[{"data":1,"prerenderedAt":992},["ShallowReactive",2],{"layer:shape-and-space:deepen":3},{"layer":4,"contentHash":968,"dependencyHashes":969,"approval":985,"releaseId":991},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":963,"reviewStatus":964,"authoring":965},1,"shape-and-space","en","deepen","Why shapes behave as they do","Proofs, edge cases and history: diagonals, angle sums, inequality, Euler and symmetry","Turn patterns into proofs: the diagonal formula, why angles add to 180° and (n − 2) × 180°, the triangle inequality, quadrilateral inheritance, why wheels are round, a sketch proof of Euler’s formula and where it fails, cube-net rules, symmetry orders, and the history of π.",[13,14,15,16,17],"Prove the diagonal formula n(n − 3) ÷ 2 in two ways and use it backwards.","Prove the triangle angle sum and use exterior angles to find regular polygons.","Apply the triangle inequality to find possible side lengths, and reason about the quadrilateral family.","Explain why Euler’s formula holds for polyhedra, where it fails, and why no polyhedron has 7 edges.","Connect line and rotational symmetry, and describe key moments in the history of geometry and π.",50,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Go deeper",{"label":26,"value":27},"Reading time","≈ 50 minutes",{"label":29,"value":30},"Prior knowledge","Understand; parallel lines and angles",{"label":32,"value":33},"Chapters","10",{"label":35,"value":36},"Labs","Angle match, regular-polygon and Euler sorts, F\u002FE\u002FV count",{"label":38,"value":39},"Big ideas","Proof, counting twice, invariants",[41,45,51,57,60,73,116,127,142,147,153,158,186,191,194,214,223,261,320,325,328,338,348,353,365,370,373,396,400,425,430,433,437,442,483,488,491,514,519,523,532,542,606,625,630,633,642,647,652,655,686,690,695,705,710,715,751,755,796,919,923,940,944,948],{"id":42,"type":43,"markdown":44},"intro-deepen","prose","In Investigate you **discovered** patterns by trying examples: diagonals grow 2, 3, 4, 5 …; triangle angles always make 180°; F + V − E came out as 2 every time. But a few examples, however many, never **prove** that something is always true. The 1,000th case might break the rule.\n\nIn this layer you will find out **why** the patterns must hold, using arguments that work for every case at once: counting in two ways, cutting shapes into triangles, walking around a polygon, and squashing a solid flat. You will also meet the edge cases where the rules bend or break, and the long history, much of it Indian, of how people came to understand shape.",{"id":46,"type":47,"variant":48,"title":49,"markdown":50},"intro-proof","callout","definition","What counts as a proof?","A **proof** is a chain of reasons, each following from definitions or from facts already proved, that shows a statement is true in **every** case. Checking 10 triangles is evidence. Showing that *any* triangle's angles can be rearranged onto a straight line, using properties of parallel lines, is a proof.",{"id":52,"type":53,"title":54,"eyebrow":55,"navLabel":56},"ch01","chapter","Counting diagonals: a formula and its proof","Chapter 01","1 Diagonal formula",{"id":58,"type":43,"markdown":59},"diag-proof","**Claim:** a polygon with n sides has **n(n − 3) ÷ 2** diagonals.\n\n**Proof by counting ends.** Each of the n vertices can be joined to the other n − 1 vertices. Two of those joins are sides (to its two neighbours), so each vertex is the end of **n − 3** diagonals. Counting from every vertex gives n(n − 3) diagonal-ends. Every diagonal has exactly **two** ends, so it has been counted twice. Therefore the number of diagonals is n(n − 3) ÷ 2. ∎\n\n**A second proof, by handshakes.** Joining every pair of vertices is like everyone at a party shaking hands with everyone else once: n people make n(n − 1) ÷ 2 handshakes. Among all these segments, exactly n are sides. The rest are diagonals: n(n − 1) ÷ 2 − n, which simplifies to n(n − 3) ÷ 2. Two different arguments reaching the same formula is strong confirmation.",{"id":61,"type":62,"items":63},"formulas-diag","formulas",[64,67,70],{"expression":65,"caption":66},"diagonals = n(n − 3) ÷ 2","n vertices, n − 3 diagonals from each, every diagonal counted twice.",{"expression":68,"caption":69},"segments = n(n − 1) ÷ 2","All joins between n vertices (the handshake count).",{"expression":71,"caption":72},"diagonals = segments − n","Remove the n sides from all the joins.",{"id":74,"type":75,"caption":76,"columns":77,"rows":82},"table-diag-big","table","Diagonals of larger polygons from the formula",[78,79,80,81],"Sides n","n − 3","n(n − 3)","Diagonals",[83,87,92,96,101,106,111],[33,84,85,86],"7","70","35",[88,89,90,91],"12","9","108","54",[93,88,94,95],"15","180","90",[97,98,99,100],"16","13","208","104",[102,103,104,105],"20","17","340","170",[107,108,109,110],"50","47","2350","1175",[112,113,114,115],"100","97","9700","4850",{"id":117,"type":118,"title":119,"problem":120,"steps":121},"we-reverse-diag","worked_example","Working backwards: which polygon has 35 diagonals?","A polygon has 35 diagonals. How many sides does it have? Could a polygon have exactly 100 diagonals?",[122,123,124,125,126],"We need n(n − 3) ÷ 2 = 35, so n(n − 3) = 70.","Look for two whole numbers that differ by 3 and multiply to 70: 10 × 7 = 70. So **n = 10**: a decagon.","For 100 diagonals we would need n(n − 3) = 200.","Try nearby values: 15 × 12 = 180 (90 diagonals) and 16 × 13 = 208 (104 diagonals).","200 falls between them, and n(n − 3) only grows as n grows, so no whole number works. **No polygon has exactly 100 diagonals.**",{"id":128,"type":129,"itemId":130,"prompt":131,"check":132,"hints":136,"feedback":139},"pr-54-diag","practice","shape-and-space.deepen-54-diagonals","A polygon has 54 diagonals. How many sides does it have?",{"kind":133,"answer":134,"tolerance":135},"number",12,0,[137,138],"Solve n(n − 3) = 108.","Look for two numbers that differ by 3 and multiply to 108.",{"correct":140,"incorrect":141},"Yes: 12 × 9 = 108, so n = 12, a dodecagon.","n(n − 3) = 2 × 54 = 108 = 12 × 9, so n = 12.",{"id":143,"type":47,"variant":144,"title":145,"markdown":146},"nuance-concave-diag","nuance","Diagonals of concave polygons","The formula counts every segment joining two non-adjacent vertices, whether it lies inside the polygon or not. In a **concave** polygon (one with a dent), some diagonals pass **outside** the shape. The star-shaped decagon still has 35 diagonals by the formula, but many of them run outside it. In a convex polygon every diagonal lies inside.",{"id":148,"type":149,"conceptId":150,"relation":151,"explanation":152},"conn-patterns-deep","connection","patterns","related_to","The diagonal numbers 0, 2, 5, 9, 14, … grow by 2, 3, 4, 5, …, a pattern closely related to the triangular numbers.",{"id":154,"type":53,"title":155,"eyebrow":156,"navLabel":157},"ch02","Why angles add up the way they do","Chapter 02","2 Angle proofs",{"id":159,"type":160,"title":161,"items":162},"steps-180-proof","steps","Proof that the angles of a triangle add to 180°",[163,167,171,175,178,182],{"title":164,"tag":165,"text":166},"Draw","triangle ABC","Take any triangle ABC with angles a at A, b at B and c at C.",{"title":168,"tag":169,"text":170},"Add a parallel line","through A","Through A draw the line parallel to BC.",{"title":172,"tag":173,"text":174},"Alternate angles","left of A","The angle between this line and AB equals b (alternate angles between parallel lines).",{"title":172,"tag":176,"text":177},"right of A","The angle between the line and AC equals c, for the same reason.",{"title":179,"tag":180,"text":181},"Straight line","at A","Along the line at A we now have angles b, a and c side by side, making a straight angle.",{"title":183,"tag":184,"text":185},"Conclude","a + b + c = 180°","So the three angles of the triangle add to 180°, for every triangle. ∎",{"id":187,"type":149,"conceptId":188,"relation":189,"explanation":190},"conn-angles-deep","angles","helps_understand","This proof rests on alternate angles formed when a transversal cuts parallel lines, and on angles on a straight line adding to 180°, both from the Angles topic.",{"id":192,"type":43,"markdown":193},"exterior-angle","**Exterior angle property.** Extend one side of a triangle past a vertex. The angle between the extension and the next side is an **exterior angle**. It equals the **sum of the two interior angles opposite it**. Why? The exterior angle and the interior angle next to it together make a straight line (180°). The three interior angles also make 180°. Take away the shared interior angle from both, and what is left must match: exterior angle = the other two interior angles.\n\n**Angle sum of any polygon.** From one vertex of an n-sided (convex) polygon, draw all n − 3 diagonals. They cut it into **n − 2** triangles, whose angles together make up exactly the polygon's angles. So the angle sum is **(n − 2) × 180°**.\n\n**Exterior angles always total 360°.** Walk round a convex polygon. At each vertex you turn through the exterior angle. By the time you are back where you started, facing the same way, you have turned through one full turn: **360°**, whatever the number of sides.",{"id":195,"type":62,"items":196},"formulas-angles",[197,199,202,205,208,211],{"expression":184,"caption":198},"The angles of any triangle.",{"expression":200,"caption":201},"exterior = sum of opposite interiors","The exterior angle property of a triangle.",{"expression":203,"caption":204},"angle sum = (n − 2) × 180°","Interior angles of any n-sided polygon.",{"expression":206,"caption":207},"sum of exterior angles = 360°","For any convex polygon: one full turn.",{"expression":209,"caption":210},"regular exterior = 360° ÷ n","A regular polygon shares the 360° equally.",{"expression":212,"caption":213},"regular interior = 180° − 360° ÷ n","Interior and exterior make a straight line.",{"id":215,"type":118,"title":216,"problem":217,"steps":218},"we-regular-from-angle","Which regular polygon has interior angles of 140°?","Each interior angle of a regular polygon is 140°. How many sides does it have? Is there a regular polygon with interior angles of 100°?",[219,220,221,222],"Interior + exterior = 180°, so each exterior angle is 180° − 140° = 40°.","The exterior angles total 360°, so the number of sides is 360° ÷ 40° = **9**: a regular nonagon.","Check: (9 − 2) × 180° = 1,260°, and 1,260° ÷ 9 = 140°. ✓","For 100°: the exterior angle would be 80°, and 360 ÷ 80 = 4.5, not a whole number. **No regular polygon has 100° angles.**",{"id":224,"type":225,"component":226,"componentVersion":5,"config":227,"objective":255,"textAlternative":256,"help":257},"lab-match-regular","interactive","match-pairs",{"prompt":228,"mode":229,"pairs":230},"Match each regular polygon with its interior angle.","connect",[231,234,237,240,243,246,249,252],{"a":232,"b":233},"Equilateral triangle","60°",{"a":235,"b":236},"Square","90°",{"a":238,"b":239},"Regular pentagon","108°",{"a":241,"b":242},"Regular hexagon","120°",{"a":244,"b":245},"Regular octagon","135°",{"a":247,"b":248},"Regular nonagon","140°",{"a":250,"b":251},"Regular decagon","144°",{"a":253,"b":254},"Regular dodecagon (12 sides)","150°","Connect each regular polygon to the size of each of its interior angles, using 180° − 360° ÷ n.","Eight regular polygons on the left and eight angles on the right. Using interior angle = 180° − 360° ÷ n: triangle 180 − 120 = 60°; square 180 − 90 = 90°; pentagon 180 − 72 = 108°; hexagon 180 − 60 = 120°; octagon 180 − 45 = 135°; nonagon 180 − 40 = 140°; decagon 180 − 36 = 144°; dodecagon 180 − 30 = 150°.\n\nAs n grows, the angles creep towards 180° but never reach it: the polygon looks more and more like a circle.",{"hints":258},[259,260],"Work out the exterior angle first: 360° ÷ n.","Then subtract it from 180°.",{"id":262,"type":225,"component":263,"componentVersion":5,"config":264,"objective":314,"textAlternative":315,"help":316},"lab-sort-regular-exists","sort-game",{"prompt":265,"bins":266,"items":273,"seconds":135},"Could a regular polygon have interior angles of this size?",[267,270],{"id":268,"label":269},"yes","Yes: such a polygon exists",{"id":271,"label":272},"no","No: impossible",[274,278,282,286,290,294,298,302,306,310],{"id":275,"label":276,"bin":268,"why":277},"r156","156°","Exterior 24°, and 360 ÷ 24 = 15: a regular 15-gon.",{"id":279,"label":280,"bin":271,"why":281},"r100","100°","Exterior 80°, and 360 ÷ 80 = 4.5, not a whole number.",{"id":283,"label":284,"bin":268,"why":285},"r160","160°","Exterior 20°, 360 ÷ 20 = 18 sides.",{"id":287,"label":288,"bin":271,"why":289},"r130","130°","Exterior 50°, 360 ÷ 50 = 7.2.",{"id":291,"label":292,"bin":268,"why":293},"r165","165°","Exterior 15°, 360 ÷ 15 = 24 sides.",{"id":295,"label":296,"bin":268,"why":297},"r170","170°","Exterior 10°, 360 ÷ 10 = 36 sides.",{"id":299,"label":300,"bin":271,"why":301},"r180","180°","Exterior 0°: you would never turn, so you could never get back to the start.",{"id":303,"label":304,"bin":271,"why":305},"r45","45°","The smallest possible interior angle of a regular polygon is 60° (the triangle).",{"id":307,"label":308,"bin":268,"why":309},"r175","175°","Exterior 5°, 360 ÷ 5 = 72 sides.",{"id":311,"label":312,"bin":271,"why":313},"r125","125°","Exterior 55°, 360 ÷ 55 ≈ 6.5.","Use the exterior angle to decide whether a regular polygon with a given interior angle can exist.","Ten angle cards. For each, find the exterior angle (180° minus the interior angle) and check whether 360° divided by it is a whole number of at least 3.\n\n**Exists:** 156° (15 sides), 160° (18 sides), 165° (24 sides), 170° (36 sides), 175° (72 sides).\n\n**Impossible:** 100° (would need 4.5 sides), 130° (7.2), 125° (about 6.5), 180° (exterior angle 0°: the walk never turns), 45° (smaller than 60°, the angle of the equilateral triangle, which is the smallest possible).",{"hints":317},[318,319],"Exterior angle = 180° − interior angle.","The number of sides is 360° ÷ exterior angle, and must be a whole number of at least 3.",{"id":321,"type":53,"title":322,"eyebrow":323,"navLabel":324},"ch03","The triangle inequality","Chapter 03","3 Triangle inequality",{"id":326,"type":43,"markdown":327},"ti-why","**Claim:** in any triangle, each side is shorter than the other two added together.\n\n**Why:** the side BC is a straight path from B to C. Going from B to A and then on to C is a detour. The straight path between two points is the shortest of all paths, so BC is shorter than BA + AC. The same holds for every side. ∎\n\nThis gives a precise rule for the **third side**: if two sides are a and b (with a ≥ b), the third side must be **longer than a − b and shorter than a + b**. Too short and the two short sides cannot reach; too long and the other two cannot stretch across it.",{"id":329,"type":118,"title":330,"problem":331,"steps":332},"we-third-side","How many whole-number third sides?","Two sides of a triangle are 7 cm and 12 cm. What lengths can the third side have? How many whole-number lengths are possible?",[333,334,335,336,337],"The third side must be more than 12 − 7 = 5 cm.","It must be less than 12 + 7 = 19 cm.","So the third side is between 5 cm and 19 cm, not equal to either. In whole centimetres: 6, 7, 8, …, 18.","Count: 18 − 6 + 1 = **13 possible whole-number lengths**.","Check the ends: 5, 7, 12 fails (5 + 7 = 12, flat); 19, 7, 12 fails (7 + 12 = 19, flat).",{"id":339,"type":118,"title":340,"problem":341,"steps":342},"we-perimeter-12","All triangles with whole-number sides and perimeter 12","How many different triangles have whole-number sides (in cm) and a perimeter of 12 cm?",[343,344,345,346,347],"List the sides in order a ≤ b ≤ c with a + b + c = 12, and keep only those with a + b > c.","Because a + b > c and a + b + c = 12, we need c \u003C 6. And c is the largest, so c ≥ 4.","c = 5: a + b = 7 with a ≤ b ≤ 5 gives (2, 5) and (3, 4). Triangles **2, 5, 5** and **3, 4, 5**.","c = 4: a + b = 8 with a ≤ b ≤ 4 gives only (4, 4). Triangle **4, 4, 4**.","Total: **3 triangles**: one isosceles, one scalene (the famous 3-4-5 right triangle) and one equilateral.",{"id":349,"type":47,"variant":350,"title":351,"markdown":352},"aha-biggest-side","aha","Biggest side faces biggest angle","In any triangle, the **longest side is opposite the largest angle**, and the shortest side opposite the smallest angle. Open a pair of compasses wider (a bigger angle) and the gap between the points (the opposite side) grows. So in a right-angled triangle the side opposite the 90° angle, the **hypotenuse**, is always the longest side.",{"id":354,"type":129,"itemId":355,"prompt":356,"check":357,"hints":359,"feedback":362},"pr-third-side","shape-and-space.deepen-third-side","Two sides of a triangle are 9 cm and 4 cm. The third side is a whole number of centimetres. What is the **largest** it can be?",{"kind":133,"answer":134,"tolerance":135,"unit":358},"cm",[360,361],"The third side must be less than 9 + 4.","It cannot equal 13: that would be flat.",{"correct":363,"incorrect":364},"Yes: it must be less than 13, so the largest whole number is 12.","The third side must be less than 9 + 4 = 13 cm (13 would make a flat triangle). So the largest whole-number length is 12 cm.",{"id":366,"type":53,"title":367,"eyebrow":368,"navLabel":369},"ch04","Reasoning with the quadrilateral family","Chapter 04","4 Quadrilateral logic",{"id":371,"type":43,"markdown":372},"quad-logic","Why do mathematicians insist that a square **is** a rectangle? Because of **inheritance**. Anything proved for all rectangles, such as *the diagonals are equal*, then automatically holds for squares, with no extra work. If squares were excluded from rectangles, every theorem would need a separate square version.\n\nHere are three short arguments that show how the family's properties follow from the definitions:\n\n1. **Parallelogram: neighbouring angles add to 180°.** Two neighbouring angles sit between a pair of parallel sides, cut by the side joining them. They are co-interior angles, which add to 180°. It follows that **opposite angles are equal** (each is 180° minus the same neighbour).\n2. **Rectangle: one right angle forces four.** A rectangle is a parallelogram with one right angle. Its neighbours are 180° − 90° = 90°, and so on round the shape.\n3. **Rhombus: diagonals are perpendicular.** In a rhombus ABCD, B and D are each the same distance from A and from C (all sides equal). Points equally far from A and C lie on the **perpendicular bisector** of AC. So BD is the perpendicular bisector of AC: the diagonals cross at right angles.",{"id":374,"type":75,"caption":375,"columns":376,"rows":381},"table-diag-key","A diagonal key: identify a quadrilateral from its diagonals alone",[377,378,379,380],"Diagonals bisect each other?","Diagonals equal?","Diagonals perpendicular?","Shape",[382,386,388,390,391,394],[383,384,384,385],"Yes","No","Parallelogram",[383,383,384,387],"Rectangle",[383,384,383,389],"Rhombus",[383,383,383,235],[392,384,383,393],"Only one bisects the other","Kite",[384,383,384,395],"Could be an isosceles trapezium",{"id":397,"type":47,"variant":144,"title":398,"markdown":399},"nuance-venn","Square = rectangle ∩ rhombus","Picture two overlapping circles, one holding all rectangles and one holding all rhombuses. The overlap holds shapes that are **both**: four right angles and four equal sides. Those are exactly the squares. Both circles sit inside a bigger one, the parallelograms, which sits inside the trapeziums (under the inclusive definition), inside all quadrilaterals. Kites form a separate circle that overlaps the parallelograms in exactly the rhombuses.",{"id":401,"type":129,"itemId":402,"prompt":403,"check":404,"hints":420,"feedback":422},"pr-quad-proof","shape-and-space.deepen-quad-reason","A parallelogram has one angle of 70°. What are its other three angles?",{"kind":405,"options":406,"correct":419},"choice",[407,410,413,416],{"id":408,"label":409},"a","70°, 70°, 70°",{"id":411,"label":412},"b","110°, 70°, 110°",{"id":414,"label":415},"c","110°, 110°, 70°",{"id":417,"label":418},"d","90°, 90°, 110°",[411],[421],"Neighbouring angles of a parallelogram add to 180°.",{"correct":423,"incorrect":424},"Right. Going round: 70°, 110°, 70°, 110°. Opposite angles are equal; neighbours add to 180°.","Neighbours add to 180°, so the angles next to 70° are 110°. The angle opposite 70° is 70° again. In order round the shape: 70°, 110°, 70°, 110°.",{"id":426,"type":53,"title":427,"eyebrow":428,"navLabel":429},"ch05","Circles: why wheels are round","Chapter 05","5 Why wheels are round",{"id":431,"type":43,"markdown":432},"wheels","A wheel works because the axle is at the **centre**, always exactly one radius above the road. Two other facts make circles special:\n\n**Constant width.** Measure a circle's width in any direction and you get the diameter. That is why a round manhole cover cannot fall into its hole whichever way you turn it: a square cover can, if you tip it so its side slips down the longer diagonal of the hole.\n\n**Finding the centre from a chord.** The perpendicular bisector of any chord passes through the centre. Archaeologists use this to find the size of a whole plate or pot from one broken piece: draw two chords on the rim fragment, construct their perpendicular bisectors, and where they cross is the centre.",{"id":434,"type":47,"variant":350,"title":435,"markdown":436},"aha-reuleaux","Round is not the only constant width","Draw an equilateral triangle, then, with the compass point on each vertex, draw an arc joining the other two. The curved triangle you get, a **Reuleaux triangle**, has the same width in every direction, just like a circle. It rolls smoothly under a flat board! But it cannot be a wheel on an axle, because its centre bobs up and down as it rolls. Special drills based on it can cut almost-square holes, and some coins around the world use Reuleaux-like shapes with 7 curved sides so that vending machines can measure them in any direction.",{"id":438,"type":47,"variant":439,"title":440,"markdown":441},"misc-circle-polygon","misconception","“A circle is a polygon with infinitely many sides”","It is a helpful picture, a regular polygon with more and more sides gets closer and closer to a circle, and Archimedes used exactly this idea to trap π. But a circle is **not** a polygon: it has no straight sides and no vertices at all. The polygons approach the circle, but never become it.",{"id":443,"type":444,"title":445,"items":446},"timeline-pi","timeline","Pinning down π",[447,451,455,459,463,467,471,475,479],{"time":448,"title":449,"text":450},"c. 1900 BCE","Babylon","A clay tablet uses a value equivalent to 3⅛ = 3.125.",{"time":452,"title":453,"text":454},"c. 1650 BCE","Egypt","The Rhind papyrus uses a rule equivalent to π ≈ 256\u002F81 ≈ 3.16.",{"time":456,"title":457,"text":458},"800–500 BCE","Sulba Sutras","Indian altar-builders’ manuals give rules for turning a square into a circle of equal area. Different rules imply different values of π, spread from about 3.0 to about 3.2; one well-known construction gives 3.088.",{"time":460,"title":461,"text":462},"c. 250 BCE","Archimedes","Trapping the circle between polygons of 96 sides, he shows 223\u002F71 \u003C π \u003C 22\u002F7 (between 3.1408 and 3.1429).",{"time":464,"title":465,"text":466},"499 CE","Aryabhata","Gives circumference 62,832 for diameter 20,000: π ≈ 3.1416, and calls it an approximation.",{"time":468,"title":469,"text":470},"c. 480 CE","Zu Chongzhi","In China, finds π ≈ 355\u002F113 = 3.1415929…, correct to 6 decimal places.",{"time":472,"title":473,"text":474},"c. 1400","Madhava","In Kerala, Madhava of Sangamagrama uses an infinite series; its first 21 terms give π correct to 11 decimal places.",{"time":476,"title":477,"text":478},"1706","The symbol π","William Jones first uses the Greek letter π; Euler makes it popular.",{"time":480,"title":481,"text":482},"Today","Trillions of digits","Computers have calculated π to more than 100 trillion digits. It never repeats.",{"id":484,"type":53,"title":485,"eyebrow":486,"navLabel":487},"ch06","Euler’s formula: why F + V − E = 2","Chapter 06","6 Euler’s formula",{"id":489,"type":43,"markdown":490},"euler-prism","**For every prism and pyramid.** We can check Euler's formula for whole families at once using algebra.\n\n- A prism with an n-sided base has F = n + 2, V = 2n, E = 3n. Then F + V − E = (n + 2) + 2n − 3n = **2**, whatever n is.\n- A pyramid with an n-sided base has F = n + 1, V = n + 1, E = 2n. Then F + V − E = (n + 1) + (n + 1) − 2n = **2**.\n\nThat proves the formula for infinitely many solids in two lines. But there are polyhedra that are neither prisms nor pyramids. For those, we need a cleverer argument.",{"id":492,"type":160,"title":493,"items":494},"steps-euler-proof","Why Euler’s formula holds for any (simple) polyhedron: a sketch",[495,499,503,506,510],{"title":496,"tag":497,"text":498},"Remove one face","F − 1","Imagine the polyhedron made of rubber. Cut out one face and stretch the rest flat, like a map. We now need V − E + F = 1 for this flat network.",{"title":500,"tag":501,"text":502},"Cut faces into triangles","no change","Draw diagonals to split every face into triangles. Each new edge adds one face too, so V − E + F does not change.",{"title":504,"tag":501,"text":505},"Peel triangles off","Remove boundary triangles one at a time. Losing one outside edge removes 1 edge and 1 face. Losing two outside edges removes 2 edges, 1 vertex and 1 face. Either way V − E + F stays the same.",{"title":507,"tag":508,"text":509},"One triangle left","3 − 3 + 1 = 1","At the end a single triangle remains: V − E + F = 3 − 3 + 1 = 1.",{"title":511,"tag":512,"text":513},"Put the face back","1 + 1 = 2","So the flat network had V − E + F = 1. Adding back the face we removed gives F + V − E = 2. ∎",{"id":515,"type":47,"variant":516,"title":517,"markdown":518},"model-limit-euler","model_limit","When the formula breaks","Euler's formula is about polyhedra **without holes** (mathematicians say *simple* or *convex* polyhedra). Build a square **picture frame** out of four cuboid bars: it has a hole through the middle. Counting carefully gives 16 faces, 16 vertices and 32 edges, so F + V − E = 16 + 16 − 32 = **0**, not 2. Each hole lowers the total by 2. The number F + V − E is called the **Euler characteristic**, and it is one of the first ideas of *topology*, the study of properties that survive stretching and bending.",{"id":520,"type":47,"variant":144,"title":521,"markdown":522},"nuance-curved-euler","Curved solids are not polyhedra","Try a cylinder, counting its curved surface as a face: F = 3, V = 0, E = 2, so F + V − E = 1. A cone gives 2 + 1 − 1 = 2, a sphere 1 + 0 − 0 = 1. These answers depend on counting conventions and do not follow a rule, because curved surfaces are not polygons. Euler's formula is a theorem about polyhedra only.",{"id":524,"type":118,"title":525,"problem":526,"steps":527},"we-euler-unknown","Using Euler’s formula to find a missing count","A polyhedron has 12 faces, all of them pentagons (a dodecahedron). How many edges and vertices does it have?",[528,529,530,531],"Each pentagon has 5 edges: 12 × 5 = 60 edge-sides.","Every edge is shared by exactly 2 faces, so E = 60 ÷ 2 = **30 edges**.","Euler: F + V − E = 2, so 12 + V − 30 = 2, and V = **20 vertices**.","Check another way: 3 pentagons meet at each vertex, so V = 60 ÷ 3 = 20. ✓",{"id":533,"type":118,"title":534,"problem":535,"steps":536},"we-seven-edges","Can a polyhedron have exactly 7 edges?","Prove that no polyhedron has exactly 7 edges.",[537,538,539,540,541],"Every face has at least 3 edges, and every edge belongs to exactly 2 faces. So 2E ≥ 3F.","Every vertex has at least 3 edges, and every edge has 2 ends. So 2E ≥ 3V.","If E = 7, then 3F ≤ 14 and 3V ≤ 14, so F ≤ 4 and V ≤ 4. That means F + V ≤ 8.","But Euler says F + V = E + 2 = 9.","8 cannot be 9: a contradiction. **No polyhedron has 7 edges.** (Every other whole number of edges from 6 upwards is possible.)",{"id":543,"type":225,"component":263,"componentVersion":5,"config":544,"objective":600,"textAlternative":601,"help":602},"lab-sort-euler",{"prompt":545,"bins":546,"items":551,"seconds":135},"Faces, vertices, edges: could a polyhedron (without holes) have these counts?",[547,549],{"id":268,"label":548},"Possible",{"id":271,"label":550},"Impossible",[552,556,560,564,568,572,576,580,584,588,592,596],{"id":553,"label":554,"bin":268,"why":555},"e1","F 6, V 8, E 12","6 + 8 − 12 = 2. This is a cube or cuboid.",{"id":557,"label":558,"bin":268,"why":559},"e2","F 5, V 6, E 9","5 + 6 − 9 = 2: a triangular prism.",{"id":561,"label":562,"bin":268,"why":563},"e3","F 4, V 4, E 6","4 + 4 − 6 = 2: a tetrahedron (triangular pyramid).",{"id":565,"label":566,"bin":268,"why":567},"e4","F 6, V 6, E 10","6 + 6 − 10 = 2: a pentagonal pyramid.",{"id":569,"label":570,"bin":271,"why":571},"e5","F 5, V 5, E 9","5 + 5 − 9 = 1, not 2. Euler’s formula fails.",{"id":573,"label":574,"bin":271,"why":575},"e6","F 4, V 5, E 7","Euler gives 2, but no polyhedron has 7 edges: 2E ≥ 3F and 2E ≥ 3V would force F + V ≤ 8.",{"id":577,"label":578,"bin":268,"why":579},"e7","F 8, V 12, E 18","8 + 12 − 18 = 2: a hexagonal prism.",{"id":581,"label":582,"bin":268,"why":583},"e8","F 20, V 12, E 30","20 + 12 − 30 = 2: the icosahedron.",{"id":585,"label":586,"bin":271,"why":587},"e9","F 6, V 6, E 11","6 + 6 − 11 = 1, not 2.",{"id":589,"label":590,"bin":268,"why":591},"e10","F 10, V 10, E 18","10 + 10 − 18 = 2: a pyramid on a 9-sided base.",{"id":593,"label":594,"bin":271,"why":595},"e11","F 16, V 16, E 32","16 + 16 − 32 = 0. These are the counts of a picture-frame solid, which has a hole.",{"id":597,"label":598,"bin":271,"why":599},"e12","F 3, V 3, E 4","Euler gives 2, but a solid needs at least 4 faces: three flat faces cannot enclose space.","Use Euler’s formula, and edge-counting limits, to decide which face, vertex and edge counts belong to a real polyhedron.","Twelve cards give F, V and E.\n\n**Possible:** F 6, V 8, E 12 (cube); F 5, V 6, E 9 (triangular prism); F 4, V 4, E 6 (tetrahedron); F 6, V 6, E 10 (pentagonal pyramid); F 8, V 12, E 18 (hexagonal prism); F 20, V 12, E 30 (icosahedron); F 10, V 10, E 18 (pyramid on a 9-gon).\n\n**Impossible:** F 5, V 5, E 9 and F 6, V 6, E 11 (F + V − E = 1); F 16, V 16, E 32 (total 0: a frame with a hole); F 4, V 5, E 7 (satisfies Euler, but no polyhedron has 7 edges); F 3, V 3, E 4 (fewer than 4 faces cannot enclose a solid).\n\nThe lesson: Euler's formula is **necessary** but not **sufficient**. Passing it does not guarantee a real solid exists.",{"hints":603},[604,605],"First test F + V − E = 2.","Then check 2E ≥ 3F and 2E ≥ 3V, and that there are at least 4 faces.",{"id":607,"type":225,"component":608,"componentVersion":5,"config":609,"objective":620,"textAlternative":621,"help":622},"lab-count-deepen","shape-explorer",{"solids":610,"polygons":617,"modes":618},[611,612,613,614,615,616],"triangular-pyramid","square-pyramid","triangular-prism","cube","pentagonal-prism","hexagonal-prism",[],[619],"count","Type F, E and V for each solid shown and confirm the prism rule (n + 2, 3n, 2n), the pyramid rule (n + 1, 2n, n + 1) and Euler’s formula.","A counting game with six polyhedra: each round shows a solid and you type its faces, edges and vertices.\n\nPyramids: triangular (n = 3) 4 faces, 6 edges, 4 vertices; square (n = 4) 5, 8, 5. These fit n + 1, 2n, n + 1.\n\nPrisms: triangular (n = 3) 5, 9, 6; cube (n = 4) 6, 12, 8; pentagonal (n = 5) 7, 15, 10; hexagonal (n = 6) 8, 18, 12. These fit n + 2, 3n, 2n.\n\nIn every case F + V − E = 2. Use the rules to answer before you count, then predict a heptagonal prism (9, 21, 14) or an octagonal pyramid (9, 16, 9).",{"hints":623},[624],"Identify n, the number of sides of the base, first.",{"id":626,"type":53,"title":627,"eyebrow":628,"navLabel":629},"ch07","Nets and dice","Chapter 07","7 Nets and dice",{"id":631,"type":43,"markdown":632},"nets-deep","Why do some six-square shapes fold into a cube and others not? Three reasoned rules settle almost every case:\n\n1. **No 2 by 2 block.** Four squares round one point would all need to meet at one vertex of the cube, but only **three** faces meet at each vertex of a cube. So the fourth overlaps.\n2. **No row of 5.** A row of squares wraps round the cube's middle, and the ring round a cube has only 4 faces, so the 5th lands on the 1st.\n3. **Opposite faces.** In a working net, two squares separated by exactly one square in a straight row (or at the ends of a line of three) become **opposite** faces. Each face must have exactly one opposite.\n\nThe net also tells you the **surface area** of a solid: it is simply the area of the net. A cube of side 5 cm has a net of 6 squares, each 25 cm², so its surface area is 6 × 25 = **150 cm²**.",{"id":634,"type":118,"title":635,"problem":636,"steps":637},"we-dice","Is this a proper dice?","A cross-shaped cube net has a row of four squares numbered 1, 2, 6, 5 from left to right, with 3 above the 2 and 4 below the 2. On a proper dice, opposite faces add up to 7. Is this a proper dice?",[638,639,640,641],"In the row of four, squares 1st and 3rd are opposite, and 2nd and 4th are opposite: 1 is opposite 6, and 2 is opposite 5.","1 + 6 = 7 ✓ and 2 + 5 = 7 ✓.","The squares above and below the row close the two ends of the ring, so they are opposite each other: 3 opposite 4, and 3 + 4 = 7 ✓.","All three pairs add to 7, so **yes, it folds into a proper dice**.",{"id":643,"type":47,"variant":644,"title":645,"markdown":646},"careful-dice-hand","careful","Two different proper dice","Even with every opposite pair adding to 7, there are **two** different ways to arrange the numbers, mirror images of each other. Look at the corner where 1, 2 and 3 meet: going round it, they run either clockwise or anticlockwise. Different makers choose different arrangements (many Western dice run 1, 2, 3 anticlockwise; many Chinese dice run the other way), and both are equally fair.",{"id":648,"type":53,"title":649,"eyebrow":650,"navLabel":651},"ch08","Symmetry: reflections and turns","Chapter 08","8 Symmetry deeper",{"id":653,"type":43,"markdown":654},"sym-deep","**Why a regular n-gon has exactly n lines of symmetry.** Each mirror line must pass through the centre. If n is **odd**, each line runs from a vertex to the midpoint of the opposite side: one line per vertex, n lines. If n is **even**, the lines come in two kinds: n ÷ 2 through pairs of opposite vertices and n ÷ 2 through midpoints of opposite sides. Again n in total.\n\n**Rotational symmetry.** A shape has rotational symmetry if it fits onto itself after a turn of less than 360° about its centre. The number of positions in one full turn where it looks the same is its **order**. A regular n-gon has order **n** (turns of 360° ÷ n). A parallelogram has order 2 but no mirror lines; the letter Z has order 2 but no mirror lines.\n\n**Mirrors make turns.** Reflect a shape in one mirror line and then in a second line that crosses the first at angle θ: the result is a **rotation** through **2θ** about the crossing point. That is why a shape with two mirror lines at right angles (like a rectangle, θ = 90°) always has half-turn (180°) symmetry too.",{"id":656,"type":75,"caption":657,"columns":658,"rows":662},"table-sym-order","Line symmetry and rotational symmetry together",[380,659,660,661],"Lines of symmetry","Order of rotation","Smallest turn",[663,668,671,673,675,676,677,679,680,682],[664,665,666,667],"Scalene triangle","0","1 (none)","360°",[669,670,666,667],"Isosceles triangle","1",[232,672,672,242],"3",[385,665,674,300],"2",[387,674,674,300],[389,674,674,300],[235,678,678,236],"4",[393,670,666,667],[241,681,681,233],"6",[683,684,684,685],"Ashoka Chakra (24 spokes, ignoring detail)","24","15°",{"id":687,"type":47,"variant":144,"title":688,"markdown":689},"nuance-order-one","Order 1 means no rotational symmetry","Every shape looks the same after a full 360° turn, so every shape trivially has rotational order at least 1. We say a shape *has rotational symmetry* only if its order is 2 or more.",{"id":691,"type":53,"title":692,"eyebrow":693,"navLabel":694},"ch09","Views: minimum and maximum","Chapter 09","9 Views puzzles",{"id":696,"type":118,"title":697,"problem":698,"steps":699},"we-min-max","How many cubes could there be?","A stack of cubes has a top view of 4 squares in a 2 by 2 block. The front view shows two columns: the left one 2 cubes tall, the right one 1 cube tall. What are the fewest and the most cubes it could have?",[700,701,702,703,704],"The top view says there are 4 columns: left-front, left-back, right-front, right-back. Each has at least 1 cube.","The front view shows the **tallest** column in each left-right position. On the left the tallest is 2; on the right the tallest is 1.","Right side: both columns must be exactly 1 cube: 2 cubes.","Left side: at least one column is 2 tall; the other can be 1 or 2. Fewest: 2 + 1 = 3. Most: 2 + 2 = 4.","Total: fewest **5**, most **6**. A side view would settle which.",{"id":706,"type":47,"variant":707,"title":708,"markdown":709},"example-maps","example","Views in maps and buildings","A town map is a top view drawn to **scale**: 1 cm on the map might stand for 100 m on the ground. Architects' drawings for a temple or a house include the **plan** (top view of each floor), **elevations** (front, back and side views) and **sections** (views of an imaginary slice through the building). Survey of India topographic maps add **contour lines**, lines joining points at the same height, so that a flat map can also show the shape of hills and valleys.",{"id":711,"type":53,"title":712,"eyebrow":713,"navLabel":714},"ch10","A short history of shape","Chapter 10","10 History",{"id":716,"type":444,"title":717,"items":718},"timeline-geometry","Shape through the ages",[719,723,725,729,733,735,739,743,747],{"time":720,"title":721,"text":722},"c. 2500 BCE","Harappan cities","Mohenjo-daro and Harappa are laid out in grids with baked bricks in a fixed ratio of 1 : 2 : 4.",{"time":456,"title":457,"text":724},"Baudhayana and others give rope-and-peg rules for building fire altars of exact shapes and areas, including a statement of the result we call Pythagoras’ theorem.",{"time":726,"title":727,"text":728},"c. 360 BCE","Plato","Plato links the five regular solids to earth, air, fire, water and the heavens.",{"time":730,"title":731,"text":732},"c. 300 BCE","Euclid’s Elements","Euclid builds geometry from definitions and axioms, and proves that exactly five regular solids exist.",{"time":464,"title":465,"text":734},"The Aryabhatiya gives rules for areas and π ≈ 3.1416.",{"time":736,"title":737,"text":738},"628 CE","Brahmagupta","Finds a formula for the area of a quadrilateral whose vertices lie on a circle.",{"time":740,"title":741,"text":742},"1619","Kepler","Studies which polygons tile the plane and finds the semi-regular tilings.",{"time":744,"title":745,"text":746},"1750","Euler","Writes about F + V − E = 2 for polyhedra twice in 1750 and publishes in 1752. Maurolico had stated it for the regular solids in 1537.",{"time":748,"title":749,"text":750},"1999","Thomas Hales","Proves the honeycomb conjecture: hexagons divide a plane into equal areas with the least perimeter.",{"id":752,"type":47,"variant":707,"title":753,"markdown":754},"example-sulba","Geometry with ropes and pegs","The **Sulba Sutras** (sulba means rope) were practical manuals for building Vedic fire altars. Altars of different shapes, such as squares, circles, and a falcon made of bricks, had to have exactly the same area. The builders used a stretched rope and pegs to make right angles (a rope knotted in the ratio 3 : 4 : 5 makes a right-angled triangle), to double a square, and to turn a square into a circle of nearly the same area. This is some of the oldest recorded geometry in the world.",{"id":756,"type":757,"title":758,"terms":759},"glossary-deepen","glossary","Words for reasoning about shape",[760,763,767,770,772,775,778,781,784,787,790,793],{"term":761,"meaning":762},"Proof","A chain of reasons showing a statement is true in every case.",{"term":764,"meaning":765,"example":766},"Theorem","A statement that has been proved.","The angle sum theorem: a triangle’s angles add to 180°.",{"term":768,"meaning":769},"Exterior angle","The angle between one side of a polygon and the extension of the next side.",{"term":172,"meaning":771},"Angles on opposite sides of a line crossing two parallel lines, between them; they are equal.",{"term":773,"meaning":774},"Co-interior angles","Angles on the same side of a line crossing two parallel lines, between them; they add to 180°.",{"term":776,"meaning":777},"Perpendicular bisector","The line that cuts a segment in half at a right angle; every point on it is equally far from both ends.",{"term":779,"meaning":780},"Hypotenuse","The side opposite the right angle in a right-angled triangle; always the longest side.",{"term":782,"meaning":783},"Constant width","Having the same width measured in every direction, like a circle or a Reuleaux triangle.",{"term":785,"meaning":786},"Euler characteristic","The number F + V − E; it is 2 for polyhedra without holes and 0 for a solid with one hole.",{"term":788,"meaning":789},"Topology","The study of properties that do not change when a shape is stretched or bent without tearing.",{"term":791,"meaning":792},"Order of rotational symmetry","How many times a shape fits onto itself during one full turn.",{"term":794,"meaning":795},"Necessary \u002F sufficient","A necessary condition must hold; a sufficient one guarantees the result. Euler’s formula is necessary but not sufficient for a polyhedron.",{"id":797,"type":798,"title":799,"questions":800},"quiz-deepen","quiz","Reasons, not just answers",[801,812,825,837,847,860,873,886,896,906],{"itemId":802,"prompt":803,"options":804,"correct":408,"why":811},"shape-and-space.deepen-q-diag20","How many diagonals does a 20-sided polygon have?",[805,806,808,809],{"id":408,"label":105},{"id":411,"label":807},"190",{"id":414,"label":104},{"id":417,"label":810},"200","20 × 17 ÷ 2 = 170. (190 is the total number of joins, including the 20 sides.)",{"itemId":813,"prompt":814,"options":815,"correct":417,"why":824},"shape-and-space.deepen-q-proof180","In the proof that a triangle’s angles add to 180°, which fact about parallel lines is used?",[816,818,820,822],{"id":408,"label":817},"Corresponding angles are supplementary",{"id":411,"label":819},"Parallel lines meet at infinity",{"id":414,"label":821},"Vertically opposite angles add to 180°",{"id":417,"label":823},"Alternate angles are equal","A line through one vertex parallel to the opposite side creates two pairs of equal alternate angles, which then sit on a straight line with the third angle.",{"itemId":826,"prompt":827,"options":828,"correct":411,"why":836},"shape-and-space.deepen-q-ext","In a triangle, two interior angles are 50° and 65°. What is the exterior angle at the third vertex?",[829,831,833,834],{"id":408,"label":830},"65°",{"id":411,"label":832},"115°",{"id":414,"label":288},{"id":417,"label":835},"245°","The exterior angle equals the sum of the two opposite interior angles: 50° + 65° = 115°.",{"itemId":838,"prompt":839,"options":840,"correct":411,"why":846},"shape-and-space.deepen-q-regular","Each exterior angle of a regular polygon is 24°. How many sides does it have?",[841,842,843,845],{"id":408,"label":88},{"id":411,"label":93},{"id":414,"label":844},"18",{"id":417,"label":684},"Exterior angles total 360°: 360 ÷ 24 = 15.",{"itemId":848,"prompt":849,"options":850,"correct":414,"why":859},"shape-and-space.deepen-q-third","Two sides of a triangle are 5 cm and 11 cm. Which could be the third side?",[851,853,855,857],{"id":408,"label":852},"5 cm",{"id":411,"label":854},"6 cm",{"id":414,"label":856},"10 cm",{"id":417,"label":858},"16 cm","It must be between 11 − 5 = 6 and 11 + 5 = 16, not equal to either. Only 10 cm works.",{"itemId":861,"prompt":862,"options":863,"correct":408,"why":872},"shape-and-space.deepen-q-euler-prism","Why does every prism satisfy Euler’s formula?",[864,866,868,870],{"id":408,"label":865},"Because (n + 2) + 2n − 3n = 2 for every n",{"id":411,"label":867},"Because prisms have flat faces",{"id":414,"label":869},"Because prisms have an even number of vertices",{"id":417,"label":871},"It has only been checked for small prisms","The general counts F = n + 2, V = 2n, E = 3n give 2 for every value of n. That is a proof for all prisms.",{"itemId":874,"prompt":875,"options":876,"correct":417,"why":885},"shape-and-space.deepen-q-frame","A solid picture frame has F = 16, V = 16, E = 32. Why doesn’t it give 2?",[877,879,881,883],{"id":408,"label":878},"It was counted wrongly",{"id":411,"label":880},"It has curved faces",{"id":414,"label":882},"Frames are not solid",{"id":417,"label":884},"It has a hole through it","Euler’s formula F + V − E = 2 is for polyhedra without holes. Each hole lowers the value by 2, so the frame gives 0.",{"itemId":887,"prompt":888,"options":889,"correct":411,"why":895},"shape-and-space.deepen-q-seven","Which number of edges is impossible for a polyhedron?",[890,891,892,894],{"id":408,"label":681},{"id":411,"label":84},{"id":414,"label":893},"8",{"id":417,"label":89},"6 is a tetrahedron, 8 a square pyramid, 9 a triangular prism. 7 is impossible: it would need F + V = 9 but F, V ≤ 4.",{"itemId":897,"prompt":898,"options":899,"correct":414,"why":905},"shape-and-space.deepen-q-rot","Which shape has rotational symmetry of order 2 but no lines of symmetry?",[900,901,902,904],{"id":408,"label":387},{"id":411,"label":393},{"id":414,"label":903},"Parallelogram (not a rectangle or rhombus)",{"id":417,"label":669},"A general parallelogram fits onto itself after a half turn but no fold matches its halves.",{"itemId":907,"prompt":908,"options":909,"correct":411,"why":918},"shape-and-space.deepen-q-reuleaux","Why can a Reuleaux triangle not be used as a wheel on a fixed axle?",[910,912,914,916],{"id":408,"label":911},"It does not have constant width",{"id":411,"label":913},"Its centre moves up and down as it rolls",{"id":414,"label":915},"It has straight sides",{"id":417,"label":917},"It is too heavy","It has constant width, so a board on top rolls smoothly, but no single point stays at a constant height, so an axle would bob.",{"id":920,"type":921,"prompt":922},"reflect-proof","reflection","Pick one pattern you discovered in Investigate. Write, in your own words, a proof that it always holds. Then explain what the proof does that checking ten examples could not.",{"id":924,"type":925,"title":926,"points":927},"cheat-deepen","summary","Cheat sheet",[928,929,930,931,932,933,934,935,936,937,938,939],"**Diagonals** = n(n − 3) ÷ 2: each vertex has n − 3, and every diagonal is counted twice. Equivalently, all n(n − 1) ÷ 2 joins minus n sides.","Triangle angles total **180°** (proof: parallel line and alternate angles). Exterior angle = sum of the two opposite interior angles.","Polygon angle sum **(n − 2) × 180°**. Exterior angles of a convex polygon total **360°**; regular exterior = 360° ÷ n.","A regular polygon with interior angle x exists only if 360 ÷ (180 − x) is a whole number ≥ 3.","**Triangle inequality:** the third side lies strictly between a − b and a + b. The longest side faces the largest angle.","Inclusive definitions let properties be **inherited**: a square gets every property of rectangles and rhombuses. Square = rectangle ∩ rhombus.","Circles have **constant width** and a centre at a fixed distance from the rim: that is why wheels are round and manhole covers are circular.","π: Archimedes 223\u002F71 \u003C π \u003C 22\u002F7; **Aryabhata 3.1416** (499 CE); Madhava’s series (c. 1400).","**Euler:** F + V − E = 2 for polyhedra without holes. Prisms: (n + 2) + 2n − 3n = 2. A frame with one hole gives 0. Curved solids do not follow it.","2E ≥ 3F and 2E ≥ 3V; so **no polyhedron has 7 edges**. Euler’s formula is necessary but not sufficient.","Cube nets: no 2 by 2 block, no row of 5. Squares one apart in a line become opposite faces. Dice: opposite faces add to 7.","Regular n-gon: **n** lines of symmetry and rotational order **n**. Two mirror lines at angle θ combine to a turn of 2θ.",{"id":941,"type":149,"conceptId":942,"relation":189,"explanation":943},"conn-lines-deep","lines","Proofs about polygons use parallel lines and perpendicular bisectors, which are introduced in the Lines topic.",{"id":945,"type":149,"conceptId":946,"relation":151,"explanation":947},"conn-number-deep","properties-of-numbers","Checking that 360 ÷ (180 − x) is a whole number is a divisibility question: a regular polygon exists only when the exterior angle is a factor of 360.",{"id":949,"type":950,"sourceIds":951},"sources-deepen","sources",[952,953,954,955,956,957,958,959,960,961,962],"shape-and-space-ncert-class7","shape-and-space-ncert-class8","shape-and-space-mathsisfun-euler","shape-and-space-mathsisfun-platonic","shape-and-space-mathsisfun-circle","shape-and-space-wiki-net","shape-and-space-wiki-honeycomb","shape-and-space-khan-geometry","shape-and-space-mactutor-sulbasutras","shape-and-space-wiki-madhava","shape-and-space-eppstein-euler",[952,953,954,955,956,957,958,959,960,961,962],"needs_review",{"generatedBy":966,"notes":967},"claude-code","Draft generated locally; pending owner review.","1344aac3a85d396ffca96e85c71c6eb5a00181da472d00ad0ca8e092e5470b75",{"logic:practice":970,"component:match-pairs@1":971,"component:sort-game@1":972,"component:shape-explorer@1":973,"source:shape-and-space-eppstein-euler":974,"source:shape-and-space-khan-geometry":975,"source:shape-and-space-mactutor-sulbasutras":976,"source:shape-and-space-mathsisfun-circle":977,"source:shape-and-space-mathsisfun-euler":978,"source:shape-and-space-mathsisfun-platonic":979,"source:shape-and-space-ncert-class7":980,"source:shape-and-space-ncert-class8":981,"source:shape-and-space-wiki-honeycomb":982,"source:shape-and-space-wiki-madhava":983,"source:shape-and-space-wiki-net":984},"3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","a8965f19a8548e866e5fcd7f4fec4a9adac35ad54d43c9d5c416cdf3348e6198","95e13e2d031c58f703234cbb201c5fb309036fe3e69ac294cefe54ce2397c51f","10f385d6be5a688f5df4a8b9ea08a6c101d7bf683d87c29e7d9d2827a5e11963","0e5c982c28d1f64e30c5f97c101933eb5d018e8b77ae00379421e5fe11b107a3","ff337c822df6bd0ba5ef54450ce49945d4ef414438058441e656ee73ca685947","512952430fd4d2cad9d5ad14eb51fe077bf3637c74e2c604e6458fe725b018b6","4a5b30c2c4240904f42b67c39dc2176e4a85ade7223fe6b2b7f47916eb25e6e5","20d54bcdc9a5cc5930102f965a7435de619c7bc9731d1de2ff67ae1d0307ca5c","9516472286b1156fa1275e7ea5b6574beff3acc596b8dfe8563f5c446cb7d6b6","4336a097e06982ba6b142ea63fb6f280c2335bc502b4e3e9f754af2d63e6771f","f13478801428576816ecbb37fdf98ba4bdf12642c97c1b870cccfe4b9af4e403","d6c8b2e7ef98f5aa4726bed69ebb23d330aeb205ee01d6f8d2b426062a952f6e",{"state":986,"reviewer":987,"selfReview":988,"reviewedAt":989,"method":990},"approved","The library owner",true,"2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899597319]