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Sia 🤖 1 year, 1 month ago

We can write 10 as

{tex}10 = 9 + 1 = 3^2 + 1^2{/tex}

Draw OA = 3 units, on the number line

Draw BA = 1 unit, perpendicular to OA.

Join OB

Figure:

Now, by Pythagoras theorem,

{tex}OB^2 = AB^2 + OA^2\\ OB^2 = 1^2 + 3^2 = 10{/tex}

{tex}\Rightarrow \quad O B=\sqrt{10}{/tex}

Taking O as centre and OB as a radius, draw an arc which intersects the number line at point C.

Clearly, OC corresponds to {tex}\sqrt{10}{/tex} on the number line.

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