Thursday, July 23, 2020

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5

Ex 1.5 Class 9 Maths Question 1.
Classify the following numbers as rational or irrational:
(1) 2- \sqrt { 5 }
(2) (3+ \sqrt { 23 } )- \sqrt { 23 }
(3)  \cfrac { 2\sqrt { 7 } }{ 7\sqrt { 7 } }
(4) \cfrac { 1 }{ \sqrt { 2 } }
(5) 2π
Solution:
(1) Irrational ∵ 2 is a rational number and  \sqrt { 5 } is an irrational number.
∴ 2 –  \sqrt { 5 } is an irrational number.
(∵ The difference of a rational number and an irrational number is irrational)
(2) 3 +  \sqrt { 23 } –  \sqrt { 23 } = 3 (rational)
(3)  \cfrac { 2\sqrt { 7 } }{ 7\sqrt { 7 } }  =\cfrac { 2 }{ 7 } (rational)
(4)  \cfrac { 1 }{ \sqrt { 2 } }  (irrational)  ∵ 1 ≠ 0 is a rational number and  \sqrt { 2 }2 ≠ 0 is an irrational a/2 number.
∴   \cfrac { 1 }{ \sqrt { 2 } }  = is an irrational number.
(∵ The quotient of a non-zero rational number with an irrational number is irrational).
(5) 2π (irrational) ∵ 2 is a rational number and π is an irrational number.
∴  2π is an irrational number, (∵ The product of a non-zero rational number with an irrational number is an irrational).
Ex 1.5 Class 9 Maths Question 2.
Simplify each of the following expressions :(1) (3 +  \sqrt { 3 } ) (2 + a/2)
(2) (3 + \sqrt { 3 } ) (3- \sqrt { 3 } )
(3) ( \sqrt { 5 } +  \sqrt { 2 } )2
(4) ( \sqrt { 5 } –  \sqrt { 2 } ) ( \sqrt { 5 } +  \sqrt { 2 } )
Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems 19
Ex 1.5 Class 9 Maths 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 = \cfrac { c }{ d }. This seems to contradict the fact that it is irrational. How will you resolve this contradiction?
Solution:
Actually \cfrac { c }{ d } = \cfrac { 22 }{ 7 } which is an approximate value of π.
Ex 1.5 Class 9 Maths 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 center O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems 20
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 center B and radius BD, which intersects the number line at point E. Then, the point E represent  \sqrt { 9.3 }.
Ex 1.5 Class 9 Maths Question 5.
Rationalise the denominators of the following:
(i)  \cfrac { 1 }{ \sqrt { 7 } }
(ii) \cfrac { 1 }{ \sqrt { 7 } -\sqrt { 6 } }
(iii) \cfrac { 1 }{ \sqrt { 5 } +\sqrt { 2 } }
(iv) \cfrac { 1 }{ \sqrt { 7 } - { 2 } }
Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems 21

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