Secondary Mathematics · Worked Repair Guide 47
Pythagoras is often taught in one direction: if a triangle is right-angled, then the square of its hypotenuse equals the sum of the squares of the other two sides. The converse reverses the logic: if the side lengths of a triangle satisfy that relationship with the longest side, then the triangle must be right-angled.
This guide develops one central habit: identify the longest side before testing the squares. The largest side must be compared against the sum of squares of the other two. Using the wrong side as the proposed hypotenuse can make a valid right triangle appear invalid.
All triangles and coordinate examples below are original teaching constructions. This page complements Trigonometry, Pythagoras, Bearings and Elevation and Angles, Similarity and Geometric Reasoning.
1. The theorem and its converse are different statements
Pythagoras theorem: if a triangle is right-angled with legs a,b and hypotenuse c, then a²+b²=c².
Converse: if a triangle has side lengths a,b,c with c the longest and a²+b²=c², then the angle opposite c is90°.
Entry check: the converse uses a numerical side relationship to prove an angle condition.
2. Always place the longest side on the comparison side
For side lengths6,8,10, the longest side is10.
6²+8²=36+64=100=10².
Therefore the triangle is right-angled, with the right angle opposite the side of length10.
3. A failed equality means the triangle is not right-angled
For sides5,6,8, compare5²+6²=61 with8²=64.
Since61≠64, the triangle is not right-angled.
The converse test answers the right-angle question directly without needing to calculate any angle.
4. The square comparison also distinguishes acute and obtuse largest angles
For sides a,b,c with c longest:
- If a²+b²=c², the largest angle is90°.
- If a²+b²>c², the largest angle is acute.
- If a²+b²<c², the largest angle is obtuse.
This comparison is consistent with the cosine rule, because c²=a²+b²−2ab cos C.
5. Test an acute triangle numerically
For sides5,6,7, the longest side is7.
5²+6²=61 while7²=49.
Since61>49, the angle opposite7 is acute. Because it is the largest angle, all angles are acute.
6. Test an obtuse triangle numerically
For sides3,4,6, the longest side is6.
3²+4²=25 while6²=36.
Since25<36, the angle opposite6 is obtuse.
The triangle is feasible because3+4>6.
7. Check triangle feasibility before classifying angles
Lengths2,3,6 cannot form a triangle because2+3<6.
It would be meaningless to classify its “largest angle” using the square comparison because no non-degenerate triangle exists.
Use triangle inequalities first when feasibility is not guaranteed.
8. Pythagorean triples are integer right-triangle side sets
A Pythagorean triple consists of positive integers a,b,c satisfying a²+b²=c².
Common primitive examples include3,4,5; 5,12,13; 8,15,17; and7,24,25.
Any common positive multiple of a triple is also a triple:6,8,10 is twice3,4,5.
9. Recognising triples speeds up but does not replace reasoning
If a triangle has sides9,12,15, recognising three times3,4,5 immediately suggests a right triangle.
A written verification remains short:9²+12²=81+144=225=15².
Recognition is a shortcut to the test, not a substitute for knowing why the test works.
10. Coordinates can turn perpendicularity into a distance test
Let A=(0,0), B=(4,0), C=(4,3).
AB=4, BC=3 and AC=5.
Since3²+4²=5², triangle ABC is right-angled at B.
This independently confirms what vertical/horizontal line geometry already shows.
11. Distance formula lets the converse work on tilted coordinate triangles
Take A=(1,1), B=(5,2), C=(3,9).
AB²=(4)²+(1)²=17. BC²=(−2)²+7²=53. AC²=2²+8²=68.
Since17+53=70≠68, the triangle is not right-angled.
Using squared distances avoids unnecessary square roots when only a Pythagorean comparison is needed.
12. Squared-distance comparison can prove perpendicular diagonals
Suppose a triangle formed by half-diagonals in a quadrilateral has side lengths6,8,10.
The converse proves the angle between the two half-diagonals is90°.
This can then feed into a quadrilateral classification proof.
13. The converse can verify construction accuracy
An invented rectangular frame is intended to have side lengths12 and16 and diagonal20.
12²+16²=144+256=400=20².
Thus those three lengths are consistent with a right angle.
In practical measurement, tolerances would matter; the exact school problem assumes exact stated lengths.
14. A near miss is still not an exact right triangle
For sides12,16,19.9, we have12²+16²=400 while19.9²=396.01.
The exact stated triangle is not right-angled.
Whether a physical object is acceptably close to90° is a measurement/tolerance question, not the exact converse theorem.
15. Similarity preserves right-angle structure
If all sides of a right triangle are multiplied by k, then (ka)²+(kb)²=k²(a²+b²)=k²c²=(kc)².
Thus every positive scale multiple of a Pythagorean triple remains right-angled.
This explains why3-4-5 and30-40-50 encode the same triangle shape.
16. The cosine rule explains the acute/obtuse comparison
From c²=a²+b²−2ab cos C:
If C=90°, cosC=0 and c²=a²+b².
If C<90°, cosC>0, so c²<a²+b².
If C>90°, cosC<0, so c²>a²+b².
The square comparison is therefore a compact angle classifier.
17. Capstone: classify three triangles efficiently
Triangle A has sides8,15,17:64+225=289, so right-angled.
Triangle B has sides8,15,16:64+225=289>256, so its largest angle is acute and therefore the triangle is acute.
Triangle C has sides8,15,18:64+225=289<324, so its largest angle is obtuse.
All three satisfy the triangle inequality, so each classification is meaningful.
18. Independent practice
- State the converse of Pythagoras.
- Test whether6,8,10 form a right triangle.
- Test whether5,6,8 form a right triangle.
- Classify the largest angle of5,6,7.
- Classify the largest angle of3,4,6.
- Can2,3,6 form a triangle?
- Verify3,4,5 as a Pythagorean triple.
- Verify5,12,13.
- Verify8,15,17.
- Is9,12,15 a right-triangle triple?
- State the right-angle vertex for A=(0,0),B=(4,0),C=(4,3).
- For sides7,24,25, classify the triangle.
- For sides10,10,12, classify the largest angle.
- For sides5,5,8, classify the largest angle.
- Explain why the longest side must be used as c.
- A frame has sides12,16 and diagonal20. Is the corner right-angled?
- Are12,16,19.9 exact right-triangle lengths?
- If3,4,5 is scaled by7, give the new triple.
- For sides8,15,16, classify the triangle.
- For sides8,15,18, classify the triangle.
19. Worked answers
1. If the square of the longest side equals the sum of squares of the other two, the triangle is right-angled opposite the longest side.
2. Yes. 36+64=100.
3. No. 25+36=61≠64.
4. Acute.
5. Obtuse.
6. No. 2+3<6.
7. 9+16=25.
8. 25+144=169.
9. 64+225=289.
10. Yes. It is3 times3,4,5.
11. B.
12. Right-angled. 49+576=625.
13. Acute. 100+100>144.
14. Obtuse. 25+25<64.
15. The right angle, if present, is opposite the longest side, so its square must be isolated on the comparison side.
16. Yes. 144+256=400.
17. No. 19.9²=396.01.
18. 21,28,35.
19. Acute.
20. Obtuse.
20. Diagnose converse errors at the longest-side choice
Common failures include putting a shorter side on the c² side, applying the test before checking whether a triangle exists, confusing the theorem with its converse, or treating approximate physical measurements as exact geometric equalities.
A useful repair note says “longest side first”, “triangle inequality before angle classification”, or “equality proves right; greater/less classifies acute/obtuse”.
21. Continue through the BTT learning routes
Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Trigonometry, Pythagoras, Bearings and Elevation, Sine Rule and Cosine Rule and Coordinate Geometry.
Within Batch12, continue to Special Quadrilaterals, Data Collection and Dot Diagrams, or Circle Symmetry, Equal Chords and Tangent Pairs.
22. Sources and scope
The triangles and coordinate examples are original teaching material. Pythagorean comparisons are used in their standard Euclidean sense; measurement tolerances belong to a separate modelling layer.