A rectangular banner has a diagonal of 9sqrt(5) meters. The length of the banner is twice its width. What is...

1. TRANSLATE the problem information

• Given information:
  • Diagonal = \(9\sqrt{5}\) meters
  • Length = twice the width
  • Need to find: width

• Let width = \(\mathrm{w}\) meters, so length = \(2\mathrm{w}\) meters

2. INFER the geometric relationship

• In any rectangle, the diagonal connects opposite corners
• This diagonal, along with the length and width, forms a right triangle
• Therefore, we can apply the Pythagorean theorem: \(\mathrm{diagonal}^2 = \mathrm{length}^2 + \mathrm{width}^2\)

3. Set up the equation

• Using Pythagorean theorem: \((9\sqrt{5})^2 = (2\mathrm{w})^2 + \mathrm{w}^2\)

4. SIMPLIFY the equation step by step

• \((9\sqrt{5})^2 = (2\mathrm{w})^2 + \mathrm{w}^2\)
• \(81 \times 5 = 4\mathrm{w}^2 + \mathrm{w}^2\)
• \(405 = 5\mathrm{w}^2\)
• \(\mathrm{w}^2 = 81\)
• \(\mathrm{w} = 9\) (since width must be positive)

Answer: C. 9

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle with "length is twice the width" and set up incorrect relationships like \(\mathrm{w} = 2\mathrm{L}\) instead of \(\mathrm{L} = 2\mathrm{w}\).

This leads to the wrong equation setup: \((9\sqrt{5})^2 = \mathrm{w}^2 + (\mathrm{w}/2)^2\), giving them \(\mathrm{w}^2 + \mathrm{w}^2/4 = 405\), which results in \(5\mathrm{w}^2/4 = 405\), so \(\mathrm{w}^2 = 324\) and \(\mathrm{w} = 18\). Since 18 isn't among the choices, this leads to confusion and guessing.

Second Most Common Error:

Missing conceptual knowledge about rectangles: Students may not realize that the diagonal forms a right triangle with the sides, instead trying to use formulas for area or perimeter.

Without the Pythagorean theorem connection, they cannot establish the fundamental equation needed to solve the problem. This causes them to get stuck and randomly select an answer.

The Bottom Line:

Success on this problem hinges on translating the "twice the width" relationship correctly and recognizing the right triangle formed by a rectangle's diagonal and sides. Miss either connection, and the algebraic setup becomes impossible.