The learning goal
Label the conditions and side roles before writing the equation. Substitute the answer back and check the longest side, units and positive length.
Check the condition before arranging letters
OpenStax explains the relationship for a right triangle: the squares of its legs add to the square of the hypotenuse. The letter c has no special power by itself. What matters is whether it represents the side opposite the right angle, regardless of how the diagram has been rotated.
Two side lengths without a right-angle condition or other sufficient information do not justify taking the square root of their squared sum as the third side. A sketch that merely looks square at a corner is not the same as a stated condition or a right-angle mark.
Three statements before an equation
Condition
Does the problem establish a right triangle?
Roles
Which side is opposite the right angle, and which two form it?
Units
Are all lengths expressed in the same unit?
Find the hypotenuse from legs of 5 and 12
Original teaching case one explicitly gives a right angle with legs of 5 cm and 12 cm. Let c be the hypotenuse: 5² + 12² = c², so 25 + 144 = 169 and c = 13 cm. Take the positive length rather than treating a negative algebraic root as a physical side length.
Substitution gives 13² = 169, matching 25 + 144. Thirteen is also larger than either leg. The right-angle condition is essential: if the angle between the same two given lengths were unknown, this derivation would no longer have its required premise.
| Step | Case-one record |
|---|---|
| Identify | 5 and 12 are legs; c is the hypotenuse |
| Form the equation | 5² + 12² = c² |
| Solve | c² = 169; c = 13 cm |
| Check | 25 + 144 = 169; c is the longest side |
Why a hypotenuse of 10 and a leg of 6 require subtraction
Teaching case two states that 10 m is the hypotenuse and 6 m is a leg. If the other leg is x, the relationship remains 6² + x² = 10². Rearranging gives x² = 100 − 36 = 64, so x = 8 m. The theorem has not changed; the unknown occupies a different position.
A common designed error is √(10² + 6²), which treats both supplied numbers as legs. Write the complete relationship before rearranging so that the known hypotenuse does not silently change roles.
An intentionally incorrect solution
x = √(10² + 6²), treating 10 and 6 as legs.
6² + x² = 10², hence x = √64 = 8 m.
Do not add squares in different units
In teaching case three, the legs are 60 cm and 0.8 m. Convert 60 cm to 0.6 m first. Then c² = 0.6² + 0.8² = 0.36 + 0.64 = 1, giving c = 1 m or 100 cm. Adding 60² and 0.8² without conversion mixes square centimetres with square metres.
This is a practical extension of unit cancellation and magnitude checks. You may work entirely in metres or entirely in centimetres, but the equation cannot silently mix them. Squaring produces area units; taking the square root returns a length unit.
Make four decisions before checking the answers
These are original practice questions. State the condition and side roles before calculating. The fourth question deliberately omits a right angle: a pattern of three successful applications is not permission to apply the same method to an unsupported fourth case.
As with substituting a candidate back into a radical equation, obtaining a number is not the end of verification. It must fit the original requirements. No learning-effect percentage is claimed; the exercise observes whether the necessary conditions survive your reasoning.
| Question | Reference and reason |
|---|---|
| Right-triangle legs 9 and 12 cm; find hypotenuse | √(81 + 144) = 15 cm |
| Hypotenuse 13 and leg 5 cm; find other leg | √(169 − 25) = 12 cm |
| Right-triangle legs 60 cm and 0.8 m | Convert to 0.6 m and 0.8 m; result 1 m |
| Only sides 5 and 12 cm; included angle unknown | The third side cannot be asserted to be 13 cm |
Check each solution
- A right angle is established rather than guessed
- The hypotenuse is opposite it and is longest
- Units agree before squaring
- The result is positive and satisfies substitution
- Missing conditions are stated explicitly
