What is the least number of (fixed) parameters I can ask for, when calculating area of a triangle of unknown type? I need to calculate the area of a triangle, but I don't know, whether it is right angled, isoscele or equilateral.
What parameters do I need to calculate the area of a triangle of unknown type?
 A: The area of a triangle can also be calculated as $\frac12 rp$, where $r$ is the radius of the incircle and $p$ is the perimeter.
A: There is another general formula for the area of a triangle: $A=\frac{1}{2}ab\sin\theta$ where $\theta$ is the angle enclosed by the sides of length $a$ and $b$. This is simply a generalisation of $A=\frac{1}{2}bh$ when the concept of height doesn't correspond to a side of the triangle.
A: There are a lot of answers to this question--all* of them will involve knowing 3 parameters about the triangle; the formula you use depends on which 3 you know.
So far we've seen Heron's formula if you know three sides and the law of cosines approach if you know two sides and an angle (note that also subsumes the formula for a right triangle).
I'll add the shoelace formula if you know the coordinates $(x_i,y_i), i=1,\ldots,3$ of the vertices:
$$\frac{1}{2}|x_1y_2 + x_2y_3 + x_3y_1-y_1x_2-y_2x_3-y_3x_1|$$
*not exactly, see comment below
A: All you need is the lengths of each side of the triangle.
By Heron's Formula, we know for a triangle with sides $a,b,c$, we have
$$A=\sqrt{s(s-a)(s-b)(s-c)}\text{ ,where }s=\frac{a+b+c}2$$
Reference:
https://en.wikipedia.org/wiki/Heron%27s_formula

EDIT:
In response to suggestion by @Hurkyl , I now add the case of ASA and SsA.

(ASA)
With a known side $c$ and its neighbouring angles $\alpha,\beta$, we have:
$$A=\frac{c^2\sin\alpha\sin\beta}{2\sin{(\alpha+\beta)}}$$
Proof:
Let $\alpha,\beta,\theta$ be the corresponding angles of sides $a,b,c$ respectively, we have
$$\theta=\pi-\alpha-\beta\implies\sin\theta=\sin{(\pi-\alpha-\beta)}=\sin{(\alpha+\beta)}$$
By Sine Law,
$$\frac c{{\sin{(\pi-\alpha-\beta)}}}=\frac a{\sin\alpha}=\frac b{\sin\beta}$$
So $a=\frac{c\sin\alpha}{\sin{(\alpha+\beta)}}$, $b=\frac{c\sin\beta}{\sin{(\alpha+\beta)}}$
Therefore,
\begin{align}
A&=\frac12ab\sin\theta\\&=\frac12\cdot{\frac{c\sin\alpha}{\sin{(\alpha+\beta)}}}\cdot\frac{c\sin\beta}{\sin{(\alpha+\beta)}}\cdot\sin{(\alpha+\beta)}\\&=\frac{c^2\sin\alpha\sin\beta}{2\sin{(\alpha+\beta)}}
\end{align}

(SsA)
With 2 known sides $b,c$ and $\beta$, the corresponding angle of $b$, we have:
$$A=\frac12 c\sin\beta[\sqrt{1-(\frac{c\sin\beta}b)^2}+c\cos\beta]$$
Proof:
By Sine Law:
$$\frac b{\sin\beta}=\frac c{\sin\theta}$$
$$\sin\theta=\frac{c\sin\beta}b, \cos\theta=\sqrt{1-(\frac{c\sin\beta}b)^2}$$
By Cosine Law,
$$c^2=a^2+b^2-2ab\cos\theta$$
$$c^2-b^2+b^2\cos^2\theta=a^2-2ab\cos\theta+b^2\cos^2\theta$$
$$c^2-b^2\sin^2\theta=(a-b\cos\theta)^2$$
$$a=b\cos\theta+\sqrt{c^2-b^2\sin^2\theta}=b\cos\theta+c\cos\beta$$
So
\begin{align}
A&=\frac12ab\sin\theta\\&=\frac12(b\cos\theta+c\cos\beta)(b)\cdot\frac{c\sin\beta}b\\&=\frac12 c\sin\beta[\sqrt{1-(\frac{c\sin\beta}b)^2}+c\cos\beta]
\end{align}
A: *

*If all the sides say $a$, $b$ & $c$ of a triangle are known then the area of the triangle is calculated as follows  $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ Where, $s=\frac{a+b+c}{2}=\text{semi perimeter}$

*If we know any two sides & the angle included between them then the area of triangle is given as follows $$A=\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin A=\frac{1}{2}ac\sin B$$ 
