Bluebook SAT Test Area and Volume 6

Question 1

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?

Accepted Answer

228π

Answer

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.

Question 2

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?

Accepted Answer

3

Answer

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.

Question 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?

Accepted Answer

36π

Answer

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.

Question 4

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?

Accepted Answer

64π/3 − 16√3

Answer

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.

Question 5

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?

Accepted Answer

5

Answer

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.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Area and Volume 6