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 \(\sqrt{5}\) .