Question

Question 5
Locate $\sqrt{5},\sqrt{10}$ and $\sqrt{17}$ on the number line.

Answer 1

(i) Here $5=2^2+1^2$
So, draw a right angled ∆OAB, in which OA = 2 units and AB=1 unit and $\angle OAB={90}^{\circ }$
By using Pythagoras theorem, we get
$OB=\sqrt{O{A}^2+A{B}^2}=\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt{5}$

Taking $OB=\sqrt{5}$ as radius and point O as centre, draw an arc which meets the number line at point P on the positive side of it.
Hence, it is clear that point P represents $\sqrt{5}$ on the number line.
(ii) Here, $10=3^2+1$
So, draw a right angled $\Delta OAB$ in which OA=3 units and AB = 1 unit and $\angle OAB={90}^{\circ }$. By using Pythagoras theorem, we get
$OB=\sqrt{O{A}^2+A{B}^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}$

Taking $OB=\sqrt{10}$ as radius and point O as centre, draw an arc which meets the number line at a point P on the positive side of it. The point P represents $\sqrt{10}$ on the number line.

(iii)Here,$17=4^2+1$
So, draw a right angled $\Delta OAB$, in which OA = 4 units and AB = 1 unit and $\angle OAB={90}^{\circ }$
By using Pythagoras theorem, we get
$OB=\sqrt{O{A}^2+A{B}^2}=\sqrt{4^2+1^2}=\sqrt{16+1}=\sqrt{17}$

Taking $OB=\sqrt{17}$ as radius and point O as centre, draw an arc which meets the number line at point P on the positive side of it. The point P represents $\sqrt{17}$ on the number line.