CBSE Class 9 Mathematics Number System Assignment Set 08

Mathematics Assignment - Class IX
NUMBER SYSTEM

Q1. Find an irrational number between
(i) 1 and 2
(ii) 3 and 4
(iii) 1 and 5
Answer: To find an irrational number between two positive numbers \(a\) and \(b\), we can use the formula \(\sqrt{ab}\) provided the product is not a perfect square.
(i) For 1 and 2: \(\sqrt{1 \times 2} = \sqrt{2}\)
(ii) For 3 and 4: \(\sqrt{3 \times 4} = \sqrt{12}\)
(iii) For 1 and 5: \(\sqrt{1 \times 5} = \sqrt{5}\)
In simple words: Multiply the two given numbers and put a square root sign over the result.

Exam Tip: Always check if the product forms a perfect square; if it does, choose another method (like decimal expansion) to ensure the result is irrational.

Q2. Find two irrational numbers between 3 and 4
Answer: First irrational number = \(\sqrt{3 \times 4} = \sqrt{12}\)
Second irrational number = \(\sqrt{3 \times \sqrt{12}}\) or using decimal non-terminating non-recurring form like \(3.101001000...\)
Using the product method shown in the source key: First number is \(\sqrt{12}\) and second number is \(\sqrt{3 \times \sqrt{12}}\)
In simple words: Find successive square root products between the given limits.

Exam Tip: Between any two real numbers, there exist infinitely many irrational numbers.

Q3. Locate \(\sqrt{2}, \sqrt{3}, \sqrt{5}\) on number line
Answer:
1. For \(\sqrt{2}\): Construct a right-angled triangle with base of 1 unit and perpendicular of 1 unit on the number line. The hypotenuse will be \(\sqrt{1^2 + 1^2} = \sqrt{2}\). Using a compass with center at origin, arc it down to the number line.
2. For \(\sqrt{3}\): Taking the segment of length \(\sqrt{2}\) as base, draw a perpendicular of 1 unit. The new hypotenuse is \(\sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}\). Arc it to the number line.
3. For \(\sqrt{5}\): Construct a right-angled triangle with base of 2 units and perpendicular of 1 unit. The hypotenuse will be \(\sqrt{2^2 + 1^2} = \sqrt{5}\). Arc it to the number line.
In simple words: Use Pythagoras theorem geometrically on the number line starting from unit lengths.

Exam Tip: Always draw sharp arcs using a compass and label each point clearly.

Q4. Find \(\sqrt{3.6}\) geometrically
Answer:
1. Draw a line segment \(AB = 3.6\) units.
2. Extend \(AB\) to \(C\) such that \(BC = 1\) unit.
3. Find the midpoint \(O\) of \(AC\).
4. Draw a semi-circle with center \(O\) and radius \(OC\) (or \(OA\)).
5. Draw a perpendicular line from point \(B\) meeting the semi-circle at point \(D\).
6. The length \(BD\) represents \(\sqrt{3.6}\).
In simple words: Add 1 unit to the given number, find the center of the total length, draw a semicircle, and erect a perpendicular from the original endpoint.

Exam Tip: This construction is based on the geometric mean theorem for right triangles.

Q5. Find the value of \(\sqrt{12}\) geometrically and verify by actual calculations.
Answer:
Geometric method: Express 12 as \(3 \times 4\) or write \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\). Construct segment of length 12 units, extend by 1 unit, find midpoint, draw semicircle, and construct perpendicular.
Actual calculation: \(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\)
Using approximate value of \(\sqrt{3} \approx 1.732\):
\(2 \times 1.732 = 3.464\)
In simple words: Simplify the surd by factoring out perfect squares and compare with geometric construction.

Exam Tip: Always simplify surds completely by extracting factors that are perfect squares.

Q6. Simplify
(i) \(2\sqrt{6} \times 3\sqrt{7}\)
(ii) \(5\sqrt{7} \div 10\sqrt{35}\)
Answer:
(i) \(2\sqrt{6} \times 3\sqrt{7} = (2 \times 3)(\sqrt{6 \times 7}) = 6\sqrt{42}\)
(ii) \(5\sqrt{7} \div 10\sqrt{35} = \frac{5\sqrt{7}}{10\sqrt{35}} = \frac{1}{2} \times \sqrt{\frac{7}{35}} = \frac{1}{2} \times \sqrt{\frac{1}{5}} = \frac{1}{2\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{10}\) (or written as \(\frac{\sqrt{5}}{10}\))
In simple words: Multiply or divide the rational parts together and the radical parts together.

Exam Tip: Rationalize the denominator if required when simplifying division of surds.

Q7. Multiply
(i) \((\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})\)
(ii) \((\sqrt{5} - 2)(\sqrt{5} + 2)\)
Answer:
(i) Using identity \((a+b)(a-b) = a^2 - b^2\):
\((\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1\)
(ii) Using identity \((a-b)(a+b) = a^2 - b^2\):
\((\sqrt{5})^2 - (2)^2 = 5 - 4 = 1\)
In simple words: Use the algebraic difference of squares formula to simplify quickly.

Exam Tip: Recognizing standard algebraic identities saves time and avoids expansion errors.

Q8. Divide \(8\sqrt{15}\) by \(2\sqrt{3}\)
Answer:
\(\frac{8\sqrt{15}}{2\sqrt{3}} = \left(\frac{8}{2}\right) \times \sqrt{\frac{15}{3}}\)
\(= 4 \times \sqrt{5} = 4\sqrt{5}\)
In simple words: Divide the outer numbers together and the numbers inside square roots together.

Exam Tip: Cancel common factors in coefficients before dividing radicands.

Q9. Add \(3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}\)
Answer:
\(3\sqrt{2} + 5\sqrt{2} + 7\sqrt{2} = (3 + 5 + 7)\sqrt{2} = 15\sqrt{2}\)
In simple words: Combine like surds by adding their coefficients.

Exam Tip: Only surds with the same radicand (like terms) can be added or subtracted directly.

Q10. Subtract \(3\sqrt{5}\) from \(5\sqrt{125}\)
Answer:
First, simplify \(5\sqrt{125}\):
\(125 = 25 \times 5\)
\(\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}\)
\(5\sqrt{125} = 5 \times 5\sqrt{5} = 25\sqrt{5}\)
Now, subtract \(3\sqrt{5}\):
\(25\sqrt{5} - 3\sqrt{5} = (25 - 3)\sqrt{5} = 22\sqrt{5}\)
In simple words: Simplify the surds to have the same root part first, then subtract.

Exam Tip: Always simplify all radicals to their simplest form before performing addition or subtraction.

Q11. Sum
Answer: Since no specific additional terms are provided in the source after question 11, the heading indicates continuation of problems or general summation exercises. Based on standard worksheets, this serves as a placeholder for further practice on adding surds.
In simple words: Add the given expressions by combining like terms.

Exam Tip: Group like surds together before summing.