AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30cm, then find the distance of AB from the center of the circle.

It is given that, Diameter, AD = 34 cm, AB = 30cm,

Radiusc = AO,
$\begin{array}{l}\Rightarrow \mathrm{AO}=\frac{1}{2}\mathrm{AD}\\ \Rightarrow \mathrm{AO}=\frac{1}{2}\times 34\mathrm{cm}\\ \Rightarrow \mathrm{AO}=17\mathrm{cm}\end{array}$ 
We have, $\mathrm{ON}\perp \mathrm{AB}$ 

As we know, a perpendicular from the center to a chord bisects the chord into two equal parts,
Hence, ON  bisects AB at N, 
So, $\mathrm{AN}=\frac{1}{2}\mathrm{AB}$ 
$\begin{array}{l}\Rightarrow \mathrm{AN}=\frac{1}{2}\times 30\\ \Rightarrow \mathrm{AN}=15 \mathrm{cm}\end{array}$ 
Now, $\mathrm{AO}=17 \mathrm{cm}, \mathrm{AN}=15 \mathrm{cm},$ and $\angle \mathrm{ANO}=90°$ 
$\mathrm{\Delta AON}$ is the right triangle with hypotenuse AO.
By Pythagoras' theorem, we have,
⇒ ${\mathrm{AO}}^2-{\mathrm{AN}}^2={\mathrm{ON}}^2$ 
⇒ ${\text{ON}}^2=({17}^2-{15}^2)\mathrm{ }{\mathrm{cm}}^2$ 
⇒ $\text{ON=8\hspace{0.17em}cm}$ 
Thus, AB  is at a distance of 8 cm from the center O. 
Hence, the correct option is 8 cm.