The formula for the area, A, of a trapezoid is \(\mathrm{A = \frac{1}{2}(b_1 + b_2)h}\), where b_1 and b_2 are...

1. TRANSLATE the problem information

• Given: Area formula \(\mathrm{A = \frac{1}{2}(b_1 + b_2)h}\)
• Goal: Solve for \(\mathrm{b_1}\) in terms of A, \(\mathrm{b_2}\), and h

2. INFER the solution strategy

• We need to "undo" the operations applied to \(\mathrm{b_1}\) in the original formula
• The original formula applies: addition \(\mathrm{(b_1 + b_2)}\), multiplication by h, then multiplication by \(\mathrm{\frac{1}{2}}\)
• We'll use inverse operations in reverse order: multiply by 2, divide by h, subtract \(\mathrm{b_2}\)

3. SIMPLIFY through algebraic manipulation

• Start with: \(\mathrm{A = \frac{1}{2}(b_1 + b_2)h}\)
• Eliminate the fraction by multiplying both sides by 2:
  \(\mathrm{2A = (b_1 + b_2)h}\)
• Isolate the parentheses by dividing both sides by h:
  \(\mathrm{\frac{2A}{h} = b_1 + b_2}\)
• Finally, subtract \(\mathrm{b_2}\) from both sides:
  \(\mathrm{\frac{2A}{h} - b_2 = b_1}\)

Answer: C \(\mathrm{(\frac{2A}{h} - b_2)}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students often struggle with the correct sequence of operations, particularly forgetting to multiply by 2 first to eliminate the fraction.

Many students try to divide by h first, leading to: \(\mathrm{\frac{A}{h} = \frac{1}{2}(b_1 + b_2)}\), then subtract \(\mathrm{b_2}\) to get \(\mathrm{\frac{A}{h} - b_2 = \frac{1}{2}b_1}\). They forget about the remaining factor of \(\mathrm{\frac{1}{2}}\) and incorrectly conclude \(\mathrm{b_1 = \frac{A}{h} - b_2}\).

This leads them to select Choice A \(\mathrm{(\frac{A}{h} - b_2)}\).

Second Most Common Error:

Poor algebraic organization: Students correctly multiply by 2 first but then make errors in handling the fraction \(\mathrm{\frac{2A - b_2}{h}}\) by incorrectly grouping the terms.

Some students write \(\mathrm{\frac{2A}{h} - b_2}\) as \(\mathrm{\frac{2A - b_2}{h}}\), thinking they can factor out the division by h.

This may lead them to select Choice B \(\mathrm{(\frac{2A - b_2}{h})}\).

The Bottom Line:

This problem tests careful execution of inverse operations in the correct sequence. The key insight is recognizing that eliminating the fraction must come first, followed by systematic isolation of the target variable.