Jharkhand Board JAC Class 9 Maths Solutions Chapter 1 Number Systems Ex 1.5 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 1 Number Systems Exercise 1.5

Question 1.

Classify the following numbers as rational or irrational:

(i) 2 – \(\sqrt{5}\)

(ii) (3 + \( \sqrt{23}\)) – \(\sqrt{23}\)

(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)

(iv) \(\frac{1}{\sqrt{2}}\)

(v) 2π

Answer:

(i) 2 – \(\sqrt{5}\) = 2 – 2.2360679…

= -0.2360679…

Since the number is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + \(\sqrt{23}\)) – \(\sqrt{23}\)

3 + \( \sqrt{23}\) – \(\sqrt{23}\)

= 3 = \(\frac{3}{1}\)

The number is rational number as it can represented in \(\frac{p}{q}\) form, where p, q ∈ Z, q ≠ 0

(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)

The number is rational number as it can represented in \(\frac{p}{q}\) form, where p, q ∈ Z, q ≠ 0

(iv) \(\frac{1}{\sqrt{2}}\)

= 0.7071067811…

Since the number is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…

Since the number is non-terminating non-recurring therefore, it is an irrational number.

Question 2.

Simplify each of the following expressions:

(i) (3 + \(\sqrt{3}\)) (2 + \(\sqrt{2}\))

(ii) (3 + \(\sqrt{3}\)) (3 – \(\sqrt{3}\))

(iii) (\(\sqrt{5}\) + \(\sqrt{2}\))^{2}

(iv) (\(\sqrt{5}\) – \(\sqrt{2}\))(\(\sqrt{5}\) + \(\sqrt{2}\))

Answer:

(i) (3 + \(\sqrt{3}\)) (2 + \(\sqrt{2}\))

= 3 × 2 + 2\(\sqrt{3}\) + 3\(\sqrt{2}\) + \(\sqrt{3}\) ×\(\sqrt{2}\)

= 6 + 2\(\sqrt{3}\) + 3\(\sqrt{2}\) + \(\sqrt{6}\)

(ii) (3 + \(\sqrt{3}\)) (3 – \(\sqrt{3}\))

= 3^{2}– (\(\sqrt{3}\))^{2} = 9 – 3 = 6

[∵ (a + b) (a – b) = a^{2} – b^{2}]

(iii) (\(\sqrt{5}\) + \(\sqrt{2}\))^{2}

= (\(\sqrt{5}\))^{2} + (\(\sqrt{2}\))^{2} + 2 × \(\sqrt{5}\) × \(\sqrt{2}\)

[∵ (a + b)^{2} = a^{2} + b^{2} + 2ab]

= 5 + 2 + 2 × \(\sqrt{5}\) ×\(\sqrt{2}\)

= 7 + 2\(\sqrt{10}\)

(iv) (\(\sqrt{5}\) – \(\sqrt{2}\))(\(\sqrt{5}\) + \(\sqrt{2}\))

= (\(\sqrt{5}\))^{2} – (\(\sqrt{2}\))^{2}

= 5 – 2 = 3

[∵ (a + b) (a – b) = a^{2} – b^{2}]

Question 3.

Recall, 7t is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, n = c/d. This seems to contradict the fact that n 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 n is almost equal to 22/7 or 3.14159…

Question 4.

Represent \(\sqrt{9.3}\) on the number line.

Answer:

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

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

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

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

Step 5: Now, OBD is a right angled triangle.

Here, OD = 10.3/2 (radius of semicircle),

OC = \(\frac{10.3}{2}\), BC = 1

OB = OC – BC = (\(\frac{10.3}{2}\)) – 1 = \(\frac{8.3}{2}\)

Using Pythagoras theorem

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

Thus, the length of BD is \(\sqrt{9.3}\) units.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment AC extended at the point E. The point E is at a distance of \(\sqrt{9.3}\) from B as shown in the figure.

Question 5.

Rationalise the denominators of the following:

(i) \(\frac{1}{\sqrt{7}}\)

(ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

(iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)

(iv) \(\frac{1}{\sqrt{7}-2}\)

Answer:

(i) \(\frac{1}{\sqrt{7}}\) = \(\frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}\) = \(\frac{\sqrt{7}}{7}\)

(ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

(iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)

(iv) \(\frac{1}{\sqrt{7}-2}\)