Class 9 Maths Chapter 4 Exercise 4.3 Solutions – Ganita Manjari
Welcome to Sid Classes, your trusted destination for accurate, reliable, and student-friendly academic resources! If you are working through algebraic expansions and trinomial identities, you are in the right place. In this detailed study guide, we provide complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3. Designed to reflect how top-scoring students write their answers in school examinations, this walkthrough makes complex square calculations, trinomial factorizations, and identity verifications entirely clear.
Our expert-written solutions focus on choosing the most efficient algebraic identities—such as \((a+b)^2\), \((a-b)^2\), and \((a+b+c)^2\)—to simplify complex numerical squares and multi-variable polynomial expressions. Follow along to strengthen your problem-solving skills with absolute confidence.
🔑 Key Identities Used in this Exercise:
- Square of Sum: \((a + b)^2 = a^2 + 2ab + b^2\)
- Square of Difference: \((a - b)^2 = a^2 - 2ab + b^2\)
- Square of a Trinomial: \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)
NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions
Question 1
Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier:
Solution:
(i) \(117^2\)
Choosing the subtraction identity by expressing \(117\) as \((120 - 3)\) makes the calculation simpler since squaring \(120\) is straightforward:
\(= (120 - 3)^2\)
Here, \(a = 120\) and \(b = 3\). Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (120)^2 - 2(120)(3) + (3)^2\)
\(= 14400 - 720 + 9\)
\(= \mathbf{13689}\)
(ii) \(78^2\)
Expressing \(78\) as \((80 - 2)\):
\(= (80 - 2)^2\)
Here, \(a = 80\) and \(b = 2\). Using the identity:
\(= (80)^2 - 2(80)(2) + (2)^2\)
\(= 6400 - 320 + 4\)
\(= \mathbf{6084}\)
(iii) \(198^2\)
Expressing \(198\) as \((200 - 2)\):
\(= (200 - 2)^2\)
Here, \(a = 200\) and \(b = 2\). Using the identity:
\(= (200)^2 - 2(200)(2) + (2)^2\)
\(= 40000 - 800 + 4\)
\(= \mathbf{39204}\)
(iv) \(214^2\)
Expressing \(214\) as \((210 + 4)\):
\(= (210 + 4)^2\)
Here, \(a = 210\) and \(b = 4\). Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (210)^2 + 2(210)(4) + (4)^2\)
\(= 44100 + 1680 + 16\)
\(= \mathbf{45796}\)
(v) \(1104^2\)
Expressing \(1104\) as \((1100 + 4)\):
\(= (1100 + 4)^2\)
Here, \(a = 1100\) and \(b = 4\). Using the identity:
\(= (1100)^2 + 2(1100)(4) + (4)^2\)
\(= 1210000 + 8800 + 16\)
\(= \mathbf{1218816}\)
(vi) \(1120^2\)
Expressing \(1120\) as \((1100 + 20)\):
\(= (1100 + 20)^2\)
Here, \(a = 1100\) and \(b = 20\). Using the identity:
\(= (1100)^2 + 2(1100)(20) + (20)^2\)
\(= 1210000 + 44000 + 400\)
\(= \mathbf{1254400}\)
Question 2
Factor using suitable identities:
Solution:
(i) \(16y^2 - 24y + 9\)
\(= (4y)^2 - 2(4y)(3) + (3)^2\)
Using the identity \(a^2 - 2ab + b^2 = (a - b)^2\), where \(a = 4y\) and \(b = 3\):
\(= \mathbf{(4y - 3)^2}\)
(ii) \(\frac{9}{4}s^2 + 6st + 4t^2\)
\(= \left(\frac{3}{2}s\right)^2 + 2\left(\frac{3}{2}s\right)(2t) + (2t)^2\)
Using the identity \(a^2 + 2ab + b^2 = (a + b)^2\), where \(a = \frac{3}{2}s\) and \(b = 2t\):
\(= \mathbf{\left(\frac{3}{2}s + 2t\right)^2}\)
(iii) \(\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2\)
Rearranging terms into the trinomial form \(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\):
\(= \left(\frac{m}{3}\right)^2 + \left(\frac{k}{2}\right)^2 + (3n)^2 + 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) + 2\left(\frac{k}{2}\right)(3n) + 2(3n)\left(\frac{m}{3}\right)\)
Using the identity \((a + b + c)^2\), where \(a = \frac{m}{3}\), \(b = \frac{k}{2}\), and \(c = 3n\):
\(= \mathbf{\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2}\)
(iv) \(\frac{p^2}{16} - 2 + \frac{16}{p^2}\)
\(= \left(\frac{p}{4}\right)^2 - 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2\)
Using the identity \(a^2 - 2ab + b^2 = (a - b)^2\), where \(a = \frac{p}{4}\) and \(b = \frac{4}{p}\):
\(= \mathbf{\left(\frac{p}{4} - \frac{4}{p}\right)^2}\)
(v) \(9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc\)
Notice the negative signs are associated with terms containing variable \(b\) (\(-12ab\) and \(-4bc\)), meaning \(b\) is negative:
\(= (3a)^2 + (-2b)^2 + (c)^2 + 2(3a)(-2b) + 2(-2b)(c) + 2(c)(3a)\)
Using the identity \((a + b + c)^2\), where \(a = 3a\), \(b = -2b\), and \(c = c\):
\(= \mathbf{(3a - 2b + c)^2}\)
Question 3
Expand the following using the identity \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\):
Solution:
(i) \((p + 3q + 7r)^2\)
Here, \(a = p\), \(b = 3q\), and \(c = 7r\).
\(= (p)^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)\)
\(= \mathbf{p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr}\)
(ii) \((3x - 2y + 4z)^2\)
Here, \(a = 3x\), \(b = -2y\), and \(c = 4z\).
\(= (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)\)
\(= \mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx}\)
Question 4
Is this an identity? \((a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2\)
Solution:
Let us expand each term on the Left Hand Side (LHS):
1. \((a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca\)
2. \((a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca\)
3. \((a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca\)
Adding all three expanded terms together:
\(\text{LHS} = (a^2 + a^2 + a^2) + (b^2 + b^2 + b^2) + (c^2 + c^2 + c^2) + (2ab - 2ab - 2ab) + (-2bc - 2bc + 2bc) + (-2ca + 2ca - 2ca)\)
\(\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca\)
Since \(\text{LHS} \neq 2a^2 + 2b^2 + 2c^2\) for all values of \(a, b, \text{ and } c\),
\(\mathbf{\text{No, this equation is not an identity.}}\)
Frequently Asked Questions (FAQs)
Q1: How do you choose whether to use \((a+b)^2\) or \((a-b)^2\) for numbers like \(117\)?
A: Choose the form that uses a cleaner multiple of 10. Since \(117\) is close to \(120\), writing it as \((120 - 3)\) is easier and involves smaller numbers than \((100 + 17)\).
Q2: How do you handle negative signs when factoring trinomials like \(9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc\)?
A: Look at the negative terms to find the common variable. Both \(-12ab\) and \(-4bc\) contain \(b\), which indicates that the term containing \(b\) must be negative (i.e., \(-2b\)).
Q3: What makes an equation an "identity"?
A: An equation is an identity only if it holds true for every possible numerical value of the variables involved, which can be verified by expanding and simplifying both sides.
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