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) $\frac{1}{\sqrt{7}-2}=\frac{1}{(\sqrt{7}-2)(\sqrt{7}+2)}$
$=\frac{\sqrt{7}+2}{(\sqrt{7})^{2}-(2)^{2}}$
$=\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}$
0 comments:
Post a Comment