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.?

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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.?

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Answer 1 · 2015-09-11T16:12:06

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

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

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

$x (5 x + 24) - 4 (5 x + 24) = 0$ $x(5x+24) -4(5x+24)=0$
$(x - 4) (5 x + 24) = 0$ $(x-4)(5x+24)=0$ . 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 · 2015-09-11T16:13:59

You'll find $4 c m and 5 c m$ $4cm and 5 cm$

Explanation:

Call the length of the side of the square base $x$ $x$ :
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so:
Surface area $A$ $A$ is the sum of the areas of the 4 sides plus the area of the base, i.e.:
$A = 4 [x \cdot (x + 1)] + x^{2} = 96$ $A=4[x*(x+1)]+x^2=96$
$4 x^{2} + 4 x + x^{2} - 96 = 0$ $4x^2+4x+x^2-96=0$
$5 x^{2} + 4 x - 96 = 0$ $5x^2+4x-96=0$
Using the Quadratic Formula:
$x_{1, 2} = \frac{- 4 \pm \sqrt{16 + 1920}}{10} = \frac{- 4 \pm 44}{10}$ $x_(1,2)=(-4+-sqrt(16+1920))/10=(-4+-44)/10$
The useful solution will then be:
$x = \frac{40}{10} = 4 c m$ $x=40/10=4cm$

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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.?

2 Answers

Explanation:

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