--- title: "NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3" url: https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-3 dateModified: 2026-10-07T15:43:39+00:00 --- # NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Chapter 4: Exploring Algebraic Identities. Every question from Exercise 4.3, with full working and the final answer. Free PDF (27 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-4-exploring-algebraic-identities-4d2a728afe.pdf ## Exercise Set 4.3 ### Question 1 *3 marks · Short answer* Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i) $117^2$ (ii) $78^2$ (iii) $198^2$ (iv) $214^2$ (v) $1104^2$ (vi) $1120^2$ **Part (i)** 1. Write $117^2$ as $(100 + 17)^2$ or $(120 - 3)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 100$ and $b = 17$. 2. $117^2 = (100 + 17)^2 = 100^2 + 2(100)(17) + 17^2$ 3. $117^2 = 10000 + 3400 + 289 = 13689$ Answer (i): 13689 **Part (ii)** 1. Write $78^2$ as $(80 - 2)^2$. Using $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 80$ and $b = 2$. 2. $78^2 = (80 - 2)^2 = 80^2 - 2(80)(2) + 2^2$ 3. $78^2 = 6400 - 320 + 4 = 6084$ Answer (ii): 6084 **Part (iii)** 1. Write $198^2$ as $(200 - 2)^2$. Using $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 200$ and $b = 2$. 2. $198^2 = (200 - 2)^2 = 200^2 - 2(200)(2) + 2^2$ 3. $198^2 = 40000 - 800 + 4 = 39204$ Answer (iii): 39204 **Part (iv)** 1. Write $214^2$ as $(200 + 14)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 200$ and $b = 14$. 2. $214^2 = (200 + 14)^2 = 200^2 + 2(200)(14) + 14^2$ 3. $214^2 = 40000 + 5600 + 196 = 45796$ Answer (iv): 45796 **Part (v)** 1. Write $1104^2$ as $(1100 + 4)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 1100$ and $b = 4$. 2. $1104^2 = (1100 + 4)^2 = 1100^2 + 2(1100)(4) + 4^2$ 3. $1104^2 = 1210000 + 8800 + 16 = 1218816$ Answer (v): 1218816 **Part (vi)** 1. Write $1120^2$ as $(1100 + 20)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 1100$ and $b = 20$. 2. $1104^2 = (1100 + 20)^2 = 1100^2 + 2(1100)(20) + 20^2$ 3. $1120^2 = 1210000 + 44000 + 400 = 1254400$ Answer (vi): 1254400 **Answer:** (i) 13689, (ii) 6084, (iii) 39204, (iv) 45796, (v) 1218816, (vi) 1254400 > Common mistake: Choosing numbers that lead to difficult squares of large two-digit numbers. ### Question 2 *3 marks · Short answer* Factor using suitable identities: (i) $16y^2 - 24y + 9$ (ii) $\frac{9}{4}s^2 + 6st + 4t^2$ (iii) $\frac{m^2}{9} - \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$ (iv) $\frac{p^2}{16} - 2 + \frac{16}{p^2}$ (v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$ **Part (i)** 1. $16y^2 - 24y + 9 = (4y)^2 - 2(4y)(3) + 3^2$ 2. Using the identity $a^2 - 2ab + b^2 = (a - b)^2$, we get $(4y - 3)^2$. Answer (i): $(4y - 3)^2$ **Part (ii)** 1. $\frac{9}{4}s^2 + 6st + 4t^2 = \left(\frac{3}{2}s\right)^2 + 2\left(\frac{3}{2}s\right)(2t) + (2t)^2$ 2. Using the identity $a^2 + 2ab + b^2 = (a + b)^2$, we get $\left(\frac{3}{2}s + 2t\right)^2$. Answer (ii): $\left(\frac{3}{2}s + 2t\right)^2$ **Part (iii)** 1. Rewrite the expression as $\left(\frac{m}{3}\right)^2 + \left(\frac{k}{2}\right)^2 + (3n)^2 + 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) + 2\left(\frac{k}{2}\right)(3n) + 2(3n)\left(\frac{m}{3}\right)$ 2. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$, we get $\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2$. Answer (iii): $\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2$ **Part (iv)** 1. $\frac{p^2}{16} - 2 + \frac{16}{p^2} = \left(\frac{p}{4}\right)^2 - 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2$ 2. Using $a^2 - 2ab + b^2 = (a - b)^2$, we get $\left(\frac{p}{4} - \frac{4}{p}\right)^2$. Answer (iv): $\left(\frac{p}{4} - \frac{4}{p}\right)^2$ **Part (v)** 1. $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc = (3a)^2 + (-2b)^2 + c^2 + 2(3a)(-2b) + 2(-2b)(c) + 2(c)(3a)$ 2. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$, we get $(3a - 2b + c)^2$. Answer (v): $(3a - 2b + c)^2$ **Answer:** (i) $(4y - 3)^2$, (ii) $(\frac{3}{2}s + 2t)^2$, (iii) $(\frac{m}{3} + \frac{k}{2} + 3n)^2$, (iv) $(\frac{p}{4} - \frac{4}{p})^2$, (v) $(3a - 2b + c)^2$ > Common mistake: Incorrectly assigning signs to terms with negative products in three-term expansions. ### Question 3 *3 marks · Short answer* Expand the following using the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$: (i) $(p + 3q + 7r)^2$ (ii) $(3x - 2y + 4z)^2$ **Part (i)** 1. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ with $a = p$, $b = 3q$, $c = 7r$. 2. $(p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p) = p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$ Answer (i): $p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$ **Part (ii)** 1. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ with $a = 3x$, $b = -2y$, $c = 4z$. 2. $(3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x) = 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$ Answer (ii): $9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$ **Answer:** (i) $p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$, (ii) $9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$ > Common mistake: Forgetting to multiply the cross terms by 2. ### Question 4 *3 marks · Proof* Is this an identity? $(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2$. **Solution** 1. Consider the L.H.S.: $(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2$. 2. Expand each term using $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$. 3. $(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca$ 4. $(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$ 5. $(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca$ 6. Add the three expansions: $(a^2 + b^2 + c^2 + 2ab - 2bc - 2ca) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca) + (a^2 + b^2 + c^2 - 2ab + 2bc - 2ca)$. 7. Combine like terms: $(1+1+1)a^2 + (1+1+1)b^2 + (1+1+1)c^2 + (2-2-2)ab + (-2-2+2)bc + (-2+2-2)ca$. 8. Simplify to get $3a^2 + 3b^2 + 3c^2 - 2bc - 2ca - 2ab$. Since this does not identically equal $2a^2 + 2b^2 + 2c^2$ for all values, it is not an identity in that form (note: the text prints the question as an exploration of whether it equals $2a^2 + 2b^2 + 2c^2$, and the result shows it is False). **Answer:** No, it is not an identity. > Common mistake: Mistakes in handling negative signs while squaring trinomials. ## Related pages - [All Chapter 4 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions) - [Exercise 4.1](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-1) - [Exercise 4.2](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-2) - [Exercise 4.4](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-4) - [Exercise 4.5](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-5) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. 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