# Represent √9.3 on the number line

**Solution:**

Let's look into the steps below to represent √9.3 on the number line.

**Step I:** Draw a line and take AB = 9.3 units on it.

**Step II:** From B, measure a distance of 1 unit and mark C on the number line. Mark the midpoint of AC as O.

**Step III:** With ‘O’ as center and OC as radius, draw a semicircle.

**Step IV:** At B, draw a perpendicular to cut the semicircle at D.

**Step V**: With B as a center and BD as radius draw an arc to cut the number line at E. Thus, taking B as the origin the distance BE = √9.3

Therefore, point E represents √9.3 on the number line.

Let's look at the proof shown below.

AB = 9.3, BC = 1

AC = AB + BC = 10.3

OC = AC/2 = 10.3/2 = 5.15

OC = OD = 5.15

OB = OC – BC = 5.15 - 1 = 4.15

In right-angled ∆OBD, using Pythagoras theorem we have,

BD^{2} = OD^{2} - OB^{2}

= (5.15)^{2} - (4.15)^{2}

= (5.15 + 4.15)(5.15 - 4.15) [Using a² - b² = (a + b)(a - b)]

= 9.3 × 1

= 9.3

Hence, BD = √9.3 = BE [Since they are the radii of the same circle]

Thus, we can say point E represents √9.3 on the number line.

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

**Video Solution:**

## Represent √9.3 on the number line

NCERT Solutions Class 9 Maths Chapter 1 Exercise 1.5 Question 4

**Summary:**

With the help of construction and Pythagoras theorem, we have seen the representation of √9.3 on the number line. The point E on the number line represents √9.3.

**☛ Related Questions:**

- Classify the following numbers as rational or irrational: i) 2 - √5 ii) (3 + √23) - √23 iii) 2√7 ÷ 2√7 iv) 1/√2 v) 2π.
- Simplify each of the following expressions: (i) (3 + √3)(2 + √2) (ii) (3 + √3)(3 - √3) (iii) (√5 + √2)² (iv) (√5 - √2)(√5 + √2)
- Recall, π is defined as the ratio of circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?