he dimensions of a piece of A4-size cardboard are 297 mm by 210 mm . smaller rectangular sheet of coloured paper of area \( 320 \mathrm{~cm}^{2} \) is pasted in the centre of the cardboard, ving a border of equal width around all four sides of the paper. Convert \( 320 \mathrm{~cm}^{2} \) into \( \mathrm{mm}^{2} \). Find the width of the border, giving your answer in mm .

To convert the area from square centimeters to square millimeters, we use the conversion factor that \( 1 \, \text{cm}^2 = 100 \, \text{mm}^2 \). Given the area of the colored paper is \( 320 \, \text{cm}^2 \): \[ 320 \, \text{cm}^2 = 320 \times 100 \, \text{mm}^2 = 32000 \, \text{mm}^2 \] Now, let the width of the border be \( x \) mm. The dimensions of the colored paper can be expressed in terms of \( x \): - The width of the cardboard is \( 210 \, \text{mm} \). - The height of the cardboard is \( 297 \, \text{mm} \). The dimensions of the colored paper, which is centered on the cardboard, will be: - Width: \( 210 - 2x \) - Height: \( 297 - 2x \) The area of the colored paper can be expressed as: \[ (210 - 2x)(297 - 2x) = 32000 \, \text{mm}^2 \] Expanding this equation: \[ 210 \times 297 - 2x \times 210 - 2x \times 297 + 4x^2 = 32000 \] Calculating \( 210 \times 297 \): \[ 210 \times 297 = 62370 \] Substituting this back into the equation: \[ 62370 - 420x - 594x + 4x^2 = 32000 \] Combining like terms: \[ 62370 - 1014x + 4x^2 = 32000 \] Rearranging the equation: \[ 4x^2 - 1014x + 62370 - 32000 = 0 \] This simplifies to: \[ 4x^2 - 1014x + 30370 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -1014 \), and \( c = 30370 \). Calculating the discriminant: \[ b^2 - 4ac = (-1014)^2 - 4 \times 4 \times 30370 \] Calculating \( (-1014)^2 \): \[ (-1014)^2 = 1028196 \] Calculating \( 4 \times 4 \times 30370 \): \[ 4 \times 4 \times 30370 = 486320 \] Now, substituting back into the discriminant: \[ 1028196 - 486320 = 541876 \] Now, we can find \( x \): \[ x = \frac{1014 \pm \sqrt{541876}}{8} \] Calculating \( \sqrt{541876} \): \[ \sqrt{541876} \approx 736.0 \] Now substituting back: \[ x = \frac{1014 \pm 736}{8} \] Calculating the two possible values for \( x \): 1. \( x = \frac{1014 + 736}{8} = \frac{1750}{8} = 218.75 \, \text{mm} \) (not possible since it exceeds the dimensions of the cardboard) 2. \( x = \frac{1014 - 736}{8} = \frac{278}{8} = 34.75 \, \text{mm} \) Thus, the width of the border is: \[ \boxed{34.75 \, \text{mm}} \]