Case Study Questions for Class 9

Case Study Questions for Class 9 on Number system

This page provides case study-based questions for Class 9 students focused on the Number System. Includes solved MCQs, theory explanations, and practice questions aligned with CBSE guidelines for effective exam preparation.

Case Study 1: Representing Irrational Numbers on the Number Line

Theory:

Irrational numbers like $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ cannot be expressed as exact fractions and have non-terminating, non-repeating decimal expansions. To represent $\sqrt{2}$ on the number line, we use the Pythagorean theorem. Construct a right-angled triangle with legs of 1 unit each; the hypotenuse will be $\sqrt{2}$. Using a compass, we transfer this length to the number line. Similarly, $\sqrt{3}$ is constructed using $\sqrt{2}$ as a base, forming a spiral of right triangles. This geometric method ensures accurate placement of irrational numbers.

Question:

A student is constructing $\sqrt{5}$ on the number line using the following steps:

  1. Draw a number line and mark 0 as point O.
  2. From O, measure 2 units to point A.
  3. At A, draw a perpendicular line and mark 1 unit upwards to point B.
  4. Join O to B.

Based on this, answer the following MCQs:

1. What is the length of OB?

2. To construct $\sqrt{3}$ using a similar method, which leg lengths would you use?

3. Which irrational number is constructed using legs of length 1 and $\sqrt{2}$?

4. What is the decimal approximation of $\sqrt{5}$ (rounded to three decimal places)?

5. Which theorem is used to justify this construction?

Solution/Explanation:

  • 1. (c) $\sqrt{5}$. By the Pythagorean theorem, $\text{OB} = \sqrt{(\text{OA})^2 + (\text{AB})^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$.
  • 2. (b) Base 1 unit, Perpendicular $\sqrt{2}$ units. For $\sqrt{3}$, we need $\sqrt{1^2 + (\sqrt{2})^2} = \sqrt{1 + 2} = \sqrt{3}$.
  • 3. (a) $\sqrt{3}$. The length of the hypotenuse is $\sqrt{1^2 + (\sqrt{2})^2} = \sqrt{1 + 2} = \sqrt{3}$.
  • 4. (a) 2.236. The value of $\sqrt{5}$ is approximately 2.23606…
  • 5. (b) Pythagoras theorem. This theorem is the basis for relating the two legs (2 and 1) to the hypotenuse ($\sqrt{5}$).