A rectangular piece of paper measures 17cm and 14cm. Strips of equal widru are cut off one side and one end. The area of the remaining piece is 108cm². Find the width of the strips removed

1 answer

Let the width of the strips cut off be \( x \) cm. After cutting the strips, the new dimensions of the rectangle will be:

  • Length: \( 17 - x \) cm
  • Width: \( 14 - x \) cm

The area \( A \) of the remaining rectangle is given as 108 cm². Thus, we can set up the equation:

\[ (17 - x)(14 - x) = 108 \]

Expanding this equation, we have:

\[ 238 - 17x - 14x + x^2 = 108 \]

This simplifies to:

\[ x^2 - 31x + 238 - 108 = 0 \]

\[ x^2 - 31x + 130 = 0 \]

Next, we can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -31 \), and \( c = 130 \). Plugging in these values:

\[ x = \frac{31 \pm \sqrt{(-31)^2 - 4 \cdot 1 \cdot 130}}{2 \cdot 1} \] \[ x = \frac{31 \pm \sqrt{961 - 520}}{2} \] \[ x = \frac{31 \pm \sqrt{441}}{2} \] \[ x = \frac{31 \pm 21}{2} \]

Calculating the two possible values for \( x \):

\[ x = \frac{31 + 21}{2} = \frac{52}{2} = 26 \]

\[ x = \frac{31 - 21}{2} = \frac{10}{2} = 5 \]

Since \( x = 26 \) is not a valid solution (as it exceeds both dimensions of the rectangle), we have:

\[ x = 5 \]

Thus, the width of the strips removed is \( \boxed{5} \) cm.