Overview
A solid has several different measurable features: the material inside it, the exposed area around it, and distances either through it or along its surface. These quantities use different formulas and respond differently to scaling. We will count exposed faces, use a volume formula, conserve material when spheres are fused, and unfold a prism to compare surface routes. The aim is to let the physical condition determine the mathematical model.
Count only the faces that exist
An open-top box is 6 cm long, 4 cm wide and 3 cm high. Its exterior includes one bottom of area 6·4 = 24, two long walls totaling 2·6·3 = 36, and two short walls totaling 2·4·3 = 24. The total is 84 cm².
A closed-box surface formula would include another 24 cm² for the top and give 108, but the prompt explicitly removes that face. Sketching a net or naming the face groups is safer than applying a memorized formula before reading the object’s construction. “Exterior area” here counts the exposed geometric faces, not both sides of a material sheet. In a different manufacturing problem, thickness or inside coating could change the accounting; do not add those assumptions when they are not stated. Match the actual faces and the requested surface.
Use volume to recover a missing dimension
For a cylinder with volume 45π and radius 3, the supplied formula V = πr²h becomes 45π = 9πh. Canceling the common factor 9π gives h = 5. The circular base contributes area πr², and stacking that base through height h gives the volume.
The distractor 15 divides by r instead of r², treating the base as though it were a length. Units explain the square: an area multiplied by a height produces cubic units. When solving a volume equation, identify which dimensions are known, substitute them exactly and isolate only the missing factor. Do not replace π by a decimal if it cancels; preserving exact form makes the cancellation clear and avoids arithmetic that the problem does not require.
Conserve volume while surface area changes
Eight identical spheres are fused into one without losing material. If r is the old radius and R the new radius, conservation of volume gives R³ = 8r³, hence R = 2r. Surface area does not remain constant. The original total area is 8·4πr² = 32πr², while the new area is 4π(2r)² = 16πr².
The stated decrease of 144π therefore gives 16πr² = 144π, so r = 3. The requested new radius is R = 6. The answer 3 is a real intermediate result but belongs to an original small sphere. This distinction is central to scaling: combining eight volumes doubles radius, and that doubled radius quadruples the area of one sphere. You must compare that one enlarged surface with the eight original surfaces, not with just one of them.
Unfold a surface route instead of crossing the interior
An ant moves between opposite vertices on the surface of a 4-by-6-by-8 prism. It cannot use the interior space diagonal. Unfolding adjacent faces makes a candidate route a straight diagonal in a plane. The three distinct dimension pairings give squared route lengths (4 + 6)² + 8² = 164, (4 + 8)² + 6² = 180, and (6 + 8)² + 4² = 212. The minimum is √164 = 2√41.
The option 6√5 is a valid but longer unfolding; 2√29 is the forbidden interior diagonal. For transfer, decide first whether the question concerns enclosed volume, exposed surface or an allowed path. Draw the relevant net when surface movement or face accounting is involved. Check what is conserved during a transformation and what changes with scale. These model choices come before formula selection; a correct formula for the wrong physical quantity can produce a very convincing distractor.
A straight segment is shortest within a flat unfolding. The comparison is needed because different pairs of adjacent faces unfold to different rectangles, and a route that is straight in one net need not be the shortest route available on the entire box. Start and end points stay at corresponding vertices when a face is rotated into the plane. This preserves the route length while making a two-dimensional distance calculation possible.
Apply the method to a new case
A 12-by-8-centimeter sheet is rolled into an open-ended circular cylinder without overlap. In version A, the 12-centimeter side becomes the circumference and the height is 8 centimeters; in version B, the 8-centimeter side becomes the circumference and the height is 12 centimeters. Which cylinder encloses more volume? Ignore sheet thickness. Version A has radius 12/(2π) = 6/π and volume π(6/π)²·8 = 288/π cubic centimeters. Version B has radius 4/π and volume π(4/π)²·12 = 192/π cubic centimeters. A has 3/2 times B's volume. Equal sheet areas do not imply equal enclosed volumes: increasing radius changes a squared factor while changing height changes a linear factor. Check the lateral areas, 12·8 and 8·12: both are 96 square centimeters, exactly the sheet area. The two circumference equations verify that neither construction stretches the sheet. Match each physical dimension to its role before applying the cylinder formula.