Question

# Locate $$\sqrt {3}$$ on number line

Solution

## Construct $$BD$$ of $$1$$ unit length perpendicular to $$OB$$ as fig. Join $$OD$$By Pythagoras theorem, $$OD = \sqrt {(\sqrt {2})^{2} + 1^{2}} = \sqrt {2 + 1} = \sqrt {3}$$Using a compass, with centre $$O$$ and radius $$OD$$, draw an arc which intersects the number line at the point $$L$$ right side to $$0$$. Then $$'L'$$ corresponds to $$\sqrt {3}$$. From this we can conclude that many points on the number line can be represented by irrational numbers also.  In the same way, we can locate $$\sqrt {n}$$ for any positive integers $$n$$, after $$\sqrt {n - 1}$$ has been located.Maths

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