Explain why the following statement is true: If the perpendicular distance of a chord from the centre is $d$ and the radius is $r$, then the chord length is $2\sqrt{(r^2 - d^2)}$.                                                                             [Page No. 105]

Given statement can be explained (as true statement) with following justification:

Let us consider a circle with centre O, chord AB and M be the foot of perpendicular from O to AB.

Then, $OM = d$ and $OA = r$.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

i.e., $\frac{AM}{AB} = 2$

In right triangle OMA (right-angled at M), by the Baudhāyana–Pythagoras theorem:

$$\begin{aligned} OA^2 &= OM^2 + AM^2 \\ r^2 &= d^2 + AM^2 \\ AM^2 &= r^2 - d^2 \\ AM &= \sqrt{r^2 - d^2} \end{aligned}$$

Therefore, chord length $AB = 2AM = 2\sqrt{r^2 - d^2}$.