Haryana State Board HBSE 9th Class Maths Solutions Chapter 1 Number Systems Ex 1.5 Textbook Exercise Questions and Answers.
Haryana Board 9th Class 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.
Solution :
We have (i) 2 – \(\sqrt{5}\)
Since, difference of a rational and an irrational number is an irrational number. Therefore, 2 – \(\sqrt{5}\) is an irrational number.
(ii) (3 + \(\sqrt{23}\)) – \(\sqrt{23}\)
3+ \(\sqrt{23}\) – \(\sqrt{23}\) = 3
It is a rational number.
Hence, (3 + \(\sqrt{23}\) – \(\sqrt{23}\)) is a rational number.
(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2}{7}\)
It is a rational number.
Hence, \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) is a rational number.
(iv) \(\frac{1}{\sqrt{2}}\)
Since, quotient of a rational number and an irrational number is an irrational number. Therefore, \(\frac{1}{\sqrt{2}}\) is an irrational number.
(v) 2π
Since, π is an irrational number and product of a rational number and an irrational number is an irrational number. Therefore, 2π 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) (\(\sqrt{5}\) – \(\sqrt{2}\)) (\(\sqrt{5}\) + \(\sqrt{2}\)).
Solution:
(i) We have
Question 3.
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = \(\frac {c}{d}\) This seems to contradict the fact that it is irrational. How will you resolve this contradiction?
Solution:
There is no contradiction. Remember that when we measure a length with a scale or any other device, we only get an approximate rational value. So, we may not realize that either cord is irrational Hence, it is an irrational
Question 4.
Represent \(\sqrt{9.3}\) on the number line.
Solution :
Draw a line AB of length 9.3 units. Produce AB to C such that BC = 1 unit. Find the mid-point of AC and marked that point O. Draw a semicircle with centre O and radius OA. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D, then BD = \(\sqrt{9.3}\) units. Now B as centre and BD as radius, draw an arc meeting AC produced at E, then BE = BD = \(\sqrt{9.3}\) units.
Question 5.
Rationalise the denominators of the following:
Solution :
(i) Since, rationalising factor of denominator \(\sqrt{7}\) is \(\sqrt{7}\).
Therefore, multiplying numerator and denominator by \(\sqrt{7}\), we get
∴ \(\frac{1}{\sqrt{7}}=\frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}\)
= \(\frac{\sqrt{7}}{7}\)
(ii) Since, rationalising factor of denominator (\(\sqrt{7}\) – \(\sqrt{6}\)) is (\(\sqrt{7}\) + \(\sqrt{6}\)).
Therefore, multiplying numerator and denominator by (\(\sqrt{7}\) + \(\sqrt{6}\)), we get
(iii) Since, rationalising factor of denominator (\(\sqrt{5}\) + \(\sqrt{2}\)) is (\(\sqrt{5}\) – \(\sqrt{2}\)).
Therefore, multiplying numerator and denominator by (\(\sqrt{5}\) – \(\sqrt{2}\)), we get
(iv) Since, rationalising factor of denominator (\(\sqrt{7}\) – 2) is (\(\sqrt{7}\) + 2).
Therefore, multiplying numerator and denominator by (\(\sqrt{7}\) + 2)
