Question 14 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 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
$OD^2 = OB^2 + DB^2$
OD² = $(\sqrt{2})^{2}$ + 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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