Diagonal Rods and Decorative Volumes for a Stage Set
The stage crew at a community theater is building a rectangular prop crate measuring 3 feet by 4 feet by 12 feet and needs to store a rigid curtain rod diagonally, corner to corner. They own a 13-foot rod that cannot be cut and can borrow a 13.5-foot backup. A second, damaged crate has an interior diagonal of √150 feet; the supplier sells rods only in lengths rounded to the nearest tenth of a foot. Separately, the crew is decorating two pieces that share the same base and height: a cylindrical column with radius 1.5 feet and height 8 feet, and a conical spire with radius 1.5 feet and height 8 feet. An oversized cone with radius 3 feet and height 8 feet might be substituted for the cylinder. A volunteer insists that a cone is always smaller than a cylinder with the same base, and the crew must check that claim before buying materials.
1. Show whether the 13-foot rod fits diagonally inside the 3-by-4-by-12 crate, and whether the 13.5-foot backup does. Explain why no other orientation allows a longer rod.
Read the worked answer
Work in two right triangles. The floor diagonal of the crate is √(3² + 4²) = √25 = 5 feet. That floor diagonal, the 12-foot height, and the space diagonal form a right triangle, so the space diagonal is √(5² + 12²) = √(25 + 144) = √169 = 13 feet.
The 13-foot rod is exactly the length of the space diagonal, so it fits only in the ideal case: perfectly corner to corner, with zero clearance for the rod's thickness or for maneuvering it through the crate. It is mathematically possible but practically risky.
The 13.5-foot backup exceeds 13 feet by 0.5 foot (6 inches) and cannot fit. The space diagonal is the greatest distance between any two points of a rectangular box, so tilting, angling, or laying the rod along a face cannot help - every other pairing of points inside the crate is shorter.
2. Use volumes in terms of π to evaluate the volunteer's claim that a cone is always smaller than a cylinder with the same base, testing both the matching spire and the oversized cone.
Read the worked answer
Cylinder with r = 1.5, h = 8: V = πr²h = π(2.25)(8) = 18π ft³.
Matching cone with r = 1.5, h = 8: V = (1/3)πr²h = (1/3)(18π) = 6π ft³.
So for identical bases and heights the cone is exactly one-third of the cylinder - the volunteer's rule works for that pair.
Oversized cone with r = 3, h = 8: V = (1/3)π(9)(8) = 24π ft³.
That is larger than the 18π cylinder. The claim “a cone is always smaller” is false, because volume depends on the actual dimensions, not on the shape name. Doubling the radius multiplies the base area by 4, which more than compensates for the cone's one-third factor. The reliable statement is narrower: with the same base and height, a cone has one-third the volume of a cylinder.
3. The damaged crate's interior diagonal is √150 feet. Between which two tenths does it lie, and what rod length should be ordered if the rod must be no longer than the diagonal?
Read the worked answer
Bracket the square root with perfect squares: 12² = 144 and 13² = 169, so √150 is between 12 and 13.
Refine to tenths:
- 12.2² = 148.84, which is below 150.
- 12.3² = 151.29, which is above 150.
So √150 lies between 12.2 and 12.3. To choose between them, compare the distances: 150 − 148.84 = 1.16 and 151.29 − 150 = 1.29. Since 1.16 < 1.29, the value is closer to 12.2; more precisely √150 ≈ 12.247, which rounds to 12.2 to the nearest tenth.
Because the rod must not exceed the diagonal, rounding up to 12.3 would produce a rod that is too long, and a 12-foot rod would be wastefully short. Order 12.2 feet.