Question
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
$OD^2 = OB^2 + DB^2$
$OD^2 = (\sqrt{2})^{2}+ 12$
$= 2 + 1 = 3$
$OD = \sqrt{3}$
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 $\sqrt{3}$.
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
$OD^2 = OB^2 + DB^2$
$OD^2 = (\sqrt{2})^{2}+ 12$
$= 2 + 1 = 3$
$OD = \sqrt{3}$
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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