Area and Volume Flashcards

A cone with base radius 9 and height 12 is cut by a plane parallel to the base at a height of 4 above the base, creating a frustum (the lower portion). What is the volume of the frustum?

Answer: 228π

The cut is at height 4 from the base, which is height 8 from the apex. Since the small cone and original cone are similar, the linear scale factor is 8/12 = 2/3. The small cone has radius 9 × (2/3) = 6 and height 8. Volume of original cone = (1/3)π(9²)(12) = 324π. Volume of small cone = (1/3)π(6²)(8) = 96π. Frustum volume = 324π − 96π = 228π.

A sphere is inscribed inside a right circular cone with base radius 6 and height 8 (the sphere is tangent to the base and to the lateral surface). What is the radius of the inscribed sphere?

Answer: 3

The slant height of the cone is √(6² + 8²) = 10. For a sphere of radius r inscribed in a cone, the sphere is tangent to the lateral surface and the base. Using the cross-sectional triangle (an isoceles triangle with base 12 and height 8), the inscribed circle's radius equals the area of the triangle divided by its semi-perimeter. Area = (1/2)(12)(8) = 48. Semi-perimeter = (12 + 10 + 10)/2 = 16. r = 48/16 = 3.

Two similar cylinders have volumes in the ratio 27 : 8. The total surface area of the larger cylinder is 81π. What is the total surface area of the smaller cylinder?

Answer: 36π

Since the volume ratio is 27:8 = (3)³:(2)³, the linear scale factor between the cylinders is 3:2. Surface areas scale as the square of the linear scale factor, so the surface area ratio is 3²:2² = 9:4. Surface area of smaller cylinder = (4/9) × 81π = 36π.

A circle has an area of 64π. A chord subtends a central angle of 120°. What is the area of the minor segment formed by the chord and its arc?

Answer: 64π/3 − 16√3

From the area 64π, the radius r = 8. The minor sector area = (120°/360°) × 64π = 64π/3. The triangle formed by the two radii and the chord has two sides of length 8 and an included angle of 120°, so its area = (1/2)(8)(8)sin(120°) = 32 × (√3/2) = 16√3. The minor segment area = sector area − triangle area = 64π/3 − 16√3.

A hollow cylindrical tube has an outer radius of 5, an inner radius of 3, and a height of 12. The tube is melted down and recast as a solid sphere with no material wasted. What is the radius of the sphere, rounded to the nearest whole number?

Answer: 5

Volume of hollow cylinder = π(5² − 3²)(12) = π(25 − 9)(12) = 192π. Setting the sphere's volume equal: (4/3)πr³ = 192π → r³ = 144 → r = ∛144 ≈ 5.24. Rounded to the nearest whole number, r = 5.

A square has a side length of 10. A quarter-circle of radius 5 is drawn at each corner, with each arc connecting the midpoints of the two sides adjacent to that corner. What is the area of the region inside the square but outside all four arcs?

Answer: 100 − 25π

Each arc is a quarter-circle of radius 5 (since the midpoint of a side of length 10 is exactly 5 units from each corner). The area of each quarter-circle is (1/4)π(5²) = 25π/4. The four quarter-circles together cover 4 × (25π/4) = 25π with no overlaps: arcs from adjacent corners meet tangentially at each side's midpoint (distance between adjacent centers = 10 = 5 + 5), and arcs from opposite corners are too far apart to overlap (diagonal = 10√2 > 10). Therefore, the area inside the square but outside all arcs = 100 − 25π.