Students can Download Class 9 Maths Chapter 1 Number Systems Ex 1.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 9 Maths 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
Answer:
i) \(2-\sqrt{5}\) = 2 – 2.2360679……….. = – 0.2360679
This is a non-terminating, non-recurring decimal.
∴ This is an irrational number.
ii) \((3+\sqrt{23})-\sqrt{23}\)
= \(3+\sqrt{23}-\sqrt{23}\)
= 3
⇒ \(\frac{3}{1}\). This can be written in the form of \(\frac{p}{q}\).
∴ This is a rationl number.
iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2 \sqrt{7}}{\sqrt{7}}=2 \Rightarrow \frac{2}{1}\)
This can be written in the form of \(\frac{p}{q}\)
∴ This is a rational number.
iv) \(\frac{1}{\sqrt{2}}\)
\(\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.4142}{2}\)
= 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) \((3+\sqrt{3})(2+\sqrt{2})\)
ii) \((3+\sqrt{3})(3-\sqrt{3})\)
iii) \((\sqrt{5}+\sqrt{2})^{2}\)
iv) \((\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\)
Answer:
i) \((3+\sqrt{3})(2+\sqrt{2})\)
Solution:
3(2 + √2) + √3(2 + √2)
= 6 + 3√2 + 2√3 + √6
ii) \((3+\sqrt{3})(3-\sqrt{3})\)
Solution:
it is in the form of (a + b) (a – b)
∴ (3 +√3)(3 – √3) = 32 – (√3)2
= 9 – 3 = 6
iii) \((\sqrt{5}+\sqrt{2})^{2}\)
Solution:
it is in the form of (a + b)2
∴ (√5 + √2)2 = (√5)2 + 2. √5. √2
= 5 + 2 + 2√10 = 7 + 2√10.
iv) \((\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\)
Solution:
It is in the form of (a – b) (a + b)
= a2 – b2
(√5 – √2)
Question 3.
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, \(\pi=\frac{c}{d}\). 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 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 \(\frac{c}{d}\) is irrational number.
∴ The value of π is also an irrational number.
Question 4.
Represent \(\sqrt{9.3}\) 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 = \(\sqrt{9.3}\).
Question 5.
Rationalise the denominators of the following:
Solution:
i) \(\frac{1}{\sqrt{7}}\)
Denominator’s factor is \(\sqrt{7}\) Mulitplying Numerator and denominator by \(\sqrt{7}\).
ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
Denominator’s factor is \(\sqrt{7}+\sqrt{6}\)
Multiplying numerator and denominator by \(\sqrt{7}+\sqrt{6}\),
{Multiplying on both numerator and denominator}
iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
Solution:
Denominator’s factor is \(\sqrt{5}-\sqrt{2}\)
Multiplying numerator and denominator by \(\sqrt{5}-\sqrt{2}\).
iv) \(\frac{1}{\sqrt{7}-2}\)
Solution:
Denominator’s factor is \(\sqrt{7}+2\)
Multiplying numerator and denominator by \(\sqrt{7}+2\).
