Triangles ABC and DEF are similar. Each side length of triangle ABC is 4 times the corresponding side length of...

1. TRANSLATE the problem information

• Given information:
  • Triangles ABC and DEF are similar
  • Each side length of ABC is 4 times the corresponding side length of DEF
  • Area of triangle ABC = 270 square inches
• What this tells us: Triangle ABC is the larger triangle with a scale factor of 4 compared to triangle DEF

2. INFER the area relationship

• Key insight: For similar figures, if the sides scale by factor k, then the areas scale by \(\mathrm{k^2}\)
• Since ABC's sides are 4 times DEF's sides, the scale factor \(\mathrm{k = 4}\)
• Therefore: \(\mathrm{Area\,of\,ABC = k^2 \times Area\,of\,DEF = 4^2 \times Area\,of\,DEF = 16 \times Area\,of\,DEF}\)

3. Set up and solve the equation

• From our relationship: \(\mathrm{270 = 16 \times Area\,of\,DEF}\)
• Divide both sides by 16: \(\mathrm{Area\,of\,DEF = \frac{270}{16}}\)

4. SIMPLIFY the fraction

• \(\mathrm{\frac{270}{16} = \frac{135}{8}}\) (dividing both numerator and denominator by 2)
• As a decimal: \(\mathrm{\frac{135}{8} = 16.875}\)

Answer: \(\mathrm{\frac{135}{8}}\) square inches (or \(\mathrm{16.875}\) square inches)

Why Students Usually Falter on This Problem

Most Common Error Path:

Conceptual confusion about area scaling: Students often think that if the sides are 4 times larger, the area is also 4 times larger (not 16 times larger).

Using linear scaling instead of quadratic scaling: \(\mathrm{Area\,of\,DEF = \frac{270}{4} = 67.5}\)

This leads to confusion when trying to match answer choices, or if given multiple choice options, selecting an answer around 67.5.

Second Most Common Error:

Weak SIMPLIFY execution: Students set up the problem correctly but make arithmetic errors when computing \(\mathrm{\frac{270}{16}}\).

Common calculation mistakes include:

• \(\mathrm{\frac{270}{16} = \frac{27}{16}}\) (incorrectly removing a zero)
• Not reducing to lowest terms and leaving as \(\mathrm{\frac{270}{16}}\)
• Decimal conversion errors

This causes them to get stuck on fraction simplification or arrive at incorrect decimal approximations.

The Bottom Line:

This problem tests whether students truly understand that similarity affects area quadratically, not linearly. The calculation itself is straightforward once the \(\mathrm{k^2}\) relationship is properly applied.