Recognise congruent polygons and match corresponding parts
Recognise congruent polygons through same shape and size, match corresponding parts and distinguish insufficient area, perimeter or angle evidence.
Section 4.2 · Polygons
Exam scope and assessment notes
Foundation and Higher share all seven section4.2 requirements A–G. The Higher table says See Foundation Tier; there are no additional H4.2 rows. These seven original lessons teach polygon names, convex/concave quadrilateral sums, all six named quadrilateral families, general polygon sums, regular interior/exterior angles and congruence correspondence. F:4.2C requires both the parallelogram/rectangle/rhombus/square lesson and the trapezium/kite lesson with substantive teaching, a worked example and every Guided, Independent and Transfer stage. F and G are two distinct published rows; the same substantive congruence lesson links both without inventing a third requirement. Seven lessons and28 selected seed cases are StudyForA’s design, not a syllabus-prescribed count. Rhombus/square diagonal endpoint-angle bisection, kite symmetry-diagonal endpoint-angle bisection and isosceles-trapezium equal diagonals are conservative standard-property breadth under the broad C wording, not new verbatim syllabus bullets or an assertion that the public book chart enumerates each one. Use labelled boundary order, actual interior/reflex arcs, given or justified parallel/equal-side/right-angle marks and geometric reasons before equations. Diagrams need not be to scale; do not measure them unless instructed. Use the local exactly-one-parallel-pair trapezium convention and different-length adjacent-pair kite fixtures; the sources do not settle every inclusive naming convention. Keep rectangle equal diagonals, rhombus perpendicular/bisected diagonals, square45° endpoint halves and kite one-way diagonal bisection distinct. General simple-polygon sums include checked concave triangulation; crossed boundaries, holes and curved outlines are outside these cases. Regular examples use consistent convex exterior turns and exact integer side-count checks; unknown-n full outlines follow the deduction, while impossible marked data remain condition cards. Congruence preserves full corresponding shape and size under turns/reflections, including reflex interiors; equal area, perimeter or angles alone is insufficient. A fixed reference picture and a correct number do not establish that a learner drew or justified the configuration. Written work is not automatically marked. Attempt acknowledgements and revealed references do not award marks or establish mastery. These structural links are not exhaustive subskill approval; specialist approval and formal assessment validation remain unestablished. Section4.3 symmetry,4.4 measurement/bearings,4.5 formal instrument constructions,4.11 similarity and the full H4.7A reasoning scope remain separate. Local symmetry-diagonal use is not complete symmetry coverage; no blanket formal-proof waiver is imported from unrelated circle-theorem notes. The wider Maths A course remains incomplete. Publisher book navigation is separate evidence: Book1 Shape and Space1 has literal F4.2B/C chapter and Hub19 Geometry of Shapes2 tags, not A; map60–99 and proof start60/next100 agree. Book2 Shape and Space6 has literal D/E chapter tags, with Hub59 Angles in Polygons2 for D and Hub58 Angles in Polygons1 for E; map128–160 and proof start128/next161 agree. The map associates F/G with Shape and Space7 pages223–248, while the public proof chapter starts222 and its next chapter starts248; proof contents instead place Congruent shapes129 and Similar shapes147 in Shape and Space6. Preserve that unresolved topic/page conflict rather than moving tags silently. Hub85 Similarity1 literally tags F4.2F and F4.11A, not G. A has no literal chapter/Hub tag and G has no literal Hub tag. Only selected Book1 public uncorrected proof body pages69–71 were inspected; Book2 relevant chapters are contents-only evidence in the public sample. Full final-book and Teaching Hub body/pagination alignment remains unverified; chapter links confer no full-book or modular teaching credit.
Before you begin
- Identify corresponding labelled sides and angles, including reflex interiors.
- Solve a linear equation and calculate rectangle area/perimeter with units.
- Use a supplied rigid-copy relationship without needing transformation centres or a distance formula.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A polygon is reflected as a rigid copy with every corresponding length and angle preserved. Does reversing its boundary orientation prevent congruence?
What you will learn
- Explain congruence as same shape and size, including turned, reflected and concave copies.
- Match complete vertex correspondences and use them for named lengths and angles.
- Use counterexamples to show why equal area, perimeter or angles alone does not establish congruence.
Draw or annotate pentagon ABCDE and its given rigidly turned copy PQRST, with correspondence A↔P,B↔Q,C↔R,D↔S,E↔T. For a grid layout use A(0,0),B(4,0),C(5,2),D(2,4),E(0,3) and P(9,1),Q(9,5),R(7,6),S(5,3),T(6,1), joining each list in boundary order. If AB=4 cm and PQ=(2x−2) cm, find x. Match CD and ∠ABC, then explain why the turn does not prevent congruence. The rigid-copy relationship is given; no slope or distance formula is needed.
Draw or copy each labelled boundary on paper. Keep stated side lengths, selected interior/reflex/exterior arcs, parallel arrows and equality marks. Give geometric reasons before equations and substitute back. Exact givens determine deductions; screen diagrams are illustrative and must not be measured for an exact answer.
Original ABCDE
- α1 · arc-1: given ∠ABC
- AB: 4 cm
- CD: CD
These comparison panels share one length scale, so rigid copies retain their size and any scaled comparison retains its size difference.
Given copy PQRST
- PQ: (2x−2) cm
- RS:
These comparison panels share one length scale, so rigid copies retain their size and any scaled comparison retains its size difference.
Match the whole shape and size
Two polygons are congruent when they have the same shape and size. Their corresponding side lengths and interior angles match. A rigid turn or reflection preserves those measurements, so a differently placed or mirrored copy can still be congruent. “It points the other way” is not a reason to reject congruence.
Read a given rigid-copy construction or a complete correspondence, rather than guessing from a similar outline. One matching side or one equal angle is insufficient to establish that the whole polygons match. The tasks here use complete labelled copies and polygon reasoning, not an assumed programme of triangle congruence tests.
Keep vertex order and angle notation
Write the correspondence in boundary order before transferring a length. If A↔P,B↔Q,C↔R, then AB↔PQ and BC↔QR. For ∠ABC the middle vertex is B, so the corresponding named angle is ∠PQR. Moving a label to a different vertex changes the statement.
A reflection reverses the direction in which the two boundaries are traversed. Keep the supplied matching vertices together rather than forcing both lists to run clockwise on the page. In a concave copy the reflex interior must map to the matching reflex interior, not to its smaller outside gap. Exact given lengths and copy relationships decide the result, not measurements from an unscaled drawing.
Test claims with counterexamples
Equal area does not mean the same shape: rectangles 3 by 8 and 4 by 6 have equal areas but different dimensions. Equal perimeter is also insufficient: rectangles 3 by 8 and 5 by 6 have the same perimeter but different side pairs. These examples explain the limitation instead of merely saying that a drawing looks different.
Equal angles alone do not fix size. A double-size copy preserves all angles while doubling lengths, so it is not congruent to the original. This comparison separates shape from size; it does not by itself complete a general similarity or transformation course. Explain both parts of the definition and use the actual correspondence when solving for a missing length.
Sources for this lesson
Pause and explain
A rectangle is 3 cm by 7 cm and another is 6 cm by 14 cm. All their angles are right. Are they congruent?
Worked example
Draw or annotate pentagon ABCDE and its given rigidly turned copy PQRST, with correspondence A↔P,B↔Q,C↔R,D↔S,E↔T. For a grid layout use A(0,0),B(4,0),C(5,2),D(2,4),E(0,3) and P(9,1),Q(9,5),R(7,6),S(5,3),T(6,1), joining each list in boundary order. If AB=4 cm and PQ=(2x−2) cm, find x. Match CD and ∠ABC, then explain why the turn does not prevent congruence. The rigid-copy relationship is given; no slope or distance formula is needed.
Show the worked solution
- The given rigidly turned copy preserves every corresponding length and angle. The listed correspondence pairs AB with PQ, so 2x−2=4.
- Solve 2x=6 to obtain x=3, and substitute PQ=4 cm=AB. The variable is found from the corresponding length, without calculating diagonal distances.
- C↔R and D↔S, so CD↔RS. In ∠ABC the middle vertex B maps to Q, giving ∠PQR. A turn changes the direction on the page but not the shape or size, so the polygons are congruent.
Answer x=3 and PQ=4 cm. CD↔RS and ∠ABC↔∠PQR. The given rigid turn preserves shape and size, so the polygons are congruent.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Draw or annotate quadrilaterals ABCD and PQRS as the given reflected copies, with A↔P,B↔Q,C↔R,D↔S. A grid layout is A(0,0),B(6,0),C(4,3),D(0,2) and P(2,5),Q(−4,5),R(−2,8),S(2,7). Keep those boundary orders even though their orientations are reversed. With AB=6 cm and PQ=(3y−3) cm, find y, match BC and the angle at B, and explain why reflection preserves congruence.
Give me a hint
Write the vertex correspondence first, keeping reflected boundary order intact, then match AB with PQ.
Compare my reasoning
- A reflection is a rigid copy, so every corresponding length and angle is preserved despite the reversed boundary orientation. AB corresponds to PQ.
- Set 3y−3=6, giving 3y=9 and y=3. Substitution gives PQ=6 cm as required.
- B↔Q and C↔R, so BC↔QR. The angle at B corresponds to the angle at Q, or ∠ABC↔∠PQR when fully named.
- The reversed traversal is consistent with reflection; it does not change shape or size. Therefore the two given polygons are congruent.
y=3 and PQ=6 cm. BC↔QR and angle B↔angle Q (∠ABC↔∠PQR). Reflection reverses boundary orientation but preserves shape and size, so the polygons are congruent.
Look for these in your work
- Draw each boundary in its given order and show the complete correspondence.
- Solve the matching-length equation and name the corresponding side and angle.
- Explain why a reflected orientation still gives the same shape and size.
Draw three separate rectangles with right-angle marks and exact dimensions: R is 3 cm by 8 cm, S is 5 cm by 6 cm, T is 4 cm by 6 cm. Find each area and perimeter with units. Which matches R’s perimeter and which matches its area? Is either congruent to R? Explain from the full side dimensions rather than appearance.
Give me a hint
Compare the complete dimension pairs after calculating area and perimeter, with their different units.
Compare my reasoning
- For R, area=3×8=24 cm² and perimeter=2(3+8)=22 cm. Keep area units squared and perimeter units as lengths.
- For S, area=5×6=30 cm² and perimeter=2(5+6)=22 cm. Thus S matches R’s perimeter but not its area or side dimensions.
- For T, area=4×6=24 cm² and perimeter=2(4+6)=20 cm. Thus T matches R’s area but not its perimeter or side dimensions.
- Neither unordered side pair {5,6} nor {4,6} equals R’s {3,8}. A turn cannot fix that size/shape mismatch. Neither rectangle is congruent to R; equal perimeter or equal area alone is insufficient.
R: 24 cm² and 22 cm; S: 30 cm² and 22 cm; T: 24 cm² and 20 cm. S has R’s perimeter, T has R’s area, but neither has R’s side dimensions, so neither is congruent to R.
Look for these in your work
- Calculate three areas in cm² and three perimeters in cm.
- Identify S’s perimeter match and T’s area match.
- Use unequal dimension pairs to reject both congruence claims.
Draw the concave L-shaped hexagon A(0,0),B(5,0),C(5,2),D(2,2),E(2,5),F(0,5) and its given reflected copy P(8,6),Q(13,6),R(13,4),S(10,4),T(10,1),U(8,1), joining each boundary in order. Use A↔P through F↔U. PQ=5 cm and AB=(2z+1) cm: find z. Mark the long reflex interior at D and its corresponding interior in the copy. Would a separate double-size copy of ABCDEF be congruent to it? Explain the role of size.
Give me a hint
Use the given correspondence for z and for the reflex angle; then test both shape and size for the separate comparison.
Compare my reasoning
- The given reflected copy preserves corresponding lengths. AB↔PQ gives 2z+1=5, so 2z=4 and z=2; AB=5 cm.
- At the inward corner D, the small gap is 90° but the polygon’s long interior is 270°. D↔S, so label the corresponding reflex interior at S as 270° too.
- These reflected hexagons are congruent: every corresponding length and interior angle matches, including the reflex pair.
- A separate double-size copy has twice each side length, even though its angles remain the same. Its size differs, so it is not congruent. This comparison does not require a general similarity or transformation calculation.
z=2 and AB=5 cm. D↔S, so the corresponding reflex interior at S is 270°. The reflected copy is congruent, while a separate double-size copy is not because its lengths and size change.
Look for these in your work
- Draw a legible concave boundary and its reflected copy with all six labels.
- Find z and preserve the 270° reflex interior under D↔S.
- Explain why the reflected copy is congruent but the double-size comparison is not.
Common mistakes
- Rejecting a reflected or turned copy because it faces a different direction.
- Matching the wrong middle angle vertex or replacing a reflex interior by the smaller outside sweep.
- Treating equal area, perimeter, proportions or angles alone as sufficient for same shape and size.
What can you explain now?
The correspondence is A↔P,B↔Q,C↔R. Which named angle corresponds to ∠ABC, and why does the middle letter matter?
Compare with the explanation
∠PQR corresponds to ∠ABC.
The vertex B maps to Q, so Q must be the middle letter. The bounding sides BA and BC map to QP and QR.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Define polygon congruence, explain reflected and reflex-angle correspondences, and give dimensioned counterexamples to equal area or equal perimeter being enough.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.