Introduction / Context:
Melting conserves volume. We first find the cube’s side from equal volume, then compute both surface areas and take the difference. This tests volume conservation and area formulas.
Given Data / Assumptions:
- Rectangular solid: 27 × 8 × 1 cm
- Volume is conserved.
Concept / Approach:
- V_block = l*b*h.
- Let cube side = a with a^3 = V_block.
- Compute SA_block and SA_cube and subtract.
Step-by-Step Solution:
V_block = 27*8*1 = 216 cm3 ⇒ a = 6 cm (since 6^3 = 216).SA_block = 2*(27*8 + 8*1 + 27*1) = 2*(216 + 8 + 27) = 502 cm2.SA_cube = 6*a^2 = 6*36 = 216 cm2.Difference = 502 − 216 = 286 cm2.
Verification / Alternative check:
Recompute each product to avoid arithmetic slips; sums match.
Why Other Options Are Wrong:
- 296/300/284 cm2: Small arithmetic deviations, not matching exact computations.
Common Pitfalls:
- Using perimeter-like ideas instead of surface area.
- Forgetting both small faces 8*1 and 27*1 contribute to SA.
Final Answer:
286 cm2
Discussion & Comments