A 5 m ladder has its base 3 m from a vertical wall
- With level ground, the ladder is the hypotenuse.
- Height = √(5² − 3²) = √16 = 4 m.
- The model gives a 4 m contact height.
Choose the right triangle hidden in the situation.
Translate perpendicular changes and map scales into a justified straight-line distance.
A rectangle diagonal, a taut cable or a direct route across perpendicular paths can form a hypotenuse. Mark the right angle and label known lengths with units. Use Pythagoras only when perpendicularity is given or justified. A sloping surface is not automatically a right triangle.
Predict first. Change one value, then explain the result. Diagrams show the entered dimensions; displayed decimals are rounded.
An invented route goes 6 km east and 8 km north on perpendicular paths. The travelled distance is 14 km. The straight-line distance is √(36 + 64) = 10 km. The direct route need not be an available road. Similarly, coordinate differences give horizontal and vertical lengths; use their magnitudes as leg lengths.
A model needs assumptions. For a ladder, assume a straight ladder, vertical wall and level ground; ladder length is the hypotenuse, not the height reached. Use compatible units, keep intermediate precision and state a contextual answer. A geometric calculation alone does not establish that equipment is safe to use.
A rectangular field measures 30 m by 40 m. Compare a diagonal crossing with two adjacent sides.
The two side lengths are legs of a right triangle.
Diagonal = 50 m. Two adjacent sides total 70 m, so the direct crossing is 20 m shorter.
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.
Invent a situation in which the calculated straight-line distance is accurate but cannot be used as the actual travel route. State the model assumptions.
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