NCERT Solutions Class 9 Mathematics

Exercise 1.5

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

Q.2. Simplify each of the following expressions:

Q.3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = $\frac{\mathrm{c}}{\mathrm{d}}$. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

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

Q.5. Rationalise the denominators of the following:

NCERT Solutions Class 9 Mathematics

Exercise 1.5

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

Solution:

(i) = 2 - 2.2360679… = - 0.2360679…

2 is rational and $\sqrt{5}$ is irrational.

Since the number (2 - $\sqrt{5}$) involves subtraction of a irrational number from a rational number, it is an irrational number.

(ii)

Since the number can be represented in $\frac{\mathrm{p}}{\mathrm{q}}$ form, it is a rational number.

$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7}$

Since the number can be represented in $\frac{\mathrm{p}}{\mathrm{q}}$  form, it is a rational number.

Since the number involves, division of an irrational number by a rational number, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…

π is an irrational number. Since the number involves, multiplication of an irrational number with a rational number, it is an irrational number.

Q.2. Simplify each of the following expressions:

Solution:

(i)

(ii) Using (a + b)(a - b) = a2 - b2, we have,

= 9 - 3 = 6

(iii) Using (a + b)2 = a2 + b2 + 2ab, we have,

(iv) Using (a + b) (a - b) = a2 - b2, we have,

= 5 - 2 = 3

Q.3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = $\frac{\mathrm{c}}{\mathrm{d}}$. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Solution:

There is no contradiction in here. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realise that either c or d is irrational. Hence their ratio will also be irrational.

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

Solution:

1. Draw a line segment of unit 9.3. Extend it to C so that BC = 1 unit.

2. Now, AC = 10.3 units. Find the midpoint of AC. Let it be O.

3. Draw a semi circle with radius OA = OC and centre O.

4. Draw a perpendicular to AC at point B which intersects the semicircle at D. Join OD.

5. Now, OBD is a right angled triangle. The length of BD is $\sqrt{9.3}$.

6. Taking BD as radius and B as centre draw an arc which cuts the line segment AC at E. Now BE is equal to $\sqrt{9.3}$.

Q.5. Rationalise the denominators of the following:

Solution:

$\left(\mathrm{i}\mathrm{i}\right)\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}×\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}×\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$