Question
Represent the number $\sqrt{7}$ on the number line.
✓
Answer
Let us find $\sqrt{5}$.
Draw a number line.
Mark a point $O$ representing zero.
Take point $A$ on number line such that $O A=2$
Construct $A B \perp O A$ such that $A B=1$ unit.
$\therefore \triangle OAB$ is a right triangle.
In $\triangle OAB ,( OB )^2=( OA )^2+( AB )^2 ($Pythagoras' Theorem$)$
$\therefore(O B)^2=2^2+1^2$
$\therefore( OB )^2=5$
$\Rightarrow OB =\sqrt{5}$
Now, let us find $\sqrt{6}$.
Construct $BC \perp OB$, such that $BC =1$ unit.
$\therefore \triangle OBC$ is a right triangle.
In $\triangle OBC , OC ^2= OB ^2+ BC ^2 ($Pythagoras' Theorem$)$
$\therefore O C^2=(\sqrt{5})^2+1^2$
$\therefore O C^2=6$
$\Rightarrow O C=\sqrt{6}$
Now, let us find $\sqrt{7}$.
Construct $C D \perp O C$,
such that $C D=1$ unit.
In $\triangle O C D, O D^2=O C^2+C D^2 ($Pythagoras' Theorem$)$
$\therefore O D^2=(\sqrt{6})^2+1^2$
$\therefore OD ^2=7$
$\Rightarrow \sqrt{7}$
Draw an arc of radius $O D$ and centre $O$ and let it intersect the number line at point $E$.
$\therefore \sqrt{7}$ is thus marked at point $E$ on the number line.
Draw a number line.
Mark a point $O$ representing zero.
Take point $A$ on number line such that $O A=2$
Construct $A B \perp O A$ such that $A B=1$ unit.
$\therefore \triangle OAB$ is a right triangle.
In $\triangle OAB ,( OB )^2=( OA )^2+( AB )^2 ($Pythagoras' Theorem$)$
$\therefore(O B)^2=2^2+1^2$
$\therefore( OB )^2=5$
$\Rightarrow OB =\sqrt{5}$
Now, let us find $\sqrt{6}$.
Construct $BC \perp OB$, such that $BC =1$ unit.
$\therefore \triangle OBC$ is a right triangle.
In $\triangle OBC , OC ^2= OB ^2+ BC ^2 ($Pythagoras' Theorem$)$
$\therefore O C^2=(\sqrt{5})^2+1^2$
$\therefore O C^2=6$
$\Rightarrow O C=\sqrt{6}$
Now, let us find $\sqrt{7}$.
Construct $C D \perp O C$,
such that $C D=1$ unit.
In $\triangle O C D, O D^2=O C^2+C D^2 ($Pythagoras' Theorem$)$
$\therefore O D^2=(\sqrt{6})^2+1^2$
$\therefore OD ^2=7$
$\Rightarrow \sqrt{7}$
Draw an arc of radius $O D$ and centre $O$ and let it intersect the number line at point $E$.
$\therefore \sqrt{7}$ is thus marked at point $E$ on the number line.
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