if the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid then the volume is increased by 52 cubic units find the volume of the cuboid
Answers
Answer:
x = 2
Step-by-step explanation:
Let the side of cubic box = x units
Then Volume of Cubic box = x³ cube units
Now
After increase 1 units first side becomes = x + 1
After increase 2 units second side becomes = x + 2
After increase 3 units third side becomes = x + 3
According to formula of volume and given volume in question.
Volume of new cuboid = (x + 1)(x + 2)(x + 3) = x³ + 52
⇒ (x + 1)(x + 2)(x + 3) = x³ + 52
⇒ (x² + 1x + 2x + 2)(x + 3) = x³ + 52
⇒ (x² + 3x + 2)(x + 3) = x³ + 52
⇒ x³ + 3x² + 3x² + 9x + 2x + 6 = x³ + 52
⇒ x³ - x³ + 6x² + 11x + 6 - 52 = 0
⇒ 6x² + 11x - 46 = 0
⇒ 6x² + 23x - 12x - 46 = 0
⇒ x(6x + 23) - 2(6x + 23) = 0
⇒ (6x +23)(x - 2) = 0
⇒ 6x + 23 = 0 or x - 2 = 0
⇒ 6x = -23 or x = 2
⇒ x = -23/6 or x = 2
Since length can not be negative
So
x = 2
Answer:
Step-by-step explanation:
Let the side of cubic box = x units
Then Volume of Cubic box = x³ cube units
Now
After increase 1 units first side becomes = x + 1
After increase 2 units second side becomes = x + 2
After increase 3 units third side becomes = x + 3
According to formula of volume and given volume in question.
Volume of new cuboid = (x + 1)(x + 2)(x + 3) = x³ + 52
⇒ (x + 1)(x + 2)(x + 3) = x³ + 52
⇒ (x² + 1x + 2x + 2)(x + 3) = x³ + 52
⇒ (x² + 3x + 2)(x + 3) = x³ + 52
⇒ x³ + 3x² + 3x² + 9x + 2x + 6 = x³ + 52
⇒ x³ - x³ + 6x² + 11x + 6 - 52 = 0
⇒ 6x² + 11x - 46 = 0
⇒ 6x² + 23x - 12x - 46 = 0
⇒ x(6x + 23) - 2(6x + 23) = 0
⇒ (6x +23)(x - 2) = 0
⇒ 6x + 23 = 0 or x - 2 = 0
⇒ 6x = -23 or x = 2
⇒ x = -23/6 or x = 2
Since length can not be negative
So
x = 2
Volume of the cuboid = lbh
= 2×2×2=8 cubic units
Volume of the cuboid is 8 cubic units