Question

Show the number $\sqrt{7}$ on the number line.

Answer 1

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 = $\sqrt{5}$
Taking O as centre and radius OC = $\sqrt{5}$, draw an arc cutting the number line at D.
Clearly, OC = OD = $\sqrt{5}$
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
= ($\sqrt{5}$)² + (1)²
= 5 + 1
= 6
∴ OE = $\sqrt{6}$
Taking O as centre and radius OE = $\sqrt{6}$, draw an arc cutting the number line at F.
Clearly, OE = OF = $\sqrt{6}$
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
= ($\sqrt{6}$)² + (1)²
= 6 + 1
= 7
∴ OG = $\sqrt{7}$
Taking O as centre and radius OG = $\sqrt{7}$, draw an arc cutting the number line at H.
Clearly, OG = OH = $\sqrt{7}$
Hence, H represents $\sqrt{7}$ on the number line.