Class 9 Maths Chapter 1 Exercise 1.5 Solutions​

\( 2-\sqrt{5} \)
\(=2-2.2360679 \ldots\)
\(=-0.2360679 \ldots\)
Since the number is in non-terminating non-recurring, therefore, it is an irrational number

\( (3+\sqrt{23})-\sqrt{23}=3+\sqrt{23}-\sqrt{23} \)
\( =3 \)
\( =\frac{3 }{ 1} \)
A number is a rational number as it can be represented in the form of \(\frac{\mathrm{p}}{\mathrm{q}}\).

\( \frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2 }{ 7 }\)
The number is a rational number as it can be represented in the form of \(\frac{\mathrm{p}}{\mathrm{q}}\).

\( \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \)
\(=0.7071067811 \ldots . .\)
Since the number is a non-terminating and non-recurring number.
Hence, it is an irrational number

\( 2 \pi=2 \times 3.1415 \ldots \)
\(=6.2830 \ldots \ldots\)
Since the number is a non-terminating and non-recurring number.
Hence, it is an irrational number

\((3+\sqrt{3} )(2+\sqrt{2} )\)
\(= 3(2+\sqrt{2} )+\sqrt{3}(2+\sqrt{2})\)
\(= 3 \times 2+3 \sqrt{2} +2 \sqrt{3} +\sqrt{2} \times \sqrt{3} \)
\(= 6+3 \sqrt{2} +2 \sqrt{3} +\sqrt{6}\)

\((3+\sqrt{3})(3-\sqrt{3}) \quad\) [Since, \( (a+b)(a-b)=a^{2}-b^{2}] \)
\(=3^{2}-(\sqrt{3})^{2}\)
\(=9-3\)
\(=6\)

\( (\sqrt{5}+\sqrt{2 } )^{2} \quad\) [Since,\((a+b)^{2}=a^{2}+b^{2}+2 a b\)]
\(=(\sqrt{5})^{2}+(\sqrt{2} )^{2}+2 \times \sqrt{5} \times \sqrt{2}\)
\(=5+2+2 \times\sqrt{10}\)
\(=7+2 \sqrt{ 10} \)

\((\sqrt{ } 5-\sqrt{ } 2)(\sqrt{5}+\sqrt{ } 2)\quad\) [Since, \((a+b)(a-b)=a^{2}-b^{2}\)]
\(=(\sqrt{ } 5)^{2}-(\sqrt{ } 2)^{2}\)
\(=5-2\)
\(=3\)

There is no contradiction in the statement.
When we measure any value with a scale, we only obtain an approximate value.
We never obtain an accurate value.
Hence, we cannot say that either c or d is irrational.
The value of \( \pi \) is almost equal to \( \frac{22 }{ 7 }\) or \( 3.142857 \ldots \)

Step 1: Draw a line segment \( A B \) of \(9.3\) unit. Then, extend it to \(C\) so that \( \mathrm{BC}=1 \) unit.

Step 2: Now, \( A C=10.3 \) units. Find the center of \( A C \) and mark it as \( O \)

Step 3: Draw a semi-circle with radius \(OC\) and center \(O\).

Step 4: Draw a perpendicular line \(BD\) to \(AC\) at point \(B\) intersecting the semi-circle at \(D\). And then, join \(OD\)

Step 5: Now, \(OBD\) is a right angled triangle
Here, \(OD\) (Radius of semi-circle)
\(\mathrm{OC}=\frac{10.3}{2}\)
\(\mathrm{BC}=1\)
Then, \( \mathrm{OB}=\mathrm{OC}-\mathrm{BC} \)
Using Pythagoras theorem,
\(\mathrm{OD}^{2}=\mathrm{BD}^{2}+\mathrm{OB}^{2}\)
\(\left(\frac{10.3}{2}\right)^{2}=\mathrm{BD}^{2}+\left(\frac{8.3}{2}\right)^{2}\)
\(\mathrm{BD}^{2}=\left(\frac{10.3}{2}\right)^{2}-\left(\frac{8.3}{2}\right)^{2}\)
\(\mathrm{BD}^{2}=\left(\frac{10.3}{2}-\frac{8.3}{2}\right)\left(\frac{10.3}{2}+\frac{8.3}{2}\right)\)
\(\mathrm{BD}^{2}=9.3\)
\(\mathrm{BD}=\sqrt{9.3}\)
Thus, the length of BD is \( \sqrt{9.3} \)
Step 6: Taking \(BD\) as radius and \(B\) as the center, construct an arc which touches the line segment.
Now, the point where it touches the line segment is at a distance of from \(O\) as shown in the figure below

Rationalizing the denominator means that we have to remove the irrational component from the denominator of the fraction.
\( \frac{1}{\sqrt{7}} \)
To remove the \( \sqrt{7} \) from the denominator, multiply and divide the fraction with \( \sqrt{7} \)
\(\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}\)

Rationalizing the denominator means that we have to remove the irrational component from the denominator of the fraction.
\( \frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}} \)
We multiplied and divided the equation by \( \sqrt{7}+\sqrt{6} \) so that the denominator forms the formula of
\((a+b)(a-b)=a^{2}-b^{2}\)
\(\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}\)
\(\frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6}\)

Rationalizing the denominator means that we have to remove the irrational component from the denominator of the fraction.
\( \frac{1}{\sqrt{5}+\sqrt{2}}\)
\(= \frac{1(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}\)
\(= \frac{\sqrt{5}-\sqrt{2}}{5-2}\)
\(= \frac{\sqrt{5}-\sqrt{2}}{3}\)

Rationalizing the denominator means that we have to remove the irrational component from the denominator of the fraction.
\( \frac{1}{\sqrt{7-2}}\)
\(=\frac{(\sqrt{7}+2)}{\sqrt{7} \times \sqrt{7}-2 \times 2}\)
\(=\frac{\sqrt{7}+2}{7-4}\)
\(=\frac{\sqrt{7}+2}{3}\)