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Toggle## Class 9 Maths Chapter 1 Exercise 1.5 Solutions

**Q.1. Classify the following numbers as rational or irrational: **

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

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

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

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

**(v) ****2****𝛑**

**Ans:**

**Q.2. Simplify each of the following expressions:**

**(i) ** $\left( 3+\sqrt{3} \right)\left( 2+\sqrt{2} \right)$

**(ii) ** $\left( 3+\sqrt{3} \right)\left( 3-\sqrt{3} \right)$

**(iii) ** ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}$

**(iv) ** $\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}+\sqrt{2} \right)$

**Ans:**

**Q.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?**

**Ans: **Writing $\pi \,\,as\,\,\frac{22}{7}$ is not the exact value. It is an approximate value and also a non – terminating, non-repeating decimal. Which means π is irrational number.

if we calculate the value of $\,\,\frac{22}{7}$, we get

$\,\,\frac{22}{7}=3.142857142857143$

∴ $\pi \,\,\ne \,\,\frac{22}{7}$

Therefore, π is an irrational number.

**Q.4. Represent x $\sqrt{9.3}$. on the number line.**

**Ans:**

**Q.5. Rationalize 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) $\frac{1}{\sqrt{7}-2}$**

**Ans:**