exercise 1.5 class 9 maths

Question 1:

Classify the following numbers as rational or irrational:$$

(i)${\displaystyle 2-\sqrt{5}}$

(ii)$(3+\sqrt{23})-\sqrt{23}$

(iii)${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}}$

(iv) ${\displaystyle \frac{1}{\sqrt{2}}}$

(v) $2\pi$

Solution 1:

(i)$2-\sqrt{5}$

$\sqrt{5}=2.236\dots$ ,is a non-terminating and non-repeating irrational number.

$2\sqrt{5}=2-2.236\dots$2√5 ( By substituting the value of $\sqrt{5}$ )

$=–0.236\dots$, which is also an irrational number.

Therefore, $2-\sqrt{5}$ is an irrational number.

(ii)$(3+\sqrt{23})-\sqrt{23}$

$(3+\sqrt{23})-\sqrt{23}=3+\surd \sqrt{23}-\sqrt{23}=3$ which is in the form of ${\displaystyle \frac{p}{q}}$ and it is a rational number.

Therefore, we conclude that $(3+\sqrt{23})-\sqrt{23}$ is a rational number.

(iii)${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}={\displaystyle \frac{2}{7}}}$

Which is in the form of ${\displaystyle \frac{p}{q}}$ and it is a rational number.

Therefore, we conclude that ${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}}$ is a rational number.(iv) ${\displaystyle \frac{1}{\sqrt{2}}}$

$\sqrt{2}=1.414\dots$ , is a non-terminating and non-repeating irrational number.

Therefore, we conclude that

${\displaystyle \frac{1}{\sqrt{2}}}$ is an irrational number.

(v) $2\pi =3.1415\dots$, which is an irrational number.

Observe, Rational × Irrational = Irrational

Therefore $2\pi$ will also be an irrational number.

Question 2:

Simplify each of the following expressions:

(i)(3+√3)(2+√2)

(ii)(3+√3)(3-√3) 

(iii)(√5+√2))² 

(iv)(√5-√2)(√5+√2)

Solution 2:

(i)$(3+\surd 3)(2+\surd 2)$

$(3+\surd 3)(2+\surd 2)=3(2+\surd 2)+\surd 3(2+\surd 2)$

$=6+3\surd 2+2\surd 3+\surd 6$

(ii)$(3+\surd 3)(3-\surd 3)$

Using the identity

$(a+b)(a-b)=a^2-b^2$

$\therefore (3+\surd 3)(3-\surd 3)=9-3=6$

(iii)$(\surd 5+\surd 2{)}^2=\surd 5^2+\surd 2^2+2.\surd 5.\surd 2$

$=5+2\surd 10+2$

$=7+2\surd 10$

(iv)$(\surd 5-\surd 2)(\surd 5+\surd 2)$

Using the identity

(a+b)(a-b)=a²-b²

$\therefore$ (√5-√2)(√5+√2) = 5 - 2 = 3

Question 4. Represent √9.3 on the number line.

Solution 4:

•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 point as C.

•Find the mid-point of AC and mark the point as O.

•Draw a semi-circle with centre O and radius OC = 5.15 units

$(AC=AB+BC=9.3+1=10.3$. Therefore $OC={\displaystyle \frac{AC}{2}}={\displaystyle \frac{10.3}{2}}=5.15$.

•Draw a line perpendicular at B and draw an arc with centre $B$ and let meet at semicircle AC at D

• BE = BD = $\sqrt{9.3}$ (Radius of an arc DE).

Question 5.

Rationalize the denominators of the following:

i.${\displaystyle \frac{1}{\sqrt{7}}}$

ii.${\displaystyle \frac{1}{\sqrt{7}-\sqrt{6}}}$

iii.${\displaystyle \frac{1}{\sqrt{5}+\surd 2}}$

iv.${\displaystyle \frac{1}{\sqrt{7}-2}}$

Solution 5:

(i)${\displaystyle \frac{1}{\sqrt{7}}}$

Multiply and divide by $\sqrt{7}$

${\displaystyle \frac{1}{\sqrt{7}}}={\displaystyle \frac{1\times \sqrt{7}}{\sqrt{7}\times \sqrt{7}}}={\displaystyle \frac{\sqrt{7}}{7}}$

(ii)${\displaystyle \frac{1}{\sqrt{7}+\sqrt{6}}}$

Multiply and divide by $\sqrt{7}+\sqrt{6}$,We get

${\displaystyle \frac{1\times (\sqrt{7}-\sqrt{6})}{(\sqrt{7}+\sqrt{6})(\sqrt{7}-\sqrt{6})}}$

=${\displaystyle \frac{\sqrt{7}-\sqrt{6}}{7-6}}$

=$\sqrt{7}-\sqrt{6}$

iii.${\displaystyle \frac{1}{\sqrt{5}+\sqrt{2}}}$

Multiply and divide by $\sqrt{5}-\sqrt{2}$,We get

$1\times {\displaystyle \frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}-\sqrt{2})\times (\sqrt{5}+\sqrt{2})}}$

=${\displaystyle \frac{\sqrt{5}-\sqrt{2}}{5-2}}$

=${\displaystyle \frac{\sqrt{5}-\sqrt{2}}{3}}$

iv.${\displaystyle \frac{1}{\sqrt{7}-2}}$

Multiply and divide by $\sqrt{7}-2$, We get

${\displaystyle \frac{1\times \sqrt{7}+2}{(\sqrt{7}-2)\times (\sqrt{7}+2)}}$

=${\displaystyle \frac{\sqrt{7}+2}{7-4}}$

=${\displaystyle \frac{\sqrt{7}+2}{3}}$