Question:The area of a triangle is calculated as one-half the product of the length of its base and the length...

1. TRANSLATE the problem information

• Given information:
  • Triangle area = 32 square feet
  • Base length = Height length
• What we need to find: Length of the base

2. TRANSLATE the area relationship into an equation

• Start with the triangle area formula: \(\mathrm{A = \frac{1}{2} \times base \times height}\)
• Since base = height, let's call this length "b"
• So our equation becomes: \(\mathrm{A = \frac{1}{2} \times b \times b = \frac{1}{2}b^2}\)

3. SIMPLIFY by substituting the known area

• Replace A with 32: \(\mathrm{32 = \frac{1}{2}b^2}\)
• Multiply both sides by 2 to eliminate the fraction: \(\mathrm{64 = b^2}\)

4. SIMPLIFY to find the base length

• Take the square root of both sides: \(\mathrm{b = \sqrt{64}}\)
• Calculate: \(\mathrm{b = 8}\) feet

5. APPLY CONSTRAINTS to verify our answer

• Since we're dealing with a real triangle, length must be positive
• Our answer of 8 feet makes sense in context

Answer: C) 8

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may set up the equation incorrectly by not recognizing that "base equals height" means they can substitute one variable for the other. Instead, they might try to work with two separate unknown variables (b and h) and get stuck because they only have one equation with two unknowns.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students might correctly set up \(\mathrm{32 = \frac{1}{2}b^2}\), but make an algebraic error when solving. A common mistake is forgetting to multiply both sides by 2 first, leading them to calculate \(\mathrm{b = \sqrt{32} \approx 5.66}\), which isn't among the answer choices.

This causes them to get stuck and randomly select an answer.

The Bottom Line:

This problem tests whether students can recognize when two variables are actually the same (base = height) and translate that insight into a solvable single-variable equation. The arithmetic itself is straightforward once the setup is correct.