Find interior angle sums of simple polygons
Use 180(n−2)° for simple-polygon interiors, inverse side counts and checked triangulations of convex and concave examples.
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 triangle and quadrilateral sums and solve a linear equation.
- Identify the interior of a simple convex or concave polygon.
- Read a labelled boundary and distinguish an internal diagonal from an outside segment.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A simple seven-sided polygon is triangulated without adding interior vertices. What is its interior-angle sum?
What you will learn
- Connect the interior-angle sum 180(n−2)° with (2n−4) right angles and n−2 triangles.
- Solve missing-angle, expression and inverse side-count problems with valid integer checks.
- Select reflex interiors and a genuinely internal triangulation in a concave polygon.
Draw or annotate convex pentagon ABCDE in boundary order with interiors A=(3x+12)°,B=(2x+8)°,C=(x+30)°,D=120°,E=130°. Draw valid internal AC and AD and name the three resulting triangles. Find x and all five interiors, showing both the angle-sum equation and a substituted check. Keep the fan diagonals inside the polygon and distinguish one angle from the complete five-angle total.
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.
Polygon ABCDE
- α1 · arc-1: given (3x+12)°
- α2 · arc-2: given (2x+8)°
- α3 · arc-3: given (x+30)°
- α4 · arc-4: given 120°
- α5 · arc-5: given 130°
Count triangles rather than sides
For a simple n-sided polygon, the interior-angle sum is 180(n−2)°, equivalently (2n−4) right angles. In a convex polygon, draw internal diagonals from one vertex to all non-neighbouring vertices. This fan divides it into n−2 triangles: a pentagon gives three triangles, not five.
Each triangle contributes 180°. At the original vertices the triangle angles fit together to give precisely the polygon interiors. Label the fan’s triangles or shade them lightly so the count is visible. Keep a sum of all interiors separate from the measure of one interior; the sum rule does not say the polygon is regular.
Use the sum in both directions
For a missing angle or labelled expression, calculate the correct total first and then add every interior label once. Solve the equation, substitute and check the full total. The words pentagon or hexagon supply n even when no sides or angles are equal.
If the total is given and n is unknown, solve 180(n−2)=total. A polygon side count must be an integer at least three. A fractional answer is a reason to reject the claimed data, not an instruction to round to a nearby polygon. When n is the unknown, begin with the condition rather than a resolved full polygon that would give the count away.
Choose a valid triangulation for concavity
A simple concave polygon has the same 180(n−2)° sum. At an inward vertex include the reflex interior sweep. A five-sided outline remains a pentagon even if one interior exceeds 180°, and that larger interior must not be replaced by its smaller outside gap.
An arbitrary fan from one vertex need not stay inside a concave polygon. Choose diagonals that really lie in the interior, and check the resulting triangles fill the shape without overlap or gaps. In the notched pentagon used below, DA and DB work, whereas AC partly leaves the polygon. These ordinary angle-sum statements exclude crossed boundaries, holes and curved outlines.
Sources for this lesson
Pause and explain
A simple polygon has total interior angle 1620°. Which side and triangle counts follow?
Worked example
Draw or annotate convex pentagon ABCDE in boundary order with interiors A=(3x+12)°,B=(2x+8)°,C=(x+30)°,D=120°,E=130°. Draw valid internal AC and AD and name the three resulting triangles. Find x and all five interiors, showing both the angle-sum equation and a substituted check. Keep the fan diagonals inside the polygon and distinguish one angle from the complete five-angle total.
Show the worked solution
- A pentagon has n=5, so the total is 180(5−2)=540°, or (2×5−4)=6 right angles. Internal AC and AD make triangles ABC,ACD,ADE.
- Add the labels: (3x+12)+(2x+8)+(x+30)+120+130=540. This simplifies to 6x+300=540, so x=40.
- Substitute A=132°,B=88°,C=70°,D=120°,E=130°. Their total is 132+88+70+120+130=540°, and each is a valid convex interior.
Answer x=40; angles A,B,C,D,E=132°,88°,70°,120°,130°, totaling 540° (6 right angles). The internal fan has three triangles ABC,ACD,ADE.
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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.
Draw or annotate an irregular convex hexagon ABCDEF with successive interiors A=100°,B=115°,C=130°,D=125°,E=140°,F=x°. Draw the three internal fan diagonals from A and count the triangles. Find x without assuming that any angles are equal. Label the missing angle at F and check the total using every interior exactly once.
Give me a hint
Find n−2 before subtracting the five known interiors from the total.
Compare my reasoning
- A hexagon has six sides, so a convex fan contains 6−2=4 triangles and the interior total is 4×180°=720°.
- Add the five stated interior measures before subtracting: 100+115+130+125+140=610°. With the missing interior at F, the complete equation is 610+x=720.
- Subtract to obtain x=110°. Label F=110° and check all six interiors sum 720°.
- The fan is ABC,ACD,ADE,AEF. Nothing in this reasoning says the irregular hexagon has equal sides or equal angles.
x=110°. The six interiors total 720° and the convex fan contains four triangles, not six.
Look for these in your work
- Show and count four non-overlapping triangles in the convex fan.
- Use the 720° total and find F=110°.
- Check all six angles without adding a regularity assumption.
Begin from condition cards, not a pre-drawn resolved outline. Card 1 says a simple polygon has total interior angle 1980°. Find its side count and the number of triangles in a triangulation. After the deduction, make a schematic labelled convex example with that many sides and check its triangle count; the sketch need not establish regularity. Card 2 proposes total 1900° for a simple polygon. Decide whether the proposal is possible, without rounding its side count or drawing a fictitious fractional-sided polygon.
Give me a hint
Solve for n exactly; a valid side count must be a whole number at least three.
Compare my reasoning
- For card 1 write 180(n−2)=1980. Divide by 180 to get n−2=11, hence n=13.
- Its triangulation contains 11 triangles, contributing 11×180°=1980°. Only after finding n, sketch a convex 13-gon as a schematic example; the data do not require it to be regular.
- Apply the same inverse formula to card 2 without rounding: 180(n−2)=1900 gives n=1900/180+2=113/9, which must now be tested as a side count.
- 113/9 is not an integer, so card 2 is impossible for the ordinary simple polygon described. Rounding would change the stated total rather than solve the problem.
Card 1: n=13 sides and 11 triangles. Card 2 is impossible: its equation gives n=113/9, a non-integer side count. No rounding or fictitious resolved outline is justified.
Look for these in your work
- Infer 13 sides and 11 triangles from the 1980° total before drawing the resolved example.
- Keep the inverse calculations exact.
- Reject 113/9 as a side count instead of rounding or drawing impossible data.
Draw the exact simple boundary A(0,0),B(4,0),C(4,3),D(2,1),E(0,3), joining in that order and closing to A. Mark interiors A=90°,B=90°,C=45°,E=45° and the long interior at D as d°. Find d and the other sweep at D. Draw DA and DB to explain a valid triangulation, naming its triangles. Explain why a careless fan using AC is not justified.
Give me a hint
Trace the interior region at D and choose diagonals that stay in the polygon.
Compare my reasoning
- The labelled boundary has five sides and one inward vertex, so its interior total is still 180(5−2)=540°.
- Use the long interior at D: d=540−90−90−45−45=270°. The other sweep is 360−270=90° and lies in the notch outside the polygon.
- Internal DA and DB form triangles DAB,DBC,DEA. Their interiors fill the polygon and their combined total is 3×180°=540°.
- AC partly passes through the notch outside the polygon, so it cannot simply be counted as an internal fan diagonal. Check a proposed diagonal’s location rather than extending the convex fan argument automatically.
d=270° and the other sweep is 90°. The valid triangles are DAB,DBC,DEA, made with DA and DB. AC is not entirely internal, so that careless fan would not be a valid explanation.
Look for these in your work
- Draw the notched boundary in its exact stated order.
- Label the 270° interior and separate 90° outside sweep.
- Use DA/DB to name a valid triangulation and explain why AC fails.
Common mistakes
- Using 180n rather than 180(n−2), or confusing the total with one interior angle.
- Rounding a fractional inverse side count into a supposed valid polygon.
- Replacing a reflex interior by its outside gap, or counting an outside diagonal as part of a valid triangulation.
What can you explain now?
Does making a simple pentagon concave change its total interior angle from 540°? Which sweep is included at an inward vertex?
Compare with the explanation
No. Its total remains 540°, including the reflex interior at the inward vertex.
The simple-polygon sum depends on the five sides, not convexity. A checked internal triangulation still gives three triangles; an arbitrary fan may not stay inside.
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Make it stick
State the polygon interior-sum formula in degrees and right angles. Explain the inverse integer check and give a reason why a convex fan cannot be used blindly in a concave diagram.
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.