Question
Locate $\sqrt{17}$ on the number line.
✓
Answer
Here, $17 = 4^2 + 1$ So, draw a right angled $\triangle\text{OAB}$ in which $OA = 4$ units and $AB = 1$ unit and $\angle\text{OAB}=90^\circ$ By using Pythagoras theorem, we get $\text{OB}=\sqrt{\text{OA}^2+\text{AB}^2}$
$=\sqrt{4^2+1^2}=\sqrt{16+1}=\sqrt{17}$
Taking $\text{OB}=\sqrt{17}$ as radius and point as centre, draw an arc which meets the number line at point on the positive side of it. The point Prepresents $\sqrt{17}$ on the number line.
$=\sqrt{4^2+1^2}=\sqrt{16+1}=\sqrt{17}$
Taking $\text{OB}=\sqrt{17}$ as radius and point as centre, draw an arc which meets the number line at point on the positive side of it. The point Prepresents $\sqrt{17}$ on the number line.
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