Bihar Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.5 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 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:

(i) 2 – \(\sqrt{5}\) is an irrational number being a difference between a rational and an irrational.

(ii) (3 + \(\sqrt{23}\)) – \(\sqrt{23}\) = 3 + \(\sqrt{23}\) – \(\sqrt{23}\) = 3, which is a rational number.

(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) = \(\frac { 2 }{ 7 }\), which is a rational number.

(iv) \(\frac{1}{\sqrt{2}}\) is irrational being the quotient of a rational and an irrational.

(v) 2π is irrational being the product of rational and irrational.

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:

Question 3.

Recall, it 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 n is a irrational. How will you resolve this contradiction?

Solution:

There is no contradiction as either c or d irrational and hence π is an irrational number.

Question 4.

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

Solution:

Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semi circle at D.

Then BD = \(\sqrt{9.3}\) . To represent \(\sqrt{9.3}\) On the number line. Let us treat the line BC as the number line, with B as zero, C as 1, and so on. Draw an arc with centre B and radius BD, which intersects the number line in E. Then, E represent \(\sqrt{9.3}\).

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

Solution: