In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that $CD = 2AB$? Give reasons for your answer.                              [Page No. 106]

No, we cannot conclude CD = 2AB.

Let the distance of $CD$ from the centre be $d$, so the distance of $AB$ is $2d$.

Let $r$ be the radius.

Using Baudhāyana–Pythagoras theorem:

$$AB = 2\sqrt{r^2 - (2d)^2} = 2\sqrt{r^2 - 4d^2}$$

$$CD = 2\sqrt{r^2 - d^2}$$

Since, CD = 2AB

$$\begin{aligned} 2\sqrt{r^2 - d^2} &= 4\sqrt{r^2 - 4d^2} \\ \sqrt{r^2 - d^2} &= 2\sqrt{r^2 - 4d^2} \\ r^2 - d^2 &= 4(r^2 - 4d^2) \\ \Rightarrow 15d^2 &= 3r^2 \end{aligned}$$

$$\Rightarrow r^2 = 5d^2$$

$$\Rightarrow r = d\sqrt{5}$$

This only holds for one specific relationship between $r$ and $d$.

In general (for arbitrary $r$ and $d$ with $d < \frac{r}{2}$), $CD
eq 2AB$.

Example: Let $r = 5$ and $d = 1$

Then $AB = 2\sqrt{25 - 4} = 2\sqrt{21} \approx 9.17$ cm and

$CD = 2\sqrt{25} - 1 = 2\sqrt{24} \approx 9.80$ cm.

Here, 2AB ≈ 18.34 cm ≠ CD.

Hence, the conclusion does not hold in general.