Question
Locate $\sqrt{3}$ on the number line.
✓
Answer
Let point $A$ represents $1$ as shown in Figure.
Clearly, $\text{OA}=1$ unit.
Now, draw a right triangle $\text{OAB}$ in which $\text{AB=OA}=1$ unit.
By Using Pythagoras theorem, we have
$\ce{OB^2 = OA^2 + AB^2}$
$=1^2+1^2$
$=2$
$\Rightarrow \text{OB}=\sqrt{2}$
Taking $O$ as centre and $O B$ 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
$\ce{OD^2 = OB^2 + DB^2}$
$OD^2=(\sqrt{2})^2+12$
$=2+1=3$
$OD=\sqrt{3}$
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, $\text{OA}=1$ unit.
Now, draw a right triangle $\text{OAB}$ in which $\text{AB=OA}=1$ unit.
By Using Pythagoras theorem, we have
$\ce{OB^2 = OA^2 + AB^2}$
$=1^2+1^2$
$=2$
$\Rightarrow \text{OB}=\sqrt{2}$
Taking $O$ as centre and $O B$ 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
$\ce{OD^2 = OB^2 + DB^2}$
$OD^2=(\sqrt{2})^2+12$
$=2+1=3$
$OD=\sqrt{3}$
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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