Question 15 Marks

Locate $\sqrt{3}$ on the number line.

Answer

Let point $A$ represents $1$ as shown in Figure.
Clearly$, OA = 1 \ unit.$
Now$,$ draw a right triangle $\text{OAB}$ in which $AB = OA = 1\ unit.$
By Using Pythagoras theorem, we have
$OB^2 = OA^2 + AB^2$
$= 1^2 + 1^2$
$= 2 $
$\Rightarrow O B=\sqrt{2}$

Taking $O$ as centre and $OB$ as a radius draw an arc intersecting the number line at point $P.$
Then p corresponds to $\sqrt{2}$ on the number line. Now draw $DB$ of unit length perpendicular to $OB.$
By using Pythagoras theorem, we have
$OD^2 = OB^2 + DB^2$
$OD^2 = (\sqrt{2})^{2} + 12$
$= 2 + 1 = 3$
$OD = \sqrt{3}$
Taking $O$ as centre and $OD$ as a radius draw an arc which intersects the number line at the point $Q.$
Clearly$, Q$ corresponds to $\sqrt{3}$.

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