Question:Right circular cone P is similar to right circular cone Q. The total surface area of cone P is 72pi...

1. TRANSLATE the problem information

• Given information:
  • Cone P is similar to Cone Q
  • Surface area of P = \(72\pi \mathrm{cm}^2\)
  • Surface area of Q = \(450\pi \mathrm{cm}^2\)
  • Volume of Q = \(1{,}000 \mathrm{cm}^3\)
  • Need to find: Sum of volumes of P and Q

2. INFER the scaling approach

• Since the cones are similar, we need to use scaling properties:
  • Linear dimensions scale by factor k
  • Surface areas scale by \(\mathrm{k}^2\)
  • Volumes scale by \(\mathrm{k}^3\)
• Strategy: Find the surface area ratio → determine linear scale factor → use volume scaling

3. SIMPLIFY to find the surface area ratio

• Surface area ratio P:Q = \(\frac{72\pi}{450\pi} = \frac{72}{450} = \frac{4}{25}\)

4. INFER the linear scale factor from area ratio

• Since surface areas scale by \(\mathrm{k}^2\), we have:
  \(\mathrm{k}^2 = \frac{4}{25}\)
• Taking the square root: \(\mathrm{k} = \frac{2}{5}\)

5. INFER the volume scaling relationship

• Since volumes scale by \(\mathrm{k}^3\):
  Volume ratio P:Q = \(\left(\frac{2}{5}\right)^3 = \frac{8}{125}\) (use calculator)

6. SIMPLIFY to calculate volume of cone P

• Volume of P = Volume of Q × \(\frac{8}{125}\)
• Volume of P = \(1{,}000 \times \frac{8}{125} = \frac{8{,}000}{125} = 64 \mathrm{cm}^3\) (use calculator)

7. SIMPLIFY to find the final sum

• Sum of volumes = \(64 + 1{,}000 = 1{,}064 \mathrm{cm}^3\)

Answer: 1064

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students mix up the scaling rules and use the wrong exponent

Many students remember that similar figures scale somehow, but confuse which property uses which exponent. They might use k for areas instead of \(\mathrm{k}^2\), or use \(\mathrm{k}^2\) for volumes instead of \(\mathrm{k}^3\). For example, they might think volume ratio = \(\mathrm{k}^2 = \frac{4}{25}\), leading to Volume P = \(1{,}000 \times \frac{4}{25} = 160 \mathrm{cm}^3\), and sum = \(1{,}160 \mathrm{cm}^3\).

This leads to confusion and selecting an incorrect answer.

Second Most Common Error:

Poor SIMPLIFY execution: Students set up the problem correctly but make calculation errors

Students understand the scaling concept but make arithmetic mistakes when computing \(\left(\frac{2}{5}\right)^3\) or when calculating \(\frac{8{,}000}{125}\). They might get \(\left(\frac{2}{5}\right)^3 = \frac{6}{125}\) instead of \(\frac{8}{125}\), leading to Volume P = \(48 \mathrm{cm}^3\) and sum = \(1{,}048 \mathrm{cm}^3\).

This causes them to arrive at a plausible but incorrect answer.

The Bottom Line:

This problem tests both conceptual understanding of similarity scaling and careful algebraic execution. Success requires knowing that volumes scale by \(\mathrm{k}^3\) (not k or \(\mathrm{k}^2\)) and executing multiple calculation steps without error.