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**Karnataka State Syllabus Class 9 Maths Chapter 1 Number Systems Ex 1.5**

## Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.5

Question 1.

Classify the following numbers as rational or irrational.

Answer:

i) 2−5–√

ii) (3+23−−√)−23−−√

iii) 27√77√

iv) 12√

v) 2π

Answer:

i) 2−5–√ = 2 – 2.2360679……….. = -0.2360679

This is a non-terminating, non-recurring decimal.

∴ This is an irrational number.

ii) (3+23−−√)−23−−√

= 3+23−−√−23−−√

= 3

⇒ 31. This can be written in the form of pq.

∴ This is a rationl number.

iii) 27√77√=27√7√=2⇒21

This can be written in the form of pq

∴ This is a rational number.

iv) 12√

12√×2√2√=2√2=1.41422

= 0.707106…………

This is a non-terminating, non-recurring decimal.

∴ This is an irrational number.

v) 2π

= 2 × 3.1415…..

= 6.2830…….

This is a non-terminating, non-recurring decimal.

∴ This is an irrational number.

Question 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:

Question 3.

Recall, 2π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π=cd. 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 length with a scale or any other device we get the quotient.

Therefore we cannot judge whether c is a rational number of d is an irrational number.

∴ Value of cd is irrational number.

∴ The value of π is also an irrational number.

Question 4.

Represent 9.3−−−√ on the number line.

Answer:

Construction: Mark the distance 9.5 units from a fixed point O such that OB = 9.3 units. Mark midpoint D of OC. Draw a semicircle with centre D. Draw a line perpendicular to OC passing through E and intersecting the semicircle at E. Draw an arc BE which intersect at F. Now, BF = 9.3−−−√.

Question 5.

Rationalise the denominators of the following :

Answer:

i) 17√

Denominator’s factor is 7–√ Mulitplying Numerator and denominator by 7–√.

ii) 17√−6√

Denominator’s factor is 7–√+6–√

Multiplying numerator and denominator by 7–√+6–√,

iii) 15√+2√

Denominator’s factor is 5–√−2–√

Multiplying numerator and denominator by 5–√−2–√.

iv) 17√−2

Denominator’s factor is 7–√+2

Multiplying numerator and denominator by 7–√+2.

**All Chapter KSEEB Solutions For Class 9 Maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 9**

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