What do inscribed and circumscribed mean?
When a solid sits inside another and touches it at special points, it is inscribed in it. The outer one is circumscribed about the inner. For example, a sphere inside a cube is inscribed in the cube, and the cube is circumscribed about the sphere. Touching can happen on faces (tangent), along edges or at corners. A tangent plane touches a sphere at exactly one point.
The best trick: cut through the axis and look at the flat shape, a cross-section. A sphere in a cube becomes a circle in a square. A cone with a sphere becomes a circle in a triangle.
Sphere and cube
Sphere inside a cube (touches 6 faces). The sphere's diameter is the cube's edge: 2r = a, so r = a/2. Volume ratio sphere : cube = π/6 ≈ 0.52.
Sphere around a cube (touches 8 corners). The longest line in the cube, the space diagonal, goes through the centre and is a diameter. The space diagonal is a√3 (Pythagoras twice). So 2R = a√3 and R = a√3/2. Volume ratio cube : sphere = 2/(π√3) ≈ 0.37.
Between these two there is a third sphere that touches the 12 edges: its radius is a/√2. And R : r = √3 for the first two spheres around the same cube.
Sphere and cylinder
Sphere inside a cylinder touching both lids and the curved side: the cylinder has the same radius r and its height equals the diameter, h = 2r. Then sphere volume = (4/3)πr³ and cylinder volume = 2πr³: ratio 2 : 3. Surprise: the sphere's surface area 4πr² to the cylinder's total surface area 6πr² is also 2 : 3. This was Archimedes' favourite result.
Cylinder inside a sphere (both rims on the sphere): cut through the axis. The centre of the sphere is the middle of the cylinder's axis. The radius to a rim point is the hypotenuse of a right triangle with legs r (cylinder radius) and h/2:
R² = r² + (h/2)².
Sphere and cone
Sphere inside a cone (touches the base and the slanted side). Cut along the axis: the cone becomes an isosceles triangle with base 2r, height h and equal sides l = √(r² + h²). The sphere becomes the incircle of that triangle. The incircle radius = area ÷ semi-perimeter:
ρ = (½ · 2r · h) / ((2l + 2r)/2) = rh/(r + l).
Cone inside a sphere (apex and base rim on the sphere, base radius r, height h): R = (r² + h²)/(2h). In the cross-section, the apex and the two ends of the base lie on a circle. A regular tetrahedron of edge a has an inscribed sphere of radius a√6/12 and a circumscribed sphere of radius a√6/4, so R = 3r.
Solving a problem: a 4-step plan
- Name which solid is inside (inscribed) and which is outside.
- Decide what touches what: faces, rim, corners, slanted side.
- Cut along the axis and draw the flat cross-section.
- Use a right triangle (Pythagoras), diagonal, or incircle rule. Then find the volume or area asked.
Always label which radius is which (r for the cylinder/cone, R for the outer sphere, ρ for an inner sphere) before you start.
Try it: ball in a box
Take a ball (or a round fruit) and a box that holds it snugly. Measure the ball's width with a ruler and compare with the box's edge. They match: that is r = a/2. Next, try to imagine the ball with the box's corners touching it from outside: guess what the ball's diameter would be compared to the box's diagonal, then check with the 3D.
Key formulas and definitions
- Sphere in cube (edge a): r = a/2
- Sphere around cube: R = a√3/2
- Sphere touching the 12 edges of a cube: ρ = a√2/2
- Sphere in cylinder: h = 2r; V(sphere) : V(cylinder) = 2 : 3; S(sphere) : S(total cylinder) = 2 : 3
- Cylinder in sphere: R² = r² + (h/2)²
- Sphere in cone: ρ = rh/(r + l), l = √(r² + h²)
- Cone in sphere: R = (r² + h²)/(2h)
- Regular tetrahedron (edge a): inradius a√6/12, circumradius a√6/4
Worked examples
1. A sphere is inscribed in a cube of edge 10 cm. Find the sphere's volume.
r = a/2 = 5 cm. V = (4/3)π × 125 = 500π/3 ≈ 523.6 cm³.
2. A cube of edge 6 cm is inscribed in a sphere. Find the sphere's radius and surface area.
R = a√3/2 = 3√3 ≈ 5.20 cm. Surface area = 4πR² = 4π × 27 = 108π ≈ 339.3 cm².
3. A sphere of radius 4 cm fits exactly inside a cylinder. Find the cylinder's volume and the fraction of it taken by the sphere.
h = 2r = 8 cm. Cylinder: π × 16 × 8 = 128π cm³. Sphere: (4/3)π × 64 = 256π/3 cm³. Fraction = (256/3) ÷ 128 = 2/3.
4. A cylinder of radius 3 cm and height 8 cm is inscribed in a sphere. Find the sphere's radius.
R² = r² + (h/2)² = 9 + 16 = 25, so R = 5 cm.
5. A cone has base radius 3 cm and height 4 cm. Find the radius of the largest sphere that fits inside it.
l = √(9 + 16) = 5 cm. ρ = rh/(r + l) = (3 × 4)/(3 + 5) = 12/8 = 1.5 cm.
6. A cone with base radius 4 cm and height 8 cm is inscribed in a sphere. Find the sphere's radius.
R = (r² + h²)/(2h) = (16 + 64)/16 = 5 cm.
7. A sphere is inscribed in a cube, and the cube is inscribed in a larger sphere. Find the ratio of the surface areas of the larger sphere to the smaller sphere.
Small sphere r = a/2. Large sphere R = a√3/2. R/r = √3. Surface area ratio = (R/r)² = 3. So the ratio is 3 : 1.
Common mistakes
- Using the cube's edge as the diameter of the sphere around the cube. The sphere around a cube has the space diagonal a√3 as its diameter.
- Forgetting that the cylinder height is 2r (not r) when the sphere is inscribed in it.
- Not cutting along the axis. Without the flat cross-section the right triangle or incircle cannot be seen.
- Using the cone's height h instead of the slant height l in ρ = rh/(r + l).