Use trapezium and kite properties with correct diagonals
Use marked trapezium and kite properties, distinguish their special cases and reject data that fail the advertised geometric conditions.
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
- Use co-interior angles between parallel lines and the quadrilateral 360° sum.
- Read adjacent equal-side marks and identify opposite vertices.
- Solve linear equations and add labelled segment lengths.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In kite PQRS, equal adjacent pairs meet at P,R. PR and QS meet at T, with PT=8 cm,TR=5 cm,QT=3 cm. Which statement follows?
What you will learn
- Use co-interior angle pairs in a trapezium and the additional equal-base-angle properties of an isosceles trapezium.
- Locate the equal opposite-angle pair and correctly bisected diagonal in a kite.
- Check both the angle total and the named family conditions before accepting a proposed diagram.
Use two separate labelled panels. Panel 1: right trapezium ABCD has AB∥DC,A=D=90°,B=64°. Mark the parallel pair and find C. Panel 2: kite EFGH has EF=EH and FG=GH, with those pair lengths different; E=76°,G=104°. Mark the two adjacent equal-side pairs and find F,H. Keep each angle arc inside its named polygon. In the kite draw its symmetry diagonal EG, mark its equal angle parts at E and G, and calculate those four parts.
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.
Panel 1: ABCD
- α1 · arc-1: given 90°
- α2 · arc-2: given 64°
- α3 · arc-3: given ?
- α4 · arc-4: given 90°
Panel 2: EFGH
- α1 · arc-1: given 76°
- α2 · arc-2: given ?
- α3 · arc-3: given 104°
- α4 · arc-4: given ?
Use the actual parallel pair in a trapezium
In these examples a trapezium has exactly one pair of opposite parallel sides. If ABCD is in boundary order and AB∥DC, the non-parallel side AD is a transversal, so A+D=180°. The other non-parallel side BC gives B+C=180°. Name the side that acts as the transversal rather than adding whichever angles look adjacent on the page.
A general trapezium does not have to have equal non-parallel sides or equal base angles. An isosceles trapezium adds equal non-parallel sides, equal angles at each base and equal diagonals. A right trapezium has two right angles along one non-parallel side. The extra marks or stated name justify those additional facts; the single parallel pair does not justify them on its own.
Keep a kite’s two diagonal roles distinct
The kites used here have two different-length pairs of adjacent equal sides. In kite PQRS with PQ=PS and RQ=RS, the equal pairs meet at P and R. PR is the symmetry diagonal. The opposite angles at Q and S, lying between unequal side lengths, are equal. The angles at P and R do not have to be equal.
The symmetry diagonal PR meets QS at right angles and bisects QS; it also bisects the angles at P and R. If they meet at T, then QT=TS, but PT=TR is not a general kite property. A diagram with PT=7 cm and TR=3 cm makes the distinction visible: both pieces lie on the same symmetry diagonal, yet need not be equal. Here the symmetry line explains the specific equalities in the kite; a complete line- and rotational-symmetry course is separate.
Check whether all the givens can hold
For a valid marked shape every stated property must fit, not just the total of its interior angles. A four-angle list can add to 360° while contradicting a promised parallel pair. Test the co-interior sums and any extra equality marks before attempting to construct the advertised figure.
If data contradict one another, explain the contradiction on a labelled condition card. Do not draw a distorted “solution” and attach impossible marks to it. For a valid case, sketch the true parallel pair, equal-side pairs and diagonal intersection, then attach each deduction to a named relationship. Use exact stated data rather than measuring the picture; the local naming conventions here do not settle every textbook’s inclusive taxonomy.
Sources for this lesson
Pause and explain
Trapezium ABCD has AB∥DC,A=76°,D=104°,B=83°. Which conclusion is correct?
Worked example
Use two separate labelled panels. Panel 1: right trapezium ABCD has AB∥DC,A=D=90°,B=64°. Mark the parallel pair and find C. Panel 2: kite EFGH has EF=EH and FG=GH, with those pair lengths different; E=76°,G=104°. Mark the two adjacent equal-side pairs and find F,H. Keep each angle arc inside its named polygon. In the kite draw its symmetry diagonal EG, mark its equal angle parts at E and G, and calculate those four parts.
Show the worked solution
- In panel 1, BC is a transversal between AB and DC. Its co-interior pair B+C totals 180°, so C=180−64=116°.
- The panel 1 interiors check as 90+64+116+90=360°. The two right angles belong at A and D; no isosceles condition was supplied.
- In panel 2, the equal opposite angles are F and H, because the equal adjacent pairs meet at E and G. They share 360−76−104=180°, so F=H=90°. Check 76+90+104+90=360°.
- The kite’s equal adjacent pairs meet at E and G, so EG is its symmetry diagonal and bisects those endpoint angles. Each E part is 76÷2=38°, and each G part is 104÷2=52°. Put the two 38° arcs at E and two 52° arcs at G; these are not the meeting angles of the diagonals.
Answer Panel 1: C=116° by co-interior angles on BC. Panel 2: F=H=90° because those equal opposite kite angles share the remaining 180°. Both full interior totals are 360°. The symmetry diagonal EG bisects E into 38°+38° and G into 52°+52°.
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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.
Panel 1 is isosceles trapezium ABCD with AB∥DC,AD=BC,B=(3x+4)°,C=(5x+16)°. Draw the parallel and equal-leg marks; find x and all four interiors. Panel 2 is a separate kite PQRS with PQ=PS,RQ=RS. Draw PR and QS meeting at T, with PT=7 cm,TR=3 cm,QT=4 cm. Find TS,PR,QS and the meeting angle, and explain why T is not the midpoint of PR. In the isosceles trapezium draw AC and BD: state whether AC=BD and justify the claim from the isosceles condition. Would the same conclusion follow for every general trapezium?
Give me a hint
Name each co-interior or equal-opposite pair before writing an equation; keep both diagrams separate.
Compare my reasoning
- In panel 1, B+C=180° gives (3x+4)+(5x+16)=180. Hence 8x+20=180 and x=20, so B=64° and C=116°.
- The isosceles condition gives equal base-angle pairs: A=B=64° and D=C=116°. These four angles total 360°.
- In panel 2, PR is the symmetry diagonal, so it bisects QS at right angles. Therefore TS=QT=4 cm, QS=8 cm and PR=PT+TR=10 cm.
- The meeting angle is 90°, but PT=7 cm differs from TR=3 cm. Thus T is the midpoint of QS, not of PR; kite diagonals do not generally bisect each other.
- The equal legs AD=BC establish the isosceles trapezium property that its diagonals are equal, so AC=BD. No numerical diagonal length was supplied. A general trapezium lacks that extra condition, so equal diagonals do not follow automatically from one parallel pair.
x=20; trapezium angles A,B,C,D=64°,64°,116°,116°. In the kite, TS=4 cm,PR=10 cm,QS=8 cm and the diagonals meet at 90°. T does not bisect PR because 7 cm≠3 cm. In the isosceles trapezium AC=BD; equality is not automatic for a general trapezium and no numerical diagonal length is determined.
Look for these in your work
- Show the trapezium’s actual parallel/equal-leg marks and solve its co-interior equation.
- Use equal base angles to complete all four labels.
- Explain the kite’s one-way diagonal bisection and justify AC=BD specifically from the isosceles trapezium condition, without inventing a length.
Panel 1: general trapezium ABCD has AB∥DC,A=(2x+18)°,D=(4x+6)°,B=74°,C=106°. Find x and all four angles; decide whether it is isosceles. Panel 2: kite PQRS has equal adjacent pairs meeting at P,R, with P=(4u+12)°,R=(2u+18)°,Q=S=(3u+45)°. Draw the named angle arcs and equal-side marks; find u and all four kite interiors. Draw symmetry diagonal PR in the kite and calculate its two endpoint angle parts at P and at R; mark each equal pair.
Give me a hint
Use the parallel pair for x and then the isosceles condition; in the kite ask which diagonal is the symmetry line.
Compare my reasoning
- For panel 1, A+D=180° gives (2x+18)+(4x+6)=180, hence 6x+24=180 and x=26. Therefore A=70° and D=110°.
- Together with B=74°,C=106°, both co-interior pairs total 180° and the four interiors total 360°. A and B are unequal, so the trapezium is not isosceles.
- For panel 2, add the four interiors: (4u+12)+(2u+18)+2(3u+45)=360. Thus 12u+120=360 and u=20.
- The kite angles are P=92°,Q=105°,R=58°,S=105°. The equal pair is Q,S, not P,R; all four total 360°.
- PR bisects the angles at the vertices where the equal adjacent pairs meet. Hence each P part is 92÷2=46° and each R part is 58÷2=29°. This local symmetry-diagonal use is distinct from assuming P=R or that both diagonals are bisected.
x=26; trapezium angles A,B,C,D=70°,74°,106°,110°, so it is not isosceles. u=20; kite angles P,Q,R,S=92°,105°,58°,105°, with Q=S the equal opposite pair. PR divides P into 46°+46° and R into 29°+29°.
Look for these in your work
- Solve and check both x and u equations.
- Label all four angles in each separate diagram.
- Identify Q=S and mark the PR endpoint bisections as 46° at P and 29° at R, while explaining why the trapezium is not isosceles.
A supplier’s condition card claims trapezium ABCD has AB∥DC and interiors A=85°,B=95°,C=105°,D=75°. Check whether such a marked shape can exist; do not draw a supposedly valid outline before checking. Separately draw a valid kite PQRS with PQ=PS,RQ=RS and different pair lengths, P=110°,R=70°. Find Q,S and state which diagonal is bisected, with a reason. Mark the equal parts made by symmetry diagonal PR at P and R, and calculate them.
Give me a hint
A total of 360° is necessary but not sufficient for a marked trapezium. Check each non-parallel transversal as well.
Compare my reasoning
- The proposed trapezium angles total 85+95+105+75=360°. That passes only the general quadrilateral check.
- AB∥DC also requires A+D=180° and B+C=180°. The proposed sums are 160° and 200°, which contradict both co-interior requirements. Reject the advertised marked trapezium.
- In the separate valid kite, the equal opposite pair Q,S shares 360−110−70=180°, so Q=S=90°.
- The equal adjacent pairs meet at P,R, so PR is the symmetry diagonal and bisects QS at right angles. PR itself need not be bisected. Keep this valid diagram separate from the rejected condition card.
- PR bisects its endpoint angles, giving 110÷2=55° at each P part and 70÷2=35° at each R part. Those vertex-angle bisections are separate from PR bisecting QS and do not imply PR itself is bisected.
The advertised trapezium is impossible: its total is 360°, but the required co-interior sums are 160° and 200° instead of 180°. In the valid kite Q=S=90°; PR bisects QS perpendicularly, while PR is not generally bisected. PR divides P into 55°+55° and R into 35°+35°.
Look for these in your work
- Check all advertised parallel-angle conditions and state the contradiction.
- Keep the impossible condition card separate from the valid kite diagram.
- State which diagonal is bisected, and justify the endpoint half-angles 55° and 35° with correctly placed arcs.
Common mistakes
- Applying equal base angles or equal diagonals to every trapezium.
- Assuming both kite diagonals are bisected or choosing the wrong equal opposite-angle pair.
- Accepting impossible marked data merely because the four angles add to 360°.
What can you explain now?
Kite ABCD has AB=AD and CB=CD, with the two adjacent-pair lengths different. Its diagonals meet at O. Which diagonal is bisected, and is AO=OC guaranteed?
Compare with the explanation
AC bisects BD, so BO=OD. AO=OC is not guaranteed.
The equal pairs meet at A,C, making AC the symmetry diagonal. It perpendicularly bisects the other diagonal BD; the reverse bisection is not a general kite property.
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Make it stick
Sketch a general trapezium and an isosceles trapezium with their justified marks. Then sketch a kite and explain its equal opposite angles and the different roles of its two diagonals. Describe why an angle-total check alone can miss impossible data.
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.