Represent (√9.3) on the number line.
Represent (√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 semicircle 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 semicircle), OC = 10.3/2 , BC = 1
OB = OC – BC
⟹ (10.3/2)1 = 8.3/2
Using Pythagoras theorem,
We get,
OD^{2}= BD^{2 }+ OB^{2}
⟹ (10.3/2)^{2} = BD^{2}+(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)
⟹ BD^{2} = 9.3
⟹ BD = √9.3
Thus, the length of BD is √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 √9.3 from O as shown in the figure.

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