How to draw square root of 9.3 on number line?
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Draw a line segment AB of length 9.3 units.
Extend the line 1 unit more such that BC = 1 unit.
Draw a perpendicular bisector of AC.
Mark It as P
Draw a semicircle with the center as P and radius OC
Construct 90 degrees at point B, intersecting the semicircle at point D.
Here BD = Root 9.3, Draw an arc DE such that BE=BD
E is root 9.3
Extend the line 1 unit more such that BC = 1 unit.
Draw a perpendicular bisector of AC.
Mark It as P
Draw a semicircle with the center as P and radius OC
Construct 90 degrees at point B, intersecting the semicircle at point D.
Here BD = Root 9.3, Draw an arc DE such that BE=BD
E is root 9.3
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Answer:
Hey, here's your answer
Step-by-step explanation:
Draw a line of 9.3cm and mark the points as O&A respectively. Then from O draw a line segment of 1cm and mark the point as B.
Take your rounder and stretch from O to B nd make a perpendicular bisector. Where both lines meet up, mark that point as A'. Take your rounder again and stretch it from point O till A'. Make a semi circle by keeping the tip at A'. After making it, take a scale and keep it on A and draw a vertical line upwards where it meets the surface of the semi circle and mark that poit as C. Keep the rounder at C of lenght AC. And mark an arc on the plane or the horizontal line which we made. And that arc is your √9.3
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