Question
Represent $\sqrt{6},\sqrt{7},\sqrt{8}$ on the number line.
✓
Answer
Draw a number line and mark a point O, representing zero, on it. Suppose point A represents 1 as shown. Then, $O A$
$=1$.
Draw a right triangle $O A B$ such that $A B=O A=1$.
By pythagoras theorem, we have
$(OB)^2=(OA)^2+(AB)^2$
$\Rightarrow OB{ }^2=1^2+1^2=OB 2=1+1=2$
$\Rightarrow OB=\sqrt{2}$
Now, draw a circle with centre O and radius OB . We find that the arcle cuts the number line at A .
Clearly, $A _1= OB =$ Radius of the cirde $=\sqrt{2}$
Thus, $A _1$ represents $\sqrt{2}$ on the number line.
Now, draw a right triangle $OBB _1$, such that $BB _1=2$.
Again by pythagoras theorem, we have,
$OB_1^2=OB^2+BB_1^2$
$\Rightarrow OB_1^2=(\sqrt{2})^2+(2)^2$
$\Rightarrow OB_1^2=6$
$\Rightarrow OB_1=\sqrt{6}$
Now, draw a circle with centre O and radius $OB _1$. We find that the circle cuts the number line at $A _2$.
Clearly, $OA _2= OB _2=$ Radius of circle $=\sqrt{6}$
Thus, $A_2$ represents $\sqrt{6}$ on the number line.
Now, draw a right angle triangle $OB _1 B_2$ such that $B _1 B_2=1$.
By pythagoras theorem, we have,
$OB_2^2=OB_1^2+B_1 B_2^2$
$\Rightarrow OB_2^2=(\sqrt{6})^2+(1)^2$
$\Rightarrow OB_2^2=6+1=7$
$\Rightarrow OB_2=\sqrt{7}$
Now, draw a circle with centre O and radius $OB _2$. We find that the circle cuts the number line at $A _3$.
Clearly, $OA _3= OB _2=$ Radius of circle $=\sqrt{7}$
Thus, $A_3$ represents $\sqrt{7}$ on the number line.
Now, again draw a right triangle $OB _2 B_3$ such that $B _2 B_3=1$.
By pythagoras theorem, we have,
$OB_3^2=OB_2^2+B_2 B_3^2$
$\Rightarrow OB_3^2=(\sqrt{7})^2+(1)^2$
$\Rightarrow OB_3^2=7+1=8$
$\Rightarrow OB_3=\sqrt{8}$
Now, draw a circle with centre O and radius $OB _3$. We find that the circle cuts the number line at $A _4$.
Clearly, $OA _4= OB _3=$ Radius of circle $=\sqrt{8}$
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