Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
(Hint: Is it a circumcircle of a suitable triangle?)

Chord length = $2\sqrt{r^2 - d^2}$ :

$$6 = 2\sqrt{(r^2 - 9)}$$

$$\Rightarrow 3 = \sqrt{(r^2 - 9)}$$

$$\Rightarrow 9 = r^2 - 9$$

$$\Rightarrow r^2 = 18$$

$$\Rightarrow r = \sqrt{18} = 3\sqrt{2} \text{ cm}$$

Construction steps:

1. Draw a circle with centre O and radius $3\sqrt{2}$ cm ($\approx 4.24$ cm).

2. Mark a point M inside the circle with OM = 3 cm.

3. Draw the line through M perpendicular to OM.

4. This line intersects the circle at points A and B. AB = 6 cm is the required chord.

Verification: Consider an isosceles triangle with vertex at O and base AB. The altitude from O has length 3 cm and the base = 6 cm.

The triangle's circumcircle has radius

$$r = \sqrt{3^2 + 3^2} = 3\sqrt{2} \text{ cm.}$$