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Ex 2.4
Ex 2.4, 1 (ii)
Ex 2.4, 1 (iii) Important
Ex 2.4, 1 (iv)
Ex 2.4, 1 (v)
Ex 2.4, 2 (i) Deleted for CBSE Board 2024 Exams
Ex 2.4, 2 (ii) Important
Ex 2.4, 2 (iii)
Ex 2.4, 3 (i) Deleted for CBSE Board 2024 Exams
Ex 2.4, 3 (ii) Important
Ex 2.4, 3 (iii)
Ex 2.4, 4 (i) Deleted for CBSE Board 2024 Exams
Ex 2.4, 4 (ii)
Ex 2.4, 4 (iii) Important
Ex 2.4, 4 (iv)
Ex 2.4, 4 (v)
Ex 2.4, 4 (vi)
Ex 2.4, 5 (i) Deleted for CBSE Board 2024 Exams
Ex 2.4, 5 (ii) Important
Ex 2.4, 6 (i)
Ex 2.4, 6 (ii)
Ex 2.4, 6 (iii)
Ex 2.4, 6 (iv) Important
Ex 2.4, 7 (i) Deleted for CBSE Board 2024 Exams
Ex 2.4, 7 (ii)
Ex 2.4, 7 (iii) Important
Ex 2.4, 8 (i)
Ex 2.4, 8 (ii)
Ex 2.4, 8 (iii) Important
Ex 2.4, 8 (iv) Important
Ex 2.4, 8 (v)
Ex 2.4, 9 (i)
Ex 2.4, 9 (ii)
Ex 2.4, 10 (i) Important Deleted for CBSE Board 2024 Exams
Ex 2.4, 10 (ii)
Ex 2.4, 11 Deleted for CBSE Board 2024 Exams
Ex 2.4,12 Important
Ex 2.4,13
Ex 2.4, 14 (i)
Ex 2.4, 14 (ii) Important
Ex 2.4, 15 (i)
Ex 2.4, 15 (ii) Important You are here
Ex 2.4, 16 (i) Deleted for CBSE Board 2024 Exams
Ex 2.4, 16 (ii) Important
Ex 2.4
Last updated at May 29, 2023 by Teachoo
Ex 2.4, 15 Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : 35y2 + 13y – 12 First we factorize 35y2 + 13y – 12 We factorize using the splitting the middle term method = 35y2 + 28y – 15y – 12 = 7y(5y + 4) – 3 (5y + 4) = (5y + 4) (7y – 3) Hence, Area = (5y + 4) (7y – 3) Area = Length × Breadth Hence, there are two possibilities