eduKatePunggol · Secondary 3 Mathematics
Find your next learning step
A Face Diagonal Is Not the Cuboid’s Space Diagonal. Choose the reading route closest to your question.
Route 1 · Chapters 1–2
Route 2 · Chapters 3–4
Route 3 · Chapters 5–6
Route 4 · Chapters 7–8
Full chapter index · Independent practice · Secondary 3 Mathematics support
If your child calculates five centimetres from a cuboid’s three-by-four base and stops, check which diagonal the question requests. Five is the base-face diagonal. If the cuboid is also twelve centimetres high, the straight diagonal joining opposite corners through the solid is thirteen centimetres.
Secondary 3 Mathematics tuition in Punggol can show the two right triangles separately: 3² + 4² = 5², then 5² + 12² = 13². A Punggol Secondary 3 Maths tutor should ask where the requested line begins and ends before accepting the first calculated length.
This guide uses original idealised cuboids with mutually perpendicular edges. It distinguishes face diagonals, internal space diagonals and routes along a surface. Topic timing can differ between schools. Small-group 3-pax support should help each learner identify the relevant plane and justify the right angles before calculating.
Full chapter index
Name the diagonal · Chapters 1–2
Build the two right triangles · Chapters 3–4
Check the solid and path · Chapters 5–6
Practise and continue · Chapters 7–8
A face diagonal connects opposite corners of one rectangular face. For a base measuring 3 cm by 4 cm, Pythagoras gives diagonal d = √(3² + 4²) = 5 cm. That segment stays entirely in the base plane.
A cuboid has other rectangular faces and other face diagonals. Their lengths need not all be equal if the edge lengths differ. State which face is being used. A successful calculation on one face does not establish that it is the segment requested by a three-dimensional question.
A space diagonal joins opposite vertices through the interior of the cuboid. Its endpoints differ across all three perpendicular edge directions. For dimensions 3, 4 and 12 cm, it uses both the base diagonal and the vertical height rather than the base alone.
Mark those endpoints on the diagram before drawing a supporting triangle. If the requested segment ends at the opposite top corner while the calculated segment ends at a bottom corner, one more dimension remains. The distinction is geometric, not a demand to use a more complicated formula automatically.
Chapter 3 of 8 · Build the two right triangles
3. Why is the second triangle right-angled?
In a cuboid, the vertical edge is perpendicular to the base plane. It is therefore perpendicular to the base diagonal meeting it at the corner. The base diagonal, that vertical edge and the space diagonal form a right triangle in a vertical diagonal section.
This is why Pythagoras can be applied a second time. The right angle is established by the solid’s geometry, even if perspective makes it look slanted on the page. Do not infer a right angle merely because two lines look perpendicular in a sketch; use the stated cuboid conditions.
Chapter 4 of 8 · Build the two right triangles
4. How does the 3, 4, 12 example finish?
The base diagonal has length five. The second right triangle gives D² = 5² + 12² = 25 + 144 = 169, so D = 13 cm. Since D is a length, use the positive square root. The answer includes the twelve-centimetre height that the base-only result omitted.
Substituting the first step gives D² = 3² + 4² + 12². Thus the familiar three-dimensional formula D = √(l² + w² + h²) follows from two justified right-triangle calculations. It is not a rule for adding any three lengths found on any solid.
Yes, as long as the second right triangle uses the remaining perpendicular edge correctly. Start with the 3-by-12 face to obtain √153. Then D² = 153 + 4² = 169, again giving thirteen. The intermediate value changes, but the final segment is the same.
Keep the intermediate square root exact rather than rounding it before the second calculation. Working with the squared value 153 is especially convenient. The two-stage explanation helps locate the segment; the combined expression can simplify the arithmetic once that geometry is secure.
Chapter 6 of 8 · Check the solid and path
6. What should the answer be compared with?
The space diagonal of a non-degenerate cuboid is longer than each individual edge and longer than the corresponding face diagonal. In our example, thirteen exceeds twelve and five. An answer of five for the full space diagonal would fail that basic length comparison.
A path constrained to the outside surface is a different problem. The straight internal diagonal is not automatically an allowed surface route. Likewise, a slanted prism without mutually perpendicular edges needs further reasoning. Check the endpoints, permitted path and solid assumptions before choosing the formula.
Question 1: a cuboid measures 2, 3 and 6 cm; find the diagonal of its 2-by-3 face and its space diagonal. Question 2: a cube has edge 4 cm; find its face and space diagonals exactly. Question 3: a cuboid’s base diagonal is 8 cm and height 15 cm; find its space diagonal. Question 4: is the first face diagonal always the final answer?
Answers: 1) √13 cm and seven centimetres. 2) 4√2 cm and 4√3 cm. 3) Seventeen centimetres. 4) No; it depends on the requested endpoints. Ask your child to identify the right triangle for each calculation rather than simply list lengths.
Bring the original 3D diagram and mark where the child’s calculated segment lies. A precise note is ‘finds the base diagonal but does not include the height’. In a small-group 3-pax lesson, students can draw the diagonal cross-section and explain the two perpendicular legs independently.
For parents considering Secondary 3 Mathematics tuition in Punggol, ask ‘Does this calculated line reach the endpoint the question names?’ That prompt checks the geometry without solving the arithmetic for the child. A correct path through the solid makes subsequent calculations easier to organise and verify.
