# Locate √5, √10 and √17 on the number line

**Solution:**

We have to locate √5 on the number line.

Expressing 5 as a sum of two perfect square numbers.

5 = (2)² + (1)²

4 + 1 = 5

Consider OA = 2 units

Now draw BA perpendicular to OA

Let BA = 1 units

Now join OB.

Using a compass with centre O and radius OB draw an arc that intersects the number line at the point C.

Now, C = √5

Considering triangle OAB,

OB² = OA² + AB²

OB² = (2)² + (1)²

OB² = 4 + 1

OB² = 5

Taking square root,

Therefore, OB = √5

We have to locate √10 on the number line.

Expressing 5 as a sum of two perfect square numbers.

10 = (1)² + (3)²

1 + 9 = 10

Consider OA = 3 units

Now draw BA perpendicular to OA

Let BA = 1 units

Now join OB.

Using a compass with centre O and radius OB draw an arc that intersects the number line at the point C.

Now, C = √10

Considering triangle OAB,

By pythagoras theorem,

OB² = OA² + AB²

OB² = (3)² + (1)²

OB² = 9 + 1

OB² = 10

Taking square root,

Therefore, OB = √10

We have to locate √17 on the number line.

Expressing 5 as a sum of two perfect square numbers.

17 = (4)² + (1)²

16 + 1 = 17

Consider OA = 4 units

Now draw BA perpendicular to OA

Let BA = 1 units

Now join OB.

Using a compass with centre O and radius OB draw an arc that intersects the number line at the point C.

Now, C = √5

Considering triangle OAB,

By pythagoras theorem,

OB² = OA² + AB²

OB² = (4)² + (1)²

OB² = 16 + 1

OB² = 17

Taking square root,

Therefore, OB = √17

**✦ Try This: **Locate √7 on the number line.

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 1

**NCERT Exemplar Class 9 Maths Exercise 1.3 Problem 5**

## Locate √5, √10 and √17 on the number line

**Summary:**

A visual representation of numbers on a straight line drawn either horizontally or vertically is known as a number line. The point C on the first number line represents √5. The point C on the second number line represents √10. The point C on the third number line represents √17

**☛ Related Questions:**

visual curriculum