Three solid cubes with edges 1 cm, 6 cm, and 8 cm are melted and recast into a single cube. Find half of the total surface area of the new cube.

Introduction / Context:
When solids are melted and recast, volume is conserved (neglecting loss). The new cube’s edge is obtained from the total initial volume. With the new edge, compute the total surface area and then take half, as asked. This tests linking conservation to cube formulas.

Given Data / Assumptions:

• Original edges: 1 cm, 6 cm, 8 cm.
• Volumes: 1^3 = 1, 6^3 = 216, 8^3 = 512 cm^3.
• Total volume V_total = 1 + 216 + 512 = 729 cm^3.
• New cube edge a satisfies a^3 = 729 ⇒ a = 9 cm.

Concept / Approach:
Compute TSA of the new cube: S = 6a^2. Then take half as requested. Keep arithmetic exact; numbers are chosen to be perfect cubes/squares.

Step-by-Step Solution:
a = ∛729 = 9 cmS = 6a^2 = 6 * 9^2 = 6 * 81 = 486 sq cmHalf of S = 486 / 2 = 243 sq cm

Verification / Alternative check:
Re-cubing 9 gives back 729, matching conserved volume; halving 486 is straightforward.

Why Other Options Are Wrong:
486 sq cm is the full surface area, not half; 463 and 293 are distractors without geometric basis here.

Common Pitfalls:
Adding edges instead of volumes; forgetting to square the edge when computing surface area; or halving the edge rather than halving the surface area.

Final Answer:
243 sq cm