}5$ right triangle has a height of $2.4$ if we treat the hypotenuse as the base (because $bh=bh \implies 3 \cdot 4 = 5 \cdot h$ ).

We scale it by a factor of $5$ , and try the new side lengths: $15+20+25=60$ , so we have the "right" triangle, and its area is $\dfrac{1}{2} (25)(12) = \boxed{150 \text{ cm}^2}$

$$

Slightly Longer Solution:

We label the sides of the given triangle in the traditional way (side $a$ is opposite $\angle A$ etc.). We have the following system:

$\begin{aligned} 12a &= bc \qquad &&(1)\\ a + b + c &= 60 \qquad &&(2)\\ b^2 + c^2 &= a^2 \qquad &&(3) \end{aligned}$

Rewriting $(2)$ and squaring gives us

$\begin{aligned} b + c &= 60 - a\\ b^2 + 2bc + c^2 &= 3600 - 120a + a^2 \end{aligned}$

Now substituting in $(1)$ and $(3)$ , we get

$\begin{aligned} a^2 + 24a &= 3600 - 120a + a^2 \\ 144a &= 3600 \\ a&=25 \end{aligned}$

Then, as above, the area is $\dfrac{1}{2} (25)(12) = \boxed{150 \text{ cm}^2}$