Represent the number 5 on the number line.

Let O represent 0 on the number line. Let A be a point on the number line such that O A = 2 units. Draw A B perpendicular to O A at A such that A B = 1 unit. Then, O B = A B 2 + A O 2 = 1 2 + 2 2 = 5 With O as centre and O B as radius draw an arc cutting the line at C . Therefore, C represents 5 on the number line.

Represent the number 5 on the number line.

Let O represent 0 on the number line. Let A be a point on the number line such that O A = 2 units. Draw A B perpendicular to O A at A such that A B = 1 unit. Then, O B = A B 2 + A O 2 = 1 2 + 2 2 = 5 With O as centre and O B as radius draw an arc cutting the line at C . Therefore, C represents 5 on the number line.

Represent the number 7 on the number line.

Let O represent 0 on the number line. Let A be a point on the number line such that O A = 1 unit. Draw A B perpendicular to O A at A such that A B = 1 unit. Then, by Pythagoras theorem, we get, O B = A B 2 + A O 2 = 1 2 + 1 2 = 2 units. With O as centre and O B as radius draw an arc cutting the line at C . Then, O C = O B = 2 units. Again, draw C D perpendicular to the line O C at C such that C D = 1 unit. Then, O D = O C 2 + C D 2 = 2 2 + 1 2 = 3 units Draw an arc with O as centre and O D as radius, cutting the line at E . Then, O E = O D = 3 units. Draw E F perpendicular to the number line at E such that E F = 2 units and join O F . Now, O F = O E 2 + E F 2 = 3 2 + 2 2 = 7 units. With O as centre and O F as radius, draw an arc cutting the line at G . Therefore, O G represents 7 units on the number line.

Represent the number 11 on the number line.

We are given to represent the number 11 on the number line. We can write 11 as 11 = 10 + 1 . So, 11 = ( 10 ) 2 + 1 2 . To locate 11 on the number line, we first need to construct a length of 10 . Again, we can write 10 = 9 + 1 = 3 2 + 1 2 ⇒ 10 = 3 2 + 1 2 . Steps: a) Mark O at 0 and A at 3 on the number line, and then draw AB of unit length perpendicular to O A . Then, by Pythagoras Theorem, we get, O B 2 = O A 2 + A B 2 = 3 2 + 1 2 = 9 + 1 = 10 ⇒ O B = 10 . b) Construct B D of unit length perpendicular to O B . Then, by Pythagoras theorem we get, O D 2 = O B 2 + B D 2 ⇒ O D 2 = 10 2 + 1 2 = 10 + 1 = 11 ⇒ O D = 11 . c) Using a compass with centre at O and radius as O D , draw an arc intersecting the number line at point P . Thus, P corresponds to the number 11 . Therefore, O P is the required length 11 .

Represent the number 20 on the number line.

We are given to represent the number 20 on the number line. We can write 20 as 20 = 16 + 4 . So, 20 = ( 4 ) 2 + 2 2 . To locate 11 on the number line, we first need to construct a length of 10 . Steps: a) Mark O at 0 and A at 4 on the number line, and then draw AB of 2 units perpendicular to O A . Then, by Pythagoras Theorem, we get, O B 2 = O A 2 + A B 2 = 4 2 + 2 2 = 16 + 4 = 20 ⇒ O B = 20 . b) Using a compass with centre at O and radius as O B , draw an arc intersecting the number line at point P . Thus, P corresponds to the number 20 . Therefore, O P is the required length 20 .

Represent the number 29 on the number line.

We are given to represent the number 29 on the number line. We can write 29 as 29 = 25 + 4 . So, 29 = ( 5 ) 2 + 2 2 . To locate 29 on the number line, we first need to construct a length of 29 . Steps: a) Mark O at 0 and A at 5 on the number line, and then draw AB of 2 units perpendicular to O A . Then, by Pythagoras Theorem, we get, O B 2 = O A 2 + A B 2 = 5 2 + 2 2 = 25 + 4 = 29 ⇒ O B = 29 . b) Using a compass with centre at O and radius as O B , draw an arc intersecting the number line at point P . Thus, P corresponds to the number 29 . Therefore, O P is the required length 29 .