Class 9 Maths Chapter 4 Exercise 4.1 Solutions – Ganita Manjari

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1 solutions: expanding expressions with algebraic identities, every question solved step by step.

Welcome to Sid Classes, your premier online destination for reliable, accurate, and student-friendly academic resources! If you are preparing for your school evaluations and want to master every concept from the textbook, you have come to the exact right place. In this exhaustive study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1. Designed carefully to reinforce your conceptual understanding of algebraic identities, binomial expansions, and efficient numerical evaluations, this exercise encapsulates everything you need to excel in your mathematical journey.

Our expert-written solutions are structured precisely the way high-achieving students write their answers in school notebooks—utilizing clean mathematical steps, logical proofs, and concise explanations without unnecessary fluff. Whether you are expanding complex algebraic terms or applying formulas to calculate squares quickly, this comprehensive walkthrough will guide you smoothly through every single question with absolute clarity.

🔑 Key Identity to Remember:

  • First Algebraic Identity: \((a + b)^2 = a^2 + 2ab + b^2\)
  • Notebook Method: Always explicitly state the values of \(a\) and \(b\) before substituting them into the identity to avoid calculation errors in your examination notebook.

NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1 Solutions

Question 1

Using the identity \((a + b)^2 = a^2 + 2ab + b^2\), expand the following:

Solution:

(i) \((7x + 4y)^2\)

Here, \(a = 7x\) and \(b = 4y\).
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (7x)^2 + 2(7x)(4y) + (4y)^2\)
\(= \mathbf{49x^2 + 56xy + 16y^2}\)

(ii) \(\left(\frac{7}{5}x + \frac{3}{2}y\right)^2\)

Here, \(a = \frac{7}{5}x\) and \(b = \frac{3}{2}y\).
Using the identity:
\(= \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2\)
\(= \mathbf{\frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2}\)

(iii) \((2.5p + 1.5q)^2\)

Here, \(a = 2.5p\) and \(b = 1.5q\).
Using the identity:
\(= (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2\)
\(= 6.25p^2 + 7.5pq + 2.25q^2\)
\(= \mathbf{6.25p^2 + 7.5pq + 2.25q^2}\)

(iv) \(\left(\frac{3}{4}s + 8t\right)^2\)

Here, \(a = \frac{3}{4}s\) and \(b = 8t\).
Using the identity:
\(= \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2\)
\(= \mathbf{\frac{9}{16}s^2 + 12st + 64t^2}\)

(v) \(\left(x + \frac{1}{2y}\right)^2\)

Here, \(a = x\) and \(b = \frac{1}{2y}\).
Using the identity:
\(= (x)^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2\)
\(= \mathbf{x^2 + \frac{x}{y} + \frac{1}{4y^2}}\)

(vi) \(\left(\frac{1}{x} + \frac{1}{y}\right)^2\)

Here, \(a = \frac{1}{x}\) and \(b = \frac{1}{y}\).
Using the identity:
\(= \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2\)
\(= \mathbf{\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}}\)

Question 2

Using the same identity, find the values of the following:

Solution:

(i) \((64)^2\)

Expressing \(64\) as a sum of a multiple of 10 and a number:
\(= (60 + 4)^2\)
Here, \(a = 60\) and \(b = 4\). Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (60)^2 + 2(60)(4) + (4)^2\)
\(= 3600 + 480 + 16\)
\(= \mathbf{4096}\)

(ii) \((105)^2\)

Expressing \(105\) as a sum:
\(= (100 + 5)^2\)
Here, \(a = 100\) and \(b = 5\). Using the identity:
\(= (100)^2 + 2(100)(5) + (5)^2\)
\(= 10000 + 1000 + 25\)
\(= \mathbf{11025}\)

(iii) \((205)^2\)

Expressing \(205\) as a sum:
\(= (200 + 5)^2\)
Here, \(a = 200\) and \(b = 5\). Using the identity:
\(= (200)^2 + 2(200)(5) + (5)^2\)
\(= 40000 + 2000 + 25\)
\(= \mathbf{42025}\)

Frequently Asked Questions (FAQs)

Q1: How do you expand binomial expressions like \((7x + 4y)^2\)?
A: Identify the first term as \(a\) and second term as \(b\), then substitute them directly into the standard algebraic identity \((a + b)^2 = a^2 + 2ab + b^2\).

Q2: How can algebraic identities help evaluate squares of numbers like \((105)^2\)?
A: Split the number into a convenient sum near a multiple of 10 (e.g., \(100 + 5\)) to simplify multiplication through standard algebraic formulas.

Q3: What textbook do these solutions follow?
A: These answers strictly follow the official Ganita Manjari Part I textbook for NCERT Class 9 Mathematics.

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