# JAC Class 9 Maths Solutions Chapter 1 Number Systems Ex 1.5

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
(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) ($$\sqrt{5}$$ – $$\sqrt{2}$$)($$\sqrt{5}$$ + $$\sqrt{2}$$)
(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 – $$\sqrt{3}$$)
= 32– ($$\sqrt{3}$$)2 = 9 – 3 = 6
[∵ (a + b) (a – b) = a2 – b2]

(iii) ($$\sqrt{5}$$ + $$\sqrt{2}$$)2
= ($$\sqrt{5}$$)2 + ($$\sqrt{2}$$)2 + 2 × $$\sqrt{5}$$ × $$\sqrt{2}$$
[∵ (a + b)2 = a2 + b2 + 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) = a2 – b2]

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?
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.
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
OD2 = BD2 + OB2

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