Question

Question 3
Show how $\sqrt{5}$ can be represented on the number line.

Answer 1

Step 1: Let AB be a line of length 2 unit on the number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.

Step 3: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

$A{B}^2+B{C}^2=C{A}^2\Rightarrow 2^2+1^2=C{A}^2\Rightarrow C{A}^2=5\Rightarrow CA=\sqrt{5}$

Thus, CA is a line of length $\sqrt{5}$ unit.

Step 4: Taking CA as a radius and A as a centre draw an arc touching the number line. The point at which number line get intersected by the arc is at $\sqrt{5}$ distance from 0 because it is a radius of the circle whose centre is A.

Thus, $\sqrt{5}$ is represented on the number line as shown in the figure.