Question

The height of an open box is 1 cm more than the length of a side of its square base. if the open box has a surface area of 96 cm (squared), how do you find the dimensions.?

Answer 1

The dimensions of the box would be length= width = 4cms and height = 5 cms

Explanation:

Let the side of the square base be x cms, then height would be x+1 cms.

Surface area of the open box, would be area of the base and area of its four faces, =xx +4x*(x+1)

Therefore `x^2 +4x^2 +4x=96`

`5x^2 +4x -96=0`

`5x^2 +24x-20x-96=0`

`x(5x+24) -4(5x+24)=0`
`(x-4)(5x+24)=0`. Reject negative value for x, hence x= 4 cms

The dimensions of the box would be length= width = 4cms and height = 5 cms

Answer 2

You'll find `4cm and 5 cm`

Explanation:

Call the length of the side of the square base `x`:

so:
Surface area `A` is the sum of the areas of the 4 sides plus the area of the base, i.e.:
`A=4[x*(x+1)]+x^2=96`
`4x^2+4x+x^2-96=0`
`5x^2+4x-96=0`
Using the Quadratic Formula:
`x_(1,2)=(-4+-sqrt(16+1920))/10=(-4+-44)/10`
The useful solution will then be:
`x=40/10=4cm`