# 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.?

Sep 11, 2015

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 ${x}^{2} + 4 {x}^{2} + 4 x = 96$

$5 {x}^{2} + 4 x - 96 = 0$

$5 {x}^{2} + 24 x - 20 x - 96 = 0$

$x \left(5 x + 24\right) - 4 \left(5 x + 24\right) = 0$
$\left(x - 4\right) \left(5 x + 24\right) = 0$. Reject negative value for x, hence x= 4 cms

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

Sep 11, 2015

You'll find $4 c m \mathmr{and} 5 c m$

#### 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.:
$A = 4 \left[x \cdot \left(x + 1\right)\right] + {x}^{2} = 96$
$4 {x}^{2} + 4 x + {x}^{2} - 96 = 0$
$5 {x}^{2} + 4 x - 96 = 0$
${x}_{1 , 2} = \frac{- 4 \pm \sqrt{16 + 1920}}{10} = \frac{- 4 \pm 44}{10}$
The useful solution will then be:
$x = \frac{40}{10} = 4 c m$