Represent ($\sqrt{9.3}$) on the number line.

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right angled triangle.

Here, OD =10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC

⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras theorem,

We get,

OD²=BD²+OB²

⟹ (10.3/2)² = BD²+(8.3/2)²

⟹ BD^{2 =} (10.3/2)²-(8.3/2)²

⟹ (BD)² = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD² = 9.3

⟹ BD =  $\sqrt{9.3}$

Thus, the length of BD is $\sqrt{9.3}$.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of $\sqrt{9.3}$ from O as shown in the figure.