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

**Solution:**

We can write 5 = 4 + 1 = (2)^{2} + (1)^{2}

So we will construct a right angled triangle with sides 2 and 1 and the hypotenuse will be the required measure of $\sqrt{5}$.

- Let AB be a line of length 2 unit on number line.
- At B, draw a perpendicular line BC of length 1 unit. Join CA.
- Now, ABC is a right angled triangle.
Applying Pythagoras theorem,

AB

^{2}+ BC^{2}= CA^{2}⇒ 2

^{2}+ 1^{2}= CA^{2}⇒ CA

^{2}= 5⇒ CA = $\sqrt{5}$

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

- Taking CA as a radius and A as a centre draw an arc cutting the number line at D. Now AD = AC = $\sqrt{5}$, because they are the radii of the same circle.