NCERT Solutions Chapter 1 - Number System (II), Class 9, Maths PDF Download

Exercises 1.4

1. Visualise 3.765 on the number line using successive magnification.

Answer

2. Visualise  on the number line, up to 4 decimal places.

Answer

  = 4.2626

Exercise 1.5

1. Classify the following numbers as rational or irrational:

(i) 2 - √5

(ii) (3 + √23) - √23

(iii) 2√7/7√7

(iv) 1/√2

(v) 2π

Answer

(i) 2 - √5 = 2 - 2.2360679… = - 0.2360679…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + √23) - √23 = 3 + √23 - √23 = 3 = 3/1
Since the number is rational number as it can represented in p/q form.

(iii) 2√7/7√7 = 2/7
Since the number is rational number as it can represented in p/q form.

(iv) 1/√2 = √2/2 = 0.7071067811...
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

 

2. 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)

Answer

3. Recall, π is defined as the ratio of the 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?

Answer

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realise that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857...

4. Represent √9.3 on the number line.

Answer

Step 1: Draw a line segment of unit 9.3. Extend it to C so that BC is of 1 unit.

Step 2: Now, AC = 10.3 units. Find the centre of AC and name it as O.

Step 3: Draw a semi circle with radius OC and centre O.

Step 4: Draw a perpendicular line BD to AC at point B which intersect the semicircle at D. Also, Join OD.

Step 5: Now, OBD is a right angled triangle.
Here, OD = 10.3/2 (radius of semi circle), OC = 10.3/2, BC = 1
OB = OC – BC = (10.3/2) – 1 = 8.3/2
Using Pythagoras theorem,
OD² = BD² OB²
⇒ (10.3/2)² = BD2 (8.3/2)²
⇒ BD² = (10.3/2)² - (8.3/2)²
⇒ BD² = (10.3/2 – 8.3/2) (10.3/2 8.3/2)
⇒ BD² = 9.3
⇒ BD² =  √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure

5. Rationalise the denominators of the following:

(i) 1/√7

(ii) 1/√7-√6

(iii) 1/√5 + √2

(iv) 1/√7-2

Answer

Exercise 1.6

 1. Find:
 (i) 64^1/2
 (ii) 32^1/5
 (iii) 125^1/3

Answer

3. Simplify:

(i) 2^2/3.2^1/5

(ii) (1/3³)⁷

(iii) 11^1/2/11^1/4

(iv) 7^1/2.8^1/2

Answer