A square has a side length of s inches. The sum of the square's perimeter and the length of its...

1. TRANSLATE the problem information

• Given information:
  • Square has side length \(\mathrm{s}\) inches
  • Sum of perimeter and diagonal equals \(36 + 72\sqrt{2}\) inches

• What this means mathematically:
  • Perimeter = \(4\mathrm{s}\)
  • Diagonal = \(\mathrm{s}\sqrt{2}\) (from Pythagorean theorem)
  • \(4\mathrm{s} + \mathrm{s}\sqrt{2} = 36 + 72\sqrt{2}\)

2. INFER the diagonal relationship

• For any square, the diagonal forms a right triangle with two sides
• Using Pythagorean theorem: \(\mathrm{diagonal}^2 = \mathrm{s}^2 + \mathrm{s}^2 = 2\mathrm{s}^2\)
• Therefore: \(\mathrm{diagonal} = \sqrt{2\mathrm{s}^2} = \mathrm{s}\sqrt{2}\)

3. SIMPLIFY by factoring

• Our equation: \(4\mathrm{s} + \mathrm{s}\sqrt{2} = 36 + 72\sqrt{2}\)
• Factor out \(\mathrm{s}\): \(\mathrm{s}(4 + \sqrt{2}) = 36 + 72\sqrt{2}\)
• Now we need to solve for \(\mathrm{s}\)

4. INFER a solution strategy

• Since the right side has both regular numbers and \(\sqrt{2}\) terms, assume \(\mathrm{s}\) might also have both
• Try \(\mathrm{s} = \mathrm{a} + \mathrm{b}\sqrt{2}\) where \(\mathrm{a}\) and \(\mathrm{b}\) are regular numbers
• Use coefficient matching to find \(\mathrm{a}\) and \(\mathrm{b}\)

5. SIMPLIFY through coefficient matching

• If \(\mathrm{s} = \mathrm{a} + \mathrm{b}\sqrt{2}\), then:
  \(\mathrm{s}(4 + \sqrt{2}) = (\mathrm{a} + \mathrm{b}\sqrt{2})(4 + \sqrt{2})\)
• Expanding: \((4\mathrm{a} + 2\mathrm{b}) + (\mathrm{a} + 4\mathrm{b})\sqrt{2}\)
• Comparing with \(36 + 72\sqrt{2}\):
  • Constant terms: \(4\mathrm{a} + 2\mathrm{b} = 36\) → \(2\mathrm{a} + \mathrm{b} = 18\)
  • \(\sqrt{2}\) terms: \(\mathrm{a} + 4\mathrm{b} = 72\)

6. SIMPLIFY the system of equations

• From \(2\mathrm{a} + \mathrm{b} = 18\): \(\mathrm{b} = 18 - 2\mathrm{a}\)
• Substitute into \(\mathrm{a} + 4\mathrm{b} = 72\):
  \(\mathrm{a} + 4(18 - 2\mathrm{a}) = 72\)
  \(\mathrm{a} + 72 - 8\mathrm{a} = 72\)
  \(-7\mathrm{a} = 0\)
  \(\mathrm{a} = 0\)
• Therefore: \(\mathrm{b} = 18\), so \(\mathrm{s} = 18\sqrt{2}\)

Answer: C (\(18\sqrt{2}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students often don't recognize that the diagonal of a square is \(\mathrm{s}\sqrt{2}\), instead thinking it's just \(\mathrm{s}\) or \(2\mathrm{s}\).

If they use diagonal = \(\mathrm{s}\), their equation becomes:
\(4\mathrm{s} + \mathrm{s} = 36 + 72\sqrt{2}\)
\(5\mathrm{s} = 36 + 72\sqrt{2}\)

Since the left side has no \(\sqrt{2}\) term but the right side does, they get confused and may try to ignore the \(\sqrt{2}\) term, leading to \(\mathrm{s} = 36/5\), which isn't among the choices. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{s}(4 + \sqrt{2}) = 36 + 72\sqrt{2}\) but try to solve by dividing both sides by \((4 + \sqrt{2})\), which requires rationalizing the denominator—a complex process they may execute incorrectly.

They might attempt: \(\mathrm{s} = \frac{36 + 72\sqrt{2}}{4 + \sqrt{2}}\) and make algebraic errors in the rationalization process, possibly arriving at one of the incorrect answer choices or getting stuck entirely.

The Bottom Line:

This problem combines geometric reasoning (finding the diagonal) with advanced algebraic techniques (coefficient matching or rationalization), making it challenging at multiple levels. Students need both strong conceptual knowledge of square properties and sophisticated algebraic problem-solving skills.