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