# Squares of side 3 inches are cut from each corner of a square sheet of paper. The paper is folded to form an open box with a volume of 675 cu in. How long were the sides of the sheet of paper?

Apr 18, 2017

Hence, the side-length of the original uncut paper is $21 \text{in.}$

#### Explanation:

Suppose that, the side length of the original paper is $x$ inch.

Now, to form an open box, a square of side length $3$ inch has

been cut from $2$ corners of the original paper. This means that the

length $l$ and width $w$ of the box is $\left\{x - \left(3 + 3\right)\right\} = \left(x - 6\right) ,$

whereas, its height $h$ is, $3$ inch. i.e., the side-length of the

square, cut fro the corner.

$\therefore \text{ The Volume of the box=} l \times b \times h = \left(x - 6\right) \cdot \left(x - 6\right) \cdot 3.$

Since, this volume is known to be $675 \text{cu.in.} ,$ we have,

$3 {\left(x - 6\right)}^{2} = 675 \Rightarrow {\left(x - 6\right)}^{2} = \frac{675}{3} = 225 = {15}^{2}$

$\Rightarrow x - 6 = \pm 15 ,$but, $x - 6 = - 15 ,$ is not admissible.

$\therefore x - 6 = 15 \Rightarrow x = 21.$

Hence, the side-length of the original uncut paper is $21 \text{in.}$

Enjoy Maths.!

Apr 18, 2017

The square paper had sides of $21$ inches.

#### Explanation:

Let's find the size of the box first.

It has a square base and a height of 3 inches, because that is the size of the squares cut from each corner. This will allow the sides to fold up to form the box.

Let the sides of the square base be $x$.

$V = x \times x \times h$

${x}^{2} \times 3 = 675$

${x}^{2} = \frac{675}{3} = 225$

$x = \pm \sqrt{225}$

x = +-15 " (reject -15 as a length")#

The sides of the square base are $15$ inches.

$3$ inches were cut from each side, so the length of the original paper was $15 + 3 + 3 = 21$ inches.