What is the smallest possible area for a triangle of integral side lengths whose perimeter is 15cm?

Let the sides of the triangle be $a$, $b$, and $c$. Since the perimeter is 15cm, we have $a+b+c=15$. We want to find the smallest possible area of the triangle, which is achieved when the triangle is a right triangle. Therefore, we will try to find a Pythagorean triple $(a,b,c)$ that satisfies $a+b+c=15$.

Since $a^2+b^2=c^2$, we have $a+b+c=\sqrt{a^2+b^2}+c$, which implies $c = 15 - \sqrt{a^2+b^2}$. Substituting this into the Pythagorean equation, we get

$$a^2+b^2 = (15-\sqrt{a^2+b^2})^2 = 225-30\sqrt{a^2+b^2}+a^2+b^2.$$

Simplifying, we get

$$30\sqrt{a^2+b^2}=225 \qquad\Rightarrow\qquad \sqrt{a^2+b^2}=\frac{225}{30}=\frac{15}{2}.$$

Since $a$ and $b$ are integers, we want to find Pythagorean triples with hypotenuse $\frac{15}{2}$. The only such triple is $(5,12,13)$. Therefore, the sides of the triangle are $a=5$, $b=12$, and $c=8$, and the area is

$$\frac{1}{2}ab = \frac{1}{2}(5)(12) = 30.$$

Therefore, the smallest possible area for a triangle of integral side lengths whose perimeter is 15cm is $\boxed{30}$ square centimeters.

7, 7, 1 gets an area of about 3.491.

Don't know whether thats the best though.

One way as follows: