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 2025 Exams
Ex 2.4, 2 (ii) Important
Ex 2.4, 2 (iii)
Ex 2.4, 3 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 3 (ii) Important
Ex 2.4, 3 (iii)
Ex 2.4, 4 (i) Deleted for CBSE Board 2025 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 2025 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 2025 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 2025 Exams
Ex 2.4, 10 (ii)
Ex 2.4, 11 Deleted for CBSE Board 2025 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
Ex 2.4, 16 (i) Deleted for CBSE Board 2025 Exams
Ex 2.4, 16 (ii) Important You are here
Ex 2.4
Last updated at April 16, 2024 by Teachoo
Ex 2.4, 16 What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : 12ky2 + 8ky – 20k First we factorize 12ky2 + 8ky – 20k = 4k (3y2 + 2y – 5) We factorize (3y2 + 2y – 5) separately 3y2 + 2y – 5 = 3y2 + 5y – 3y – 5 = y (3y + 5) – 1 (3y + 5) = (3y + 5) (y – 1) Thus, 12ky2 + 8ky – 20k = 4k (3y + 5) (y – 1) So, Volume = 4k (3y + 5) (y – 1) We know that Volume of cuboid = Length × Breadth × Height Possible dimensions can be Length = 4k, Breadth = 3y + 5 , Height = y – 1