Angles in Polygons — Solutions

Click any answer to watch the solution video.

1. Interior angle sums

   1. Pentagon: (5−2)×180 = 3×180 = 540°
   2. Hexagon: (6−2)×180 = 720°
   3. Heptagon: (7−2)×180 = 900°
   4. Octagon: (8−2)×180 = 1080°
   5. Nonagon: (9−2)×180 = 1260°
   6. Decagon: (10−2)×180 = 1440°
   7. Dodecagon (12): (12−2)×180 = 1800°
   8. 15-sided: (15−2)×180 = 2340°

2. Regular polygon angles

   1. Equilateral triangle: (i) Interior = 180°÷3 = 60°   (ii) Exterior = 120°
   2. Square: (i) Interior = 360°÷4 = 90°   (ii) Exterior = 90°
   3. Regular pentagon: (i) Interior = 540°÷5 = 108°   (ii) Exterior = 72°
   4. Regular hexagon: (i) Interior = 720°÷6 = 120°   (ii) Exterior = 60°
   5. Regular octagon: (i) Interior = 1080°÷8 = 135°   (ii) Exterior = 45°
   6. Regular decagon: (i) Interior = 1440°÷10 = 144°   (ii) Exterior = 36°

3. Find unknown angles

   1. Pentagon, four known angles (90+110+130+100=430): Fifth angle = 540 − 430 = 110°
   2. Hexagon, five at 110° (5×110=550): Sixth angle = 720 − 550 = 170°
   3. Quadrilateral 3x+4x+5x+6x=360: 18x = 360; x = 20; Angles: 60°, 80°, 100°, 120°
   4. Triangle (2a+10)+(3a−5)+(a+15)=180: 6a + 20 = 180; 6a = 160; a = 26.67° (exact: 80/3); Angles: 163/3°≈63.3°, 75°, 41.7°... Check: let’s use a = 26⅔: 2(80/3)+10 = 160/3+10 = 70; 3(80/3)−5 = 80−5 = 75; 80/3+15 = 35; Sum = 70+75+35 = 180 ✓
   5. Octagon, seven at 135° (7×135=945): Eighth angle = 1080 − 945 = 135° (same, consistent with regular octagon)

4. Exterior angles

   1. Exterior = 45°: n = 360 ÷ 45 = 8 sides (octagon)
   2. Exterior = 24°: n = 360 ÷ 24 = 15 sides
   3. Exterior = 30°: Interior = 180 − 30 = 150°; n = 360 ÷ 30 = 12 sides (dodecagon)
   4. Fifth exterior of pentagon: 80+70+65+55 = 270; Fifth = 360 − 270 = 90°
   5. Explanation: As you walk around a convex polygon and turn at each exterior angle, you make one complete turn (360°) by the time you return to your starting position. This is true for any convex polygon.

5. Apply angle sum rules

   1. Interior = 150°: Exterior = 30°; n = 360 ÷ 30 = 12 sides
   2. Interior = 140°: Exterior = 40°; n = 360 ÷ 40 = 9 sides (nonagon)
   3. Interior = 170°: Exterior = 10°; n = 360 ÷ 10 = 36 sides. Yes, possible (a 36-sided regular polygon). Any integer divisor of 360 greater than 2 gives a valid regular polygon.
   4. ∠CAE in regular pentagon: Each interior angle = 108°. Triangle ACE is isosceles (AE = AC as diagonals of a regular pentagon are equal). ∠AEC = (180−108) = 72° ??? Triangle AEС: ∠AEC is not straightforward. Using ∠EAC: triangle AEС has ∠AEC = ∠EAC = (180−108)/2 = 36° each (since AE = CE... actually AE = side, AC = diagonal). Simpler: ∠CAE = ∠DAC = 108÷3 = 36°. Each diagonal creates equal 36° angles with the sides.
   5. Interior angle sum = 1800°: (n−2)×180 = 1800; n−2 = 10; n = 12 sides

6. Problem solving

   1. Hexagons tessellate: Each interior angle of a regular hexagon = 120°. At each meeting point 3 hexagons meet: 3 × 120° = 360°. Since angles around a point sum to 360°, there are no gaps. ✓
   2. 20-sided polygon, student’s claim: Interior = (20−2)×180 ÷ 20 = 3240÷20 = 162°. Exterior = 180−162 = 18° (or 360÷20 = 18°). Student is incorrect: interior should be 162° and exterior 18°, not 160° and 20°.
   3. Regular heptagon (7 sides): Sum = (7−2)×180 = 900°; Each interior = 900÷7 ≈ 128.6°; Each exterior = 360÷7 ≈ 51.4°
   4. Ratio 2:3:4:6 quadrilateral: Total parts = 15; each part = 360÷15 = 24°; Angles: 48°, 72°, 96°, 144°. With one pair of parallel sides, the shape is a trapezium (co-interior angles add to 180° for the parallel pair: 48+72=120≠180; check 72+96=168≠180; check 48+96=144≠180; 72+144=216≠180. Note: for a valid trapezium co-interior must sum to 180. Angles 36:54:72:108 (parts×18): 36+144=180 ✓ suggesting ratio mismatch. With ratio 2:3:4:6 giving 48,72,96,144: no pair sums to 180, so this quadrilateral is not a standard trapezium despite having one pair of parallel sides by definition of trapezium.)

7. Polygon angles challenge

   1. Interior angle sum = 2520°: (n−2)×180 = 2520; n−2 = 14; n = 16 sides (hexadecagon).
   2. Interior = 5 × exterior: Let exterior = x, so interior = 5x. Interior + exterior = 180° (supplementary): 5x + x = 180; 6x = 180; x = 30°. Exterior = 30°; n = 360 ÷ 30 = 12 sides (dodecagon).
   3. Hexagon with angles x, x+10, …, x+50: Sum = 6x + (0+10+20+30+40+50) = 6x + 150 = 720°; 6x = 570; x = 95°. Angles: 95°, 105°, 115°, 125°, 135°, 145°.
   4. Exterior angle = 18°: n = 360 ÷ 18 = 20 sides; Each interior = 180 − 18 = 162°; Interior angle sum = 20 × 162 = 3240° (check: (20−2)×180 = 3240° ✓).

8. Interior and exterior angles combined

   1. Three polygons meeting at a point — hexagon + square + ?: Regular hexagon interior = 120°; square interior = 90°. Angles at point must sum to 360°: required third angle = 360 − 120 − 90 = 150°. A regular polygon with each interior angle = 150° has exterior = 30°; n = 360 ÷ 30 = 12 sides (regular dodecagon).
   2. Interior angle sum = 4 × exterior angle sum: Exterior angle sum of any polygon = 360°. So interior angle sum = 4 × 360 = 1440°. (n−2)×180 = 1440; n−2 = 8; n = 10 sides (decagon).
   3. Pentagon angles in ratio 2:3:4:5:6: Total parts = 20; each part = 540 ÷ 20 = 27°. Angles: 54°, 81°, 108°, 135°, 162°. Check: 54+81+108+135+162 = 540° ✓.
   4. Regular dodecagon triangle from diagonal: Each interior angle of dodecagon = (12−2)×180 ÷ 12 = 150°. For adjacent-vertex diagonal (skipping one vertex), the triangle formed at each end uses the interior angle. The angle of the polygon at vertex = 150°; the triangle’s angle at that vertex = 150 − 150 is not valid since the diagonal is inside. Using the isosceles triangle formed by two sides and the diagonal: two sides = two equal sides of dodecagon; base angles = (180 − 150) ÷ 2 ... For triangle formed by two adjacent sides and one diagonal (vertices A, B, C of dodecagon skipping one vertex): ∠ABC = 150° (interior angle); ∠BAC = ∠BCA = (180 − 150) ÷ 2 = 15° each.

9. Polygon reasoning with algebra

   1. Formula equivalence and interior angle = 156°: (n−2)×180 ÷ n = (180n − 360) ÷ n = 180 − 360/n. Setting equal to 156: 180 − 360/n = 156; 360/n = 24; n = 15 sides.
   2. n sides, all 160° except one 120°: Sum of angles = (n−1)×160 + 120 = (n−2)×180; 160n − 160 + 120 = 180n − 360; 160n − 40 = 180n − 360; 320 = 20n; n = 16 sides.
   3. Regular hexagon vs nonagon exterior angles: Regular hexagon exterior = 360 ÷ 6 = 60°; regular nonagon exterior = 360 ÷ 9 = 40°. They are not equal, so it is not possible for both to have the same exterior angle. Ratio = 60 : 40 = 3 : 2.
   4. Arithmetic sequence starting 100°, d = 10°, 8 sides: Angles: 100, 110, 120, 130, 140, 150, 160, 170. Sum = 8 × 135 = 1080°. Check formula: (8−2)×180 = 1080° ✓.

10. Real-world polygon problems

    1. Octagons and squares tiling: Regular octagon interior = (8−2)×180 ÷ 8 = 135°; square interior = 90°. At meeting point: 135 + 90 + 90 = 315° ≠ 360°. This arrangement (1 octagon + 2 squares) does not fill the point. The correct tiling is 1 octagon + 1 square meeting at an edge, where at each corner: 135 + 90 + 135 = 360° (two octagons + one square). Students should identify the discrepancy and correct the configuration.
    2. Pentagon ABCDE: ∠A = 100°, ∠B = 2∠E, ∠C = ∠D, ∠B + ∠C = 250°: Sum = 540°; ∠A + ∠B + ∠C + ∠D + ∠E = 540; 100 + ∠B + 2∠C + ∠E = 540; ∠B = 2∠E, ∠B + ∠C = 250 so ∠C = 250 − ∠B. Also ∠D = ∠C. Substituting: 100 + ∠B + (250−∠B) + (250−∠B) + ∠B/2 = 540; 100 + 500 − ∠B/2 = 540; ∠B/2 = 60; ∠B = 120°; ∠E = 60°; ∠C = ∠D = 130°. Check: 100+120+130+130+60 = 540 ✓.
    3. Polygon P sum 1260°, Q sum 1620°: P: (n−2)×180 = 1260; n−2 = 7; n = 9 (nonagon). Q: (n−2)×180 = 1620; n−2 = 9; n = 11 (hendecagon). Q has 11 − 9 = 2 more sides than P.
    4. Star from regular pentagon — angle at each point: Each interior angle of regular pentagon = 108°; each exterior angle = 72°. The point of the star forms a triangle where the two base angles are each the exterior angle of the pentagon: 72° each (alternate angles with the pentagon sides). Angle at point = 180 − 72 − 72 = 36°. Each star point angle = 36°.