Question
Represent the numbers √5 and √10 on a number line.
✓
Answer
i. Draw a number line and take point $A$ at 2 .
Draw $A B$ perpendicular to the number line such that $A B=1$ unit.
In $\triangle OAB , m \angle OAB =90^{\circ}$
$\therefore(O B)^2=(O A)^2+(A B)^2 \ldots[$ [Pythagoras theorem]
$=(2)^2+(1)^2$
$\therefore( OB )^2=5$
$\therefore OB =\sqrt{ } 5$ units.... [Taking square root of both sides]
With O as centre and radius equal to OB , draw an arc to intersect the number line at C .
The coordinate of the point C is $√5 $.
ii. Draw a number line and take point Pat 3 .
Draw $P R$ perpendicular to the number line such that $P R=1$ unit.
In $\triangle O P R, m \angle O P R=90^{\circ}$
$\therefore( OR )^2=( OP )^2+( PR )^2 \ldots$ [Pythagoras theorem]
$=(3)^2+(1)^2$
$\therefore( OR )^2=10$
$\therefore OR =\sqrt{ } 10$ units. $\ldots$. [Taking square root of both sides]
With O as centre and radius equal to OR , draw an arc to intersect the number line at Q .
The coordinate of the point Q is $\sqrt{ } 10$.
The coordinate of the point Q is $√10 $.
Draw $A B$ perpendicular to the number line such that $A B=1$ unit.
In $\triangle OAB , m \angle OAB =90^{\circ}$
$\therefore(O B)^2=(O A)^2+(A B)^2 \ldots[$ [Pythagoras theorem]
$=(2)^2+(1)^2$
$\therefore( OB )^2=5$
$\therefore OB =\sqrt{ } 5$ units.... [Taking square root of both sides]
With O as centre and radius equal to OB , draw an arc to intersect the number line at C .
The coordinate of the point C is $√5 $.
ii. Draw a number line and take point Pat 3 .
Draw $P R$ perpendicular to the number line such that $P R=1$ unit.
In $\triangle O P R, m \angle O P R=90^{\circ}$
$\therefore( OR )^2=( OP )^2+( PR )^2 \ldots$ [Pythagoras theorem]
$=(3)^2+(1)^2$
$\therefore( OR )^2=10$
$\therefore OR =\sqrt{ } 10$ units. $\ldots$. [Taking square root of both sides]
With O as centre and radius equal to OR , draw an arc to intersect the number line at Q .
The coordinate of the point Q is $\sqrt{ } 10$.
The coordinate of the point Q is $√10 $.
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