Polygons · word problems

Polygon Heuristics

10 question types · PSLE Paper 1 and 2 · one figure that follows the deduction

01

Angle Sum of a Polygon

propertyCut the polygon into triangles from one vertex: (n − 2) × 180°

A hexagon has angles of 130°, 95°, 140°, 110° and 125°. Find its sixth angle.

130°95°140°110°125°?
From one vertex, three diagonals cut the hexagon into four triangles.
Diagonals from one vertex cut a hexagon into 6 − 2 = 4 triangles.
step 1 of 4

From one vertex a hexagon splits into four triangles, so its angles add to 4 × 180°. Subtract the five known angles.

  1. Diagonals from one vertex cut a hexagon into 6 − 2 = 4 triangles.
  2. Angle sum = 4 × 180° = 720°.
  3. The five known angles: 130 + 95 + 140 + 110 + 125 = 600.
  4. Sixth angle = 720° − 600° = 120°.

answer120°

Common pitfalls

  • Using 360°, the angle sum of a quadrilateral, for every polygon.
  • Counting 6 triangles instead of 4: the two vertices next to the starting one make no triangle of their own.
02

One Angle of a Regular Polygon

propertyAngle sum divided by the number of sides; then a diagonal makes an isosceles triangle

ABCDE is a regular pentagon and AC is a diagonal. Find ∠ ABC and ∠ ACD.

ABCDE
A pentagon’s angles add to 3 × 180° = 540°.
A pentagon's angles add to (5 − 2) × 180° = 540°.
step 1 of 4

Every angle of a regular pentagon is the same share of 540°. The diagonal AC cuts off an isosceles triangle ABC, and what remains of the angle at C is the second answer.

  1. A pentagon's angles add to (5 − 2) × 180° = 540°.
  2. Regular, so all five are equal: ∠ ABC = 540° ÷ 5 = 108°.
  3. BA = BC, so triangle ABC is isosceles with apex B: ∠ BCA = (180° − 108°) ÷ 2 = 36°.
  4. ∠ ACD = ∠ BCD − ∠ BCA = 108° − 36° = 72°.

answer∠ ABC = 108°; ∠ ACD = 72°

Common pitfalls

  • Dividing 360° by 5 and answering 72° for the interior angle: 72° is the exterior angle.
  • Assuming the diagonal bisects the angle at C: it cuts off 36°, not 54°.
03

Exterior Angles Add to 360°

propertyWalk round the polygon and you turn through one full circle

Each interior angle of a regular polygon is 156°. How many sides does it have? A different regular polygon has 20 sides; find each of its interior angles.

156°24°15 sides: exterior 24°, interior 156°
At each corner, the exterior angle is 180° minus the interior one.
Exterior angle at each vertex = 180° − 156° = 24°.
step 1 of 5

At each vertex the exterior angle is 180° minus the interior angle, and walking round any polygon turns you through 360° in total. So the exterior angle and the number of sides are 360° shared out.

  1. Exterior angle at each vertex = 180° − 156° = 24°.
  2. The exterior angles of any polygon add to 360°, and here they are all equal.
  3. Number of sides = 360° ÷ 24° = 15.
  4. For 20 sides: exterior angle = 360° ÷ 20 = 18°.
  5. Interior angle = 180° − 18° = 162°.

answer15 sides; 162°

Common pitfalls

  • Dividing 360° by the interior angle: 360 ÷ 156 is not a whole number, which is the sign the wrong angle was used.
  • Giving the 20-gon's exterior angle, 18°, as its interior angle.
04

Regular Hexagon as Six Equilateral Triangles

propertyLines from the centre to the vertices make six equilateral triangles, so the long diagonal is two sides

ABCDEF is a regular hexagon with centre O and sides of 6 cm. Find ∠ OAB and the length of the diagonal AD.

ABCDEFO6 cm60°
Six lines from O: 360° ÷ 6 = 60° each.
∠ AOB = 360° ÷ 6 = 60°.
step 1 of 4

The six lines from the centre split the full turn into six 60° angles, and each triangle at the centre is isosceles with a 60° apex, so equilateral. The diagonal through the centre is two of their sides.

  1. ∠ AOB = 360° ÷ 6 = 60°.
  2. OA = OB, so triangle OAB is isosceles: ∠ OAB = ∠ OBA = (180° − 60°) ÷ 2 = 60°.
  3. All three angles are 60°: triangle OAB is equilateral, so OA = AB = 6 cm.
  4. A, O and D lie on one straight line through the centre, so AD = AO + OD = 6 + 6 = 12 cm.

answer∠ OAB = 60°; AD = 12 cm

Common pitfalls

  • Halving the hexagon's 120° angle and stopping: 60° is right, but the reason is the equilateral triangle, and that is what gives OA = 6 cm.
  • Measuring AD as three sides, 18 cm, by counting the sides along the rim instead of the straight line through O.
05

Regular Octagon by Clipping a Square

propertyEach corner cut is a 45°-45°-90° triangle, so the octagon's angle is 90° + 45°

The four corners of a square are cut off by straight cuts so that the shape left is a regular octagon. Find the angle of the octagon at each cut, and the two acute angles of each triangle removed.

135°
A regular octagon: 1080° ÷ 8 = 135° at every corner.
An octagon's angles add to (8 − 2) × 180° = 1080°; regular, so each is 1080° ÷ 8 = 135°.
step 1 of 4

A regular octagon's angles are all equal, and each cut runs symmetrically across a corner, so the removed triangles are right-angled and isosceles.

  1. An octagon's angles add to (8 − 2) × 180° = 1080°; regular, so each is 1080° ÷ 8 = 135°.
  2. Each cut-off triangle has the square's corner as a right angle, and its two other angles add to 90°.
  3. The cut makes equal lengths along the two sides of the corner, so the triangle is isosceles: its acute angles are 45° each.
  4. Check at a cut: the octagon's angle is the square's straight side plus the triangle's acute angle: 90° + 45° = 135°.

answer135°; 45° and 45°

Common pitfalls

  • Answering 90° for the octagon's angle because the shape started as a square.
  • Taking the cut-off triangle as equilateral: it has a right angle from the square's corner.
06

Regular Pentagon and Square on One Side

propertyThe two shapes' angles at the shared vertex differ by the gap between them; equal sides make the gap's triangle isosceles

Square ABCD and regular pentagon ABEFG are drawn on the same side of AB. Find ∠ DAG and ∠ ADG.

ABCDEFG108°
The pentagon opens 108° from AB at A; the square opens 90°.
Each angle of a regular pentagon is 540° ÷ 5 = 108°, so ∠ BAG = 108°.
step 1 of 4

At A the pentagon's 108° and the square's 90° open from the same side AB, so the sliver between AD and AG is their difference. AD and AG both equal AB, so triangle ADG is isosceles.

  1. Each angle of a regular pentagon is 540° ÷ 5 = 108°, so ∠ BAG = 108°.
  2. ∠ BAD = 90°, and AD lies inside ∠ BAG: ∠ DAG = 108° − 90° = 18°.
  3. AD = AB (square) and AG = AB (regular pentagon), so AD = AG.
  4. Triangle ADG is isosceles with apex A: ∠ ADG = (180° − 18°) ÷ 2 = 81°.

answer∠ DAG = 18°; ∠ ADG = 81°

Common pitfalls

  • Adding 108° + 90°: the shapes are on the same side of AB, so the angles overlap and the gap is the difference.
  • Forgetting that the square and the pentagon share the side length, which is what makes ∠ ADG findable at all.
07

Irregular Polygon with a Reflex Angle

propertyThe angle sum counts a reflex angle at its full size, more than 180°

A pentagon has angles of 95°, 110°, 70° and a reflex angle of 230°. Find its fifth angle.

95°110°70°230°?
Three triangles from the dented vertex: 3 × 180° = 540°.
A pentagon splits into 5 − 2 = 3 triangles, so its angles add to 3 × 180° = 540°.
step 1 of 4

A pentagon's angles add to 540° whether or not one of them is reflex; the reflex angle simply takes a large share of the sum.

  1. A pentagon splits into 5 − 2 = 3 triangles, so its angles add to 3 × 180° = 540°.
  2. The reflex angle counts as 230°, the inside of the dent.
  3. The four known angles: 95 + 110 + 70 + 230 = 505.
  4. Fifth angle = 540° − 505° = 35°.

answer35°

Common pitfalls

  • Replacing the reflex angle by 360° − 230° = 130°, the angle outside the dent, which is not an angle of the polygon.
  • Refusing the answer because 35° looks small next to 230°: a sharp point opposite a dent is exactly what a reflex angle makes.
08

The Five Points of a Star

propertyEach point of a regular pentagram is 36°; the five together make 180°

A regular five-pointed star is drawn by joining every second vertex of a regular pentagon. Find the angle at each point of the star, and the sum of the five point angles.

108°
The star encloses a small regular pentagon: 108° at each corner.
The five lines of the star enclose a small regular pentagon; each of its angles is 108°.
step 1 of 5

The star's points sit on a regular pentagon, and each point is the apex of an isosceles triangle whose base angles are exterior angles of the small pentagon in the middle.

  1. The five lines of the star enclose a small regular pentagon; each of its angles is 108°.
  2. At a point of the star, the two base angles of the tip triangle lie on straight lines with two angles of the small pentagon: each is 180° − 108° = 72°.
  3. Point angle = 180° − 72° − 72° = 36°.
  4. Five points: 5 × 36° = 180°.
  5. Check another way: the point angle is half the pentagon's exterior angle, 72° ÷ 2 = 36°.

answer36°; 180°

Common pitfalls

  • Answering 72°, the pentagon's exterior angle, for the point angle.
  • Adding the five points to 540° as if they were the angles of a pentagon: the star's points are not a polygon's interior angles.
09

How Many Diagonals

propertyEach vertex reaches n − 3 others; halve, since every diagonal is counted from both ends

How many diagonals does a hexagon have? How many does a decagon (10 sides) have?

6 sides: 3 from each vertex, 9 diagonals
From one vertex of a hexagon: 6 − 3 = 3 diagonals (the green ones).
From each vertex of a hexagon, diagonals go to 6 − 3 = 3 other vertices.
step 1 of 4

From one vertex a diagonal can go to every vertex except itself and its two neighbours. Doing that at every vertex counts each diagonal twice.

  1. From each vertex of a hexagon, diagonals go to 6 − 3 = 3 other vertices.
  2. 6 vertices × 3 = 18 ends, and each diagonal has two ends: 18 ÷ 2 = 9 diagonals.
  3. For a decagon: 10 − 3 = 7 from each vertex.
  4. 10 × 7 ÷ 2 = 35 diagonals.

answer9; 35

Common pitfalls

  • Forgetting to halve, and answering 18 and 70.
  • Using n − 2 instead of n − 3: a vertex is not joined to itself, and the lines to its two neighbours are sides, not diagonals.
10

Fitting Polygons Round a Point

propertyThe angles meeting at a point must add to exactly 360°

Two regular hexagons and some equilateral triangles are placed round a point with no gaps and no overlaps. How many triangles are there? Can regular pentagons alone be fitted round a point in the same way?

120°120°
Two hexagon corners at the point: 120° + 120° = 240°.
Each angle of a regular hexagon is 120°: two hexagons take 240°.
step 1 of 5

Whatever shapes meet at a point, their angles there fill the full turn. Two hexagon corners leave a gap that triangle corners fill exactly; pentagon corners cannot fill 360° at all.

  1. Each angle of a regular hexagon is 120°: two hexagons take 240°.
  2. What is left: 360° − 240° = 120°.
  3. Each angle of an equilateral triangle is 60°: 120° ÷ 60° = 2 triangles.
  4. Regular pentagons: 360° ÷ 108° is not a whole number. Three pentagons take 324° and leave a 36° gap; a fourth would overlap.
  5. So pentagons alone cannot fit round a point.

answer2 triangles; no, three pentagons leave a 36° gap

Common pitfalls

  • Fitting the shapes by eye and believing a picture: only the angle sum decides.
  • Using 180° at the point, as if the shapes sat on a straight line rather than all the way round.
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