Bluebook SAT Test Area and Volume 4

Question 1

Two concentric circles have radii 5 and 13. What fraction of the larger circle's area is occupied by the annular region between the two circles?

Accepted Answer

144/169

Answer

The annulus area equals π(13² − 5²) = π(169 − 25) = 144π. The outer circle's area is π(13²) = 169π. The fraction is 144π ÷ 169π = 144/169. Note that 144 = 12² and 169 = 13², so this fraction does not simplify further.

Question 2

A sphere of radius r is perfectly inscribed in a right circular cylinder, touching the top, bottom, and lateral surface. What fraction of the cylinder's volume does the sphere occupy?

Accepted Answer

2/3

Answer

Since the sphere fits perfectly, the cylinder must have radius r and height 2r. Cylinder volume = πr²(2r) = 2πr³. Sphere volume = (4/3)πr³. The fraction is (4/3)πr³ ÷ 2πr³ = (4/3) ÷ 2 = 2/3. This is Archimedes' famous result.

Question 3

A circle has radius 6. A chord subtends a central angle of 120°. What is the area of the circular segment (the region between the chord and the arc)?

Accepted Answer

12π − 9√3

Answer

Sector area = (120/360)·π(6²) = (1/3)·36π = 12π. The triangle formed by the two radii and the chord has area = (1/2)(6)(6)sin(120°) = 18·(√3/2) = 9√3. Segment area = sector − triangle = 12π − 9√3.

Question 4

A solid is formed by placing a hemisphere of radius 4 directly on top of a right circular cylinder of radius 4 and height 6, with the flat face of the hemisphere flush against the cylinder's top. What is the total exposed surface area of the solid?

Accepted Answer

96π

Answer

The exposed surfaces are: (1) the circular bottom of the cylinder = π(4²) = 16π; (2) the lateral surface of the cylinder = 2π(4)(6) = 48π; (3) the curved surface of the hemisphere = 2π(4²) = 32π. The top circle of the cylinder and the flat base of the hemisphere are interior (they cancel each other). Total = 16π + 48π + 32π = 96π.

Question 5

A right circular cone has height 9 cm and base radius 6 cm. A plane parallel to the base cuts the cone at a point 3 cm above the base, forming a frustum. What is the volume of the frustum?

Accepted Answer

76π

Answer

The cut is 3 cm from the base, so 6 cm from the apex. By similar triangles, the small cone's radius = 6 × (6/9) = 4 cm. Volume of original cone = (1/3)π(6²)(9) = 108π. Volume of removed small cone = (1/3)π(4²)(6) = 32π. Frustum volume = 108π − 32π = 76π.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Area and Volume 4