How To Find The Length Of An Irregular Arc How would you find out the length of an irregular arc. e.g. An arc with a base length of $10$cm and a height of $5$cm  - what would be the length of that arc? Is there a specific formula I could use?

I assume you mean a circular arc, with chord length ("base length") of $w$ and distance along the perpendicular bisector of the base from the base chord to the arc of $h$ (the "height" of the arc).
Although by those definitions, the arc you describe is easy (it is in fact a semi-circle of a circle with radius 5 cm, so has a length of $5\pi$) the general case is not as easy but also tractable using elementary trigonometry:
Call the chord (the base) $AB$, and let $P$ be the midpoint of $AB$, $Q$ the meeting point on the arc, and $O$ be the circle center of the arc from $A$ to $B$.  Then $OA = OQ$ and in right triangle $OPA$, 
$$
OP = OA - h \\
PA = w/2 \\
(OA)^2 = (OP)^2+(PA)^2 = (OA)^2-2h(OA) +h^2 + \frac{w^2}{4} \\
2h(OA) = h^2 + \frac{w^2}{4}\\ 
OA = \frac{h}{2} + \frac{w^2}{8h}
\\
\cos \angle POA = \frac{OA-h}{OA} = \frac{  \frac{w^2}{8h} - \frac{h}{2} }
{ \frac{w^2}{8h}+\frac{h}{2} } 
$$ 
and the length of the arc from $A$ to $Q$ and then from $Q$ to $B$ is
$$
2 (OA) \cos^{-1}\frac{w^2-4h^2}{w^2+4h^2} = 2 \frac{4h^2+w^2}{8h}\cos^{-1}\frac{w^2-4h^2}{w^2+4h^2}
$$
In the example given, $w = 10, h=5$ and the arc length is
$$
2 \frac{4\cdot 25 +100}{8\cdot 5}\cos^{-1}\frac{100-4\cdot 25}{100-4\cdot 25}
=2\cdot \frac{200}{40} \cos^{-1}(0) = 2\cdot 5 \cdot \frac{\pi}{2} = 5\pi
$$
If, instead, the height were 3 cm then the circular arc length would be
$$
\frac{34}{3}\cos^{-1}\frac{8}{17} \approx 12.25$$
