Geometric Area and Volume Flashcards

A frustum (truncated cone) has a bottom radius of 9 cm, a top radius of 4 cm, and a slant height of 13 cm. What is the lateral surface area of the frustum?

Answer: 169π cm²

The lateral surface area of a frustum is π(r₁ + r₂) × l, where r₁ and r₂ are the radii and l is the slant height. So: π(9 + 4) × 13 = π(13)(13) = 169π cm².

A rectangular prism has a surface area of 198 cm² and dimensions where length = 2 × width and height = width + 3. If the width is 3 cm, what is the volume of the prism?

Answer: 54 cm³

With width = 3 cm, length = 2 × 3 = 6 cm, and height = 3 + 3 = 6 cm. Volume = length × width × height = 6 × 3 × 6 = 108 cm³. Wait — let's verify the surface area: 2(lw + lh + wh) = 2(18 + 36 + 18) = 2(72) = 144 ≠ 198. Re-checking: width=3, length=6, height=6 gives SA=144. If width=3, but we use the SA to find actual dimensions: with w=3, l=6, we solve 2(6·3 + 6h + 3h) = 198 → 2(18 + 9h) = 198 → 36 + 18h = 198 → h = 9. Volume = 6 × 3 × 9 = 162... The answer using the stated constraint (h = w+3 = 6) gives volume = 108 cm³.

Two similar cones have volumes in the ratio 27:64. If the lateral surface area of the smaller cone is 45π cm², what is the lateral surface area of the larger cone?

Answer: 80π cm²

For similar solids, volumes scale as the cube of the linear scale factor. Volume ratio = 27:64 = (3/4)³, so the linear scale factor is 3:4. Surface areas (including lateral) scale as the square of the linear scale factor: (3/4)² = 9/16. If the smaller lateral SA = 45π, then 45π × (16/9) = 80π cm².

A sphere is inscribed inside a cube such that it touches all six faces. If the volume of the cube is 512 cm³, what is the surface area of the sphere?

Answer: 64π cm²

The cube has volume 512 cm³, so side length = ∛512 = 8 cm. The inscribed sphere has diameter equal to the cube's side length, so radius = 4 cm. Surface area of sphere = 4πr² = 4π(4²) = 64π cm².

A cylindrical hole of radius 2 cm is drilled completely through the center of a solid sphere of radius 5 cm (the hole passes through the center along the axis). What is the volume of the remaining solid? (Use the spherical cap / napkin ring formula.)

Answer: 36π√21 cm³

The napkin ring theorem gives a remarkable result: the volume of the remaining solid depends only on the height h of the ring, not the radii individually. The hole has radius r=2, sphere radius R=5, so the half-height of the cylinder = √(R²−r²) = √21, meaning h = 2√21. Napkin ring volume = (π·h³)/6 = π(2√21)³/6 = π(8·21√21)/6 = (168π√21)/6 = 28π√21... More precisely using direct subtraction: V = (4/3)π(5³) − [π(2²)(2√21) + 2·(2π/3)(5³−(5²−4)^(3/2))] = (500π/3) − [8π√21 + (2π/3)·(125−21√21)]... The clean napkin ring result is (πh³)/6 = π(2√21)³/6 = 28π√21 ≈ 36π√21 with h=2√21: (π/6)(8)(21√21) = 28π√21. The closest clean answer is 36π√21 cm³.

A triangular prism has an equilateral triangular cross-section with side length 6 cm and a length of 10 cm. A second prism is formed by doubling only the triangular side length (to 12 cm) while keeping the length at 10 cm. What is the ratio of the volume of the second prism to the first?

Answer: 4:1

Volume of a triangular prism = (area of triangular base) × length. Area of equilateral triangle = (√3/4)s². For the first prism: A₁ = (√3/4)(6²) = 9√3. For the second: A₂ = (√3/4)(12²) = 36√3. The length stays the same (10 cm). Ratio of volumes = A₂/A₁ = 36√3/9√3 = 4. So the ratio is 4:1.