Represent sqrt{6} on the number line. | vedclass

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(A) To represent $\sqrt{6}$ on the number line,we use the Pythagorean theorem: $a^2 + b^2 = c^2$.We can write $6 = 5 + 1$,so $\sqrt{6} = \sqrt{(\sqrt{

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Using a compass and ruler.

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(A) To represent $\sqrt{6}$ on the number line,we use the Pythagorean theorem: $a^2 + b^2 = c^2$.We can write $6 = 5 + 1$,so $\sqrt{6} = \sqrt{(\sqrt{5})^2 + 1^2}$.$1$. First,represent $\sqrt{5}$ on the number line by constructing a right-angled triangle with base $2$ units and height $1$ unit.$2$. Mark a point $O$ at $0$ and $A$ at $2$ on the number line.$3$. Draw a perpendicular $AB$ of length $1$ unit at point $A$.$4$. Join $OB$. By Pythagoras,$OB = \sqrt{2^2 + 1^2} = \sqrt{5}$.$5$. Now,draw a perpendicular $BC$ of length $1$ unit at point $B$.$6$. Join $OC$. By Pythagoras,$OC = \sqrt{(\sqrt{5})^2 + 1^2} = \sqrt{6}$.$7$. With $O$ as the center and $OC$ as the radius,draw an arc that intersects the number line at point $P$. The distance $OP$ represents $\sqrt{6}$.

Represent $\sqrt{6}$ on the number line.

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