Solids

274. The Cuboid's Diagonal · Uncovering the space diagonal with Pythagoras

The space diagonal BH always satisfies BH² = l² + b² + h².

ACFGHl = 240l = 240b = 160b = 160h = 200h = 200DB (from △DBC) = √(l² + b²) = √(240² + 160²) = 288.4DB (from △DBC) = √(l² + b²) = √(240² + 160²) = 288.4BH (from △DBH) = √(DB² + h²) = √(288.4² + 200²) = 351BH (from △DBH) = √(DB² + h²) = √(288.4² + 200²) = 351BDE
The space diagonal of a cuboid runs from one corner to the opposite corner through the solid. By Pythagoras in three dimensions, BH² = l² + b² + h², so BH = √(l² + b² + h²).

Stuck? Ask Guru

Selina ICSE: Solids

What this lesson covers

Try to break it

Drag B for length l, D for breadth b, and E for height h. The space diagonal BH always measures √(l² + b² + h²) — apply Pythagoras once on the base for √(l² + b²), then again with h. Try to find dimensions where the readout disagrees with the formula; impossible.

How you build it

Construct a cuboid and draw its space diagonal.

  • Rectangle tool: drag out the front face of the cuboid (corners A, B, F, E).
  • Rectangle tool: drag an identical rectangle shifted up and to the right — the back face (corners D, C, G, H). The shift is the depth.
  • Segment tool: join the front-bottom-left corner A to the back-bottom-left corner D — a depth edge.
  • Segment tool: join the front-bottom-right corner B to the back-bottom-right corner C.
  • Segment tool: join the front-top-left corner E to the back-top-left corner H.
  • Segment tool: join the front-top-right corner F to the back-top-right corner G — the cuboid is complete.
  • Point tool: mark B at the front-bottom-right corner.
  • Point tool: mark D at the back-bottom-left corner — opposite B across the base.
  • Point tool: mark H at the back-top-left corner — directly above D.
  • Segment tool: join D to B — the base diagonal. In right triangle DCB, DB² = DC² + CB² = l² + b².
  • Segment tool: join B to H — the space diagonal. In right triangle DBH, BH² = DB² + DH² = l² + b² + h².

The proof, step by step

Prove that the space diagonal of a cuboid satisfies BH² = l² + b² + h².

  • In right triangle ABD, angle BAD = 90°. By Pythagoras theorem, BD² = AB² + AD² = l² + b².
  • In right triangle BDH, angle BDH = 90° (since DH is perpendicular to the base). By Pythagoras theorem, BH² = BD² + DH².
  • Substitute BD² = l² + b² and DH = h into the equation: BH² = (l² + b²) + h² = l² + b² + h².
  • Therefore, the length of the diagonal BH = √(l² + b² + h²).

Worked example

A cuboid has length 3 cm, breadth 4 cm, and height 12 cm. Find the length of its diagonal.

Diagonal = √(l² + b² + h²) = √(3² + 4² + 12²) = √(9 + 16 + 144) = √169 = 13 cm.

  • 13 cm — correct
  • 14 cm
  • 15 cm
  • 16 cm
Hold to talk

Subscription Status