Show the number √7 on the number line.

#### Solution

Draw a number line as shown in the figure and mark the points O, A and B on it such that OA = AB = 1 unit. The point O represents 0 and B represents 2. At B, draw CB perpendicular on the number line such that BC = 1 unit. Join OC. Now, ∆OBC is a right angled triangle.

In ∆OBC, by Pythagoras theorem

(OC)^{2} = (OB)^{2} + (BC)^{2}

= (2)^{2} + (1)^{2}

= 4 + 1

= 5

∴ OC = √5

Taking O as centre and radius OC = √5 , draw an arc cutting the number line at D.

Clearly, OC = OD = √5

At D, draw ED perpendicular on the number line such that ED = 1 unit. Join OE. Now, ∆ODE is a right angled triangle.

In ∆ODE, by Pythagoras theorem

(OE)^{2} = (OD)^{2} + (DE)^{2}

= (√5)^{2} + (1)^{2}

= 5 + 1

= 6

∴ OE = √6

Taking O as centre and radius OE = √6 , draw an arc cutting the number line at F.

Clearly, OE = OF = √6

At F, draw GF perpendicular on the number line such that GF = 1 unit. Join OG. Now, ∆OFG is a right angled triangle.

In ∆OFG, by Pythagoras theorem

(OG)^{2} = (OF)^{2} + (FG)^{2}

= (√6)^{2} + (1)^{2}

= 6 + 1

= 7

∴ OG = √7

Taking O as centre and radius OG = √7 , draw an arc cutting the number line at H.

Clearly, OG = OH = √7

Hence, H represents √7 on the number line.