A rectangular box is 2 cm by 3 cm by 6 cm
- Base diagonal squared = 2² + 3² = 13.
- Space diagonal squared = 13 + 6² = 49.
- Space diagonal = √49 = 7 cm.
Connect roots, geometry and a defensible final answer.
Connect right triangles across several steps and check the final result against the context.
Some problems need two applications of Pythagoras. In a rectangular box, first find the base diagonal from length and width. Then use that base diagonal and the perpendicular height as legs of another right triangle. The space diagonal is the final hypotenuse.
Predict first. Change one value, then explain the result. Diagrams show the entered dimensions; displayed decimals are rounded.
For a box with length 3, width 4 and height 12, the base diagonal has square 3² + 4² = 25. The space diagonal has square 25 + 12² = 169, so its length is 13. In general d = √(length² + width² + height²) for a rectangular box. This is an extension built from two right triangles, not a rule for arbitrary three-dimensional shapes.
The space diagonal must exceed each box dimension and the base diagonal. Identify each right angle, give exact intermediate expressions, and round only the final result to the requested precision. If the answer is unreasonable, revisit the geometry before recalculating. A method is complete when the reader can see why the equation fits the situation.
A box is 6 units by 8 units by 24 units. Find its space diagonal.
The base diagonal is 10; use it with the height of 24.
Space diagonal = √(100 + 576) = √676 = 26 units.
Work through eight questions. Two independent correct answers introduce a harder version; supported answers bring the next question back to the foundation. You can use hints at any time.
This round saves on this device. It is practice, not a school mark.
A square has diagonal 12 cm. Find its exact area without rounding a side length. Explain why the two equal leg-square areas together total 144 cm².
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