-Ushma Gaur, Subject Matter Expert at Edumarz

Solution:

Draw a number line and marking as point O and A at the place 0 and 1 on the drawn number line. Then draw BA of unit length perpendicular to OA. After that join O to B.

In triangle OAB, by Pythagoras theorem, OB2=OA2+AB2

OB2=12+12=2

Using the compass with Centre O and radius OB, draw an arc intersecting the number line at the point P. Then P corresponds to 2 on the number line.

Further, draw BD of unit length perpendicular to OB and join OD. Then using Pythagoras theorem, OD2=OB2+BD2

OD2=22+12=3

Using the compass with centre O and radius OD, draw an arc intersecting the number line at the point Q. Then Q corresponds to 3 on the number line.

Further, draw CD of unit length perpendicular to OD and join OC. Then using Pythagoras theorem,

OC2=OD2+CD2

OC2=32+12=4

Using the compass with centre O and radius OC, draw an arc intersecting the number line at the point R = 2 . Then R corresponds to4 on the number line.

Again, repeating the above process one more time to get 5.

Draw ED of unit length perpendicular to OC and join OE. Then using Pythagoras theorem,

OE2=OC2+CE2

OE2=42+12=5

Using the compass with centre O and radius OE, draw an arc intersecting the number line at the point S . Then S corresponds to 5 on the number line.