A cuboid has two types of diagonals: face diagonals (lying on the faces) and space diagonals (going through the interior). The longest diagonal of a cuboid is the space diagonal (or body diagonal), which connects two opposite vertices through the interior. The formula for the space diagonal of a cuboid with length l, breadth b, and height h is: d = √(l² + b² + h²).
Space diagonal of cuboid = √(l² + b² + h²).
Derived using Pythagoras twice: base diagonal = √(l²+b²), then space diagonal = √(l²+b²+h²).
A cuboid has 4 equal space diagonals.
Face diagonal (on l×b face) = √(l²+b²).
Cube space diagonal = a√3 (where a = side length).
Practical use: 'longest rod that fits in a room' = space diagonal of the room.
A cuboid has 6 faces, 12 edges, 8 vertices.
Formula: Space Diagonal (d) = √(l² + b² + h²)
Where: • l = length of the cuboid • b = breadth (width) of the cuboid • h = height of the cuboid
Derivation: Step 1: Find the face diagonal of the base rectangle (l × b) Face diagonal of base = √(l² + b²) [by Pythagoras]
Step 2: This face diagonal and height h form a right triangle Space diagonal² = (face diagonal of base)² + h² = (√(l² + b²))² + h² = l² + b² + h²
Space diagonal d = √(l² + b² + h²) ✓
For a cube (l = b = h = a): Space diagonal = √(a² + a² + a²) = √(3a²) = a√3
A cube with side a has space diagonal = a√3
A cuboid has 3 different types of faces (rectangles):
Face l × b (top/bottom face): Face diagonal = √(l² + b²)
Face b × h (front/back face): Face diagonal = √(b² + h²)
Face l × h (left/right face): Face diagonal = √(l² + h²)
Each cuboid has: • 4 space diagonals (equal in length: d = √(l²+b²+h²)) • 12 face diagonals (4 of each type above) • 12 edges (4 of each: l, b, h)
Difference between face diagonal and space diagonal: • Face diagonal: diagonal within one face (2D, on surface) • Space diagonal: diagonal through the interior (3D, body diagonal) • Space diagonal > face diagonal always
Example 1: Find the space diagonal of a cuboid 12 cm × 5 cm × 4 cm. d = √(12² + 5² + 4²) = √(144 + 25 + 16) = √185 ≈ 13.6 cm
Example 2: Find the diagonal of a cube with side 6 cm. d = a√3 = 6√3 ≈ 10.39 cm
Example 3: A cuboid has space diagonal 13 cm and base dimensions 3 cm × 4 cm. Find height. d = √(l² + b² + h²) 13 = √(3² + 4² + h²) 169 = 9 + 16 + h² h² = 169 − 25 = 144 h = 12 cm
Example 4: A room is 8m × 6m × 3m. What is the length of the longest rod that fits in the room? Longest rod = space diagonal = √(8² + 6² + 3²) = √(64 + 36 + 9) = √109 ≈ 10.44 m
Faces, Edges, Vertices of a cuboid: • Faces: 6 • Edges: 12 • Vertices: 8 • Space diagonals: 4 (all equal)
Euler's formula check: V − E + F = 8 − 12 + 6 = 2 ✓
The space diagonal of a cuboid with length l, breadth b, and height h is: d = √(l² + b² + h²). For a cube with side a, the diagonal = a√3. This is derived by applying Pythagoras theorem twice: first for the base face diagonal, then for the space diagonal.
d = √(3² + 4² + 12²) = √(9 + 16 + 144) = √169 = 13 cm.
Diagonal of cube = a√3 = 5√3 ≈ 8.66 cm.
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