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# Show how $$\sqrt{5}$$ can be represented on the number line.

We know that, $$\sqrt{5}$$ = $$\sqrt{4 + 1}$$ = $$\sqrt{2^2 + 1^2}$$.

Draw a right angled triangle, OQP, such that OQ = 2 Units and PQ = 1 Unit And $$\angle{OQP} = 90°$$

Now, by using Pythagoras theorem, we have $$OQ^2 = OP^2 + PQ^2 = 2^2 + 1^2$$.

Therefore, OP = $$\sqrt{5}$$.
Now, take O as centre OP = $$\sqrt{5}$$ as radius, draw an arc, which intersects the line at point R.
Hence, the point R represents $$\sqrt{5}$$ .