Inscribed Solids Problems with Geometric Relationships

Inscribed solids are three-dimensional figures placed inside another solid so that specified vertices, edges or curved boundaries touch the outer surface. Common cases include a cube or cuboid inside a sphere and a cylinder or cone inside a sphere. Their dimensions are connected using the Pythagorean theorem and standard mensuration formulas.

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What Are Inscribed Solids?

An inscribed solid is a solid positioned inside another solid with its defining boundary points touching the outer surface. For a polyhedron inscribed in a sphere, all vertices lie on the sphere; for a cylinder, its circular rim lies on the sphere.

The main method is to take a suitable cross-section and apply the Pythagorean theorem. For a cube of edge a inside a sphere of radius R, the cube's space diagonal is the sphere's diameter, so √3a = 2R. For a right circular cylinder of radius r and height h inside a sphere of radius R, r² + (h/2)² = R².

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Inscribed Shapes and Volume Ratios

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Inscribed Solids Formula & Tricks

Important Formulas

Cube in a sphere
2R = a√3, so R = a√3/2

The space diagonal of a cube with edge a equals the diameter of its circumscribed sphere.

Cuboid in a sphere
2R = √(l² + b² + h²)

For a cuboid with length l, breadth b and height h, its space diagonal equals the sphere's diameter.

Cylinder in a sphere
r² + (h/2)² = R²

A central cross-section forms a rectangle inside a circle. Here r is cylinder radius, h is height and R is sphere radius.

Cone in a sphere
r² = 2Rh − h²

For a cone of height h and base radius r inscribed in a sphere of radius R, a cross-section through the axis gives this relation.

Basic solid volumes
V_cube = a³, V_cuboid = lbh, V_cylinder = πr²h, V_cone = (1/3)πr²h

After finding an unknown dimension from the inscribed-solid relation, use the appropriate volume formula.

Quick Tricks

Use the diameter first

When a cube or cuboid is inscribed in a sphere, do not compare an edge with the sphere radius. Compare the space diagonal with the sphere diameter 2R.

Example: A cube of edge 6 cm has sphere diameter 6√3 cm and sphere radius 3√3 cm.
Reduce a cylinder to a right triangle

In a central section of a cylinder inside a sphere, the horizontal leg is r, the vertical leg is h/2 and the hypotenuse is R.

Example: If R = 5 cm and h = 6 cm, then r² = 25 − 3² = 16, so r = 4 cm.
Check dimensions before calculating

Every length in an inscribed-solid equation must use the same unit. Convert diameters to radii or radii to diameters consistently before substitution.

Example: If a sphere has diameter 14 cm, use R = 7 cm in r² + (h/2)² = R².

Inscribed Solids Concepts

Cube Inscribed in a Sphere

For a cube inscribed in a sphere, the cube's space diagonal equals the sphere's diameter: a√3 = 2R.

A cube has space diagonal √(a² + a² + a²) = a√3. Therefore, if the edge is known, R = a√3/2. If the sphere radius is known, the cube edge is a = 2R/√3.

Example: If the sphere radius is 6 cm, the cube edge is a = 12/√3 = 4√3 cm.

Cuboid Inscribed in a Sphere

For a cuboid inscribed in a sphere, its space diagonal is equal to the sphere's diameter: √(l² + b² + h²) = 2R.

The four space diagonals of a cuboid have the same length. If two dimensions and the sphere radius are known, the third dimension can be found by rearranging: h = √(4R² − l² − b²), provided the expression is non-negative.

Example: For R = 5 cm, l = 6 cm and b = 8 cm, h = √(100 − 36 − 64) = 0 cm. Thus these values form a degenerate limiting case, not a proper cuboid. A proper example is R = 6 cm, l = 6 cm and b = 8 cm: h = √(144 − 36 − 64) = √44 = 2√11 cm.

Cylinder Inscribed in a Sphere

For a right circular cylinder inscribed in a sphere, r² + h²/4 = R².

Take the cross-section through the cylinder axis. The cylinder's half-height h/2, radius r and sphere radius R form a right triangle. Hence r = √(R² − h²/4), and h = 2√(R² − r²).

Example: For a sphere of radius 5 cm and an inscribed cylinder of height 6 cm, r = √(25 − 9) = 4 cm. Its volume is πr²h = π × 16 × 6 = 96π cm³.

Cone Inscribed in a Sphere

For a cone of height h and base radius r inscribed in a sphere of radius R, r² = 2Rh − h².

Place the cone's apex on the sphere and let its base plane be at distance R − h from the sphere's centre. The base radius satisfies r² = R² − (R − h)², which simplifies to r² = 2Rh − h².

Example: If R = 5 cm and h = 6 cm, then r² = 2 × 5 × 6 − 36 = 24, so r = 2√6 cm.

Inscribed Solids Visual Guide & Data

Use the table and chart below to understand the data better.

Common Inscribed-Solid Relations

The defining dimension relation for standard solids inscribed in a sphere.

Inscribed solid Relation Meaning
Cube a√3 = 2R Space diagonal equals sphere diameter
Cuboid √(l² + b² + h²) = 2R Space diagonal equals sphere diameter
Cylinder r² + h²/4 = R² Radius and half-height form a right triangle with R
Cone r² = 2Rh − h² Base radius follows from cone height and sphere radius

Note: All lengths must be measured in the same unit.

Cube Inscribed in a Sphere

Visual guide: Cube Inscribed in a Sphere
Cube edge: a Sphere radius: R Space diagonal: a√3 = 2R
The cube's space diagonal joins opposite vertices on the sphere and equals its diameter.

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Quick Revision Notes

Inscribed Solids Revision Points

Use the boundary relation first, then apply the required area or volume formula.

  • For a cube in a sphere, a√3 = 2R.
  • For a cuboid in a sphere, √(l² + b² + h²) = 2R.
  • For a cylinder in a sphere, r² + h²/4 = R².
  • For a cone in a sphere, r² = 2Rh − h².
  • The sphere diameter, not radius, equals the space diagonal of an inscribed cube or cuboid.
  • Use consistent units for all dimensions.

Inscribed Solids FAQs

What is the formula for a cube inscribed in a sphere?

If the cube edge is a and the sphere radius is R, then a√3 = 2R. Therefore, a = 2R/√3 and R = a√3/2.

A cube of edge 8 cm is inscribed in a sphere. What is the sphere radius?

The radius is a√3/2 = 8√3/2 = 4√3 cm.

What is the relation between the dimensions of a cuboid and its circumscribed sphere?

The cuboid's space diagonal equals the sphere diameter: √(l² + b² + h²) = 2R.

A cylinder of height 8 cm is inscribed in a sphere of radius 5 cm. Find its radius.

r² = 5² − (8/2)² = 25 − 16 = 9, so r = 3 cm.

What is the cone formula for a cone inscribed in a sphere?

For sphere radius R, cone height h and base radius r, r² = 2Rh − h².

Can the cylinder height be greater than the diameter of the sphere?

No. Since r² + h²/4 = R² and r² is non-negative, h/2 ≤ R. Therefore, h ≤ 2R.

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