EXERCISE 1.5
EVIDYARTHIQ 1, Ex 1.5, Page No 24, Number Systems, Class 9th NCERT
MKR
Class - 9th, Ex - 1.5, Q 1 ( NUMBER SYSTEM ) CBSE NCERT
Page No 24:
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…
As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number.
(ii) (3+\sqrt{23})-\sqrt{23}=3=\frac{3}{1}
As it can be represented in \frac{p}{q} form, therefore, it is a rational number.
(iii) \frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2}{7}
As it can be represented in \frac{p}{q} form, therefore, it is a rational number.
(iv) \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}
=0.7071067811...
As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number.
(v) 2π = 2(3.1415 …)
= 6.2830 …
As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number.
EVIDYARTHI
Q 2, Ex 1.5, Page No 24, Number Systems, CBSE Maths Class 9th
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})=3(2+\sqrt{2})+\sqrt{3}(2+\sqrt{2})
=6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}
(ii) (3+\sqrt{3})(3-\sqrt{3})
=(3)^{2}-(\sqrt{3})^{2}
= 9 − 3 = 6
(iii) (\sqrt{5}+\sqrt{2})^{2}
=(\sqrt{5})^{2}+(\sqrt{2})^{2}+2(\sqrt{5})(\sqrt{2})
=5+2+2 \sqrt{10}
=7+2 \sqrt{10}
(iv) (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})
=(\sqrt{5})^{2}-(\sqrt{2})^{2}
= 5 − 2 = 3
MKR
Class - 9th, Ex - 1.5, Q 3 ( NUMBER SYSTEM ) CBSE NCERT
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 π is irrational.
How will you resolve this contradiction?
Answer:
There is no contradiction. When we measure a length with scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value. For this reason, we may not realise that either c or d is irrational. Therefore, the fraction \frac{c}{d} is irrational. Hence, π is irrational.EVIDYARTHIQ 4, Ex 1.5, Page No 24, Number Systems, Class 9th Maths
MKR
Class - 9th, Ex - 1.5, Q 4 ( NUMBER SYSTEM ) CBSE NCERT show Root 9.3 on number line
Question 4:
Represent \sqrt{9.3} on the number line.Answer:
Mark a line segment OB = 9.3 on number line. Further, take BC of 1 unit. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre. Draw a perpendicular to line OC passing through point B. Let it intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at F. BF is \sqrt{9.3}.
EVIDYARTHIQ 5, Ex 1.5, Page No 24, Number Systems, Class 9th Maths
MKR
How to Rationalize the Denominators
Class - 9th, Ex - 1.5, Q 5 ( NUMBER SYSTEM ) CBSE NCERT
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}
Answer:
(i) \frac{1}{\sqrt{7}}=\frac{1 \times \sqrt{7}}{1 \times \sqrt{7}}=\frac{\sqrt{7}}{7}(ii) \frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}
=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}
=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\frac{\sqrt{7}+\sqrt{6}}{1}
=\sqrt{7}+\sqrt{6}
(iii) \frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}
=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}
=\frac{\sqrt{5}-\sqrt{2}}{3}
(iv) \frac{1}{\sqrt{7}-2}=\frac{1}{(\sqrt{7}-2)(\sqrt{7}+2)}
=\frac{\sqrt{7}+2}{(\sqrt{7})^{2}-(2)^{2}}
=\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}
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