Solid Geometry Proofs and Calculations 1 Flashcards

A rectangular box (cuboid) has dimensions 2 × 4 × 4. What is the radius of its circumscribed sphere?

Answer: 3

The space diagonal of a cuboid with dimensions a × b × c is √(a²+b²+c²) = √(4+16+16) = √36 = 6. The circumscribed sphere's diameter equals the space diagonal, so R = 6/2 = 3.

A regular tetrahedron has edge length 2. What is its volume?

Answer: 2√2/3

The volume of a regular tetrahedron with edge length a is V = a³√2/12. Substituting a = 2: V = 8√2/12 = 2√2/3.

A regular quadrilateral (square) pyramid has a base side length of 4 and a height of 2√2. What is the angle between a lateral edge and the base?

Answer: 45°

The half-diagonal of the square base is (4√2)/2 = 2√2. The lateral edge length = √(height² + half-diagonal²) = √(8 + 8) = 4. Thus cos θ = (2√2)/4 = √2/2, giving θ = 45°.

A sphere of radius r is inscribed in a right circular cylinder. What is the ratio of the sphere's total surface area to the cylinder's total surface area?

Answer: 2/3

The inscribed sphere has radius r, so the cylinder has radius r and height 2r. Sphere SA = 4πr². Cylinder total SA = 2πr² + 2πr(2r) = 6πr². The ratio is 4πr²/6πr² = 2/3.

A right circular cone has a base circumference of 6π and a slant height of 5. What is its lateral surface area?

Answer: 15π

From 2πr = 6π we get base radius r = 3. The lateral surface area formula is πrl = π(3)(5) = 15π.

A frustum (truncated cone) has a top base radius of 2, a bottom base radius of 4, and a height of 3. What is its volume?

Answer: 28π

Using the frustum volume formula V = (πh/3)(R² + Rr + r²) = (π·3/3)(16 + 8 + 4) = π(28) = 28π.