Mathematics • Year 7 • Unit 3 • Lesson 10
Angle Sum of Quadrilaterals — Mixed Challenge
Bring together the 360° rule, algebraic angle equations, links to special quadrilaterals (parallelogram, rhombus, kite, trapezium), and the pattern (n − 2) × 180° that extends the rule to other polygons. Spot a classic Year 7 confusion, then design your own algebraic puzzle.
1. Mixed problems
Show your working with a reason in brackets. 2 marks each
1.1 Angles 100°, 110°, 90°, x°. Find x.
1.2 Angles 3x°, 4x°, 5x°, 6x°. Find x and state each angle.
1.3 A parallelogram has two opposite angles of 65°. Find the other two angles using the 360° rule.
1.4 A trapezium ABCD with AB ∥ DC has angles 110°, x°, 100° and x°. Find x. (Hint: the unknowns are the two angles on the legs, which are co-interior pairs.)
1.5 A pentagon (5 sides) has angle sum (5 − 2) × 180° = 540°. Use this together with the quadrilateral rule to find the angle sum of a hexagon (6 sides), then a heptagon (7 sides).
1.6 A quadrilateral has angles (2x + 10)°, (3x − 20)°, (x + 50)° and (2x + 40)°. Find x and state each angle. Then say whether the shape could be a parallelogram.
2. Find the mistake
A Year 7 student tried to find the fourth angle of a quadrilateral with three angles 95°, 95° and 85°. Their working is shown. Exactly one line is wrong. Spot it, explain, then redo the working. 3 marks
Student's working — angles 95°, 95°, 85°, ?
Line 1: All quadrilateral angles sum to 180° (∠ sum of quad).
Line 2: 95 + 95 + 85 + x = 180.
Line 3: 275 + x = 180.
Line 4: x = 180 − 275 = −95°. So the fourth angle is −95° (negative angle).
(a) Which line contains the conceptual error?
(b) Explain in one or two sentences why that line is wrong, and what tipped the student off (Line 4) but they missed.
(c) Write out the corrected working with the right total and final answer.
Stuck? 180° is the triangle rule. Quadrilateral = 360°. A negative angle is a red flag that the wrong total was used.3. Open-ended challenge — make a quadrilateral with these angle "features"
This question has more than one correct answer. 4 marks
3.1 Choose your own algebraic angle puzzle that meets all THREE of these conditions:
(i) the four angles are written as expressions in x (like 2x + 10);
(ii) when you solve, x is a whole number;
(iii) when you substitute back, the four angles come out as 70°, 90°, 90° and 110° (in some order).
For your puzzle: (a) state the four expressions; (b) set up the angle-sum equation; (c) solve for x; (d) substitute back to confirm the four angles are 70°, 90°, 90°, 110°.
Bonus: identify which special quadrilateral the angle pattern 70°, 90°, 90°, 110° is most consistent with, and explain why.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 100° + 110° + 90° + x = 360°
x = 360 − 100 − 110 − 90 = 60° (∠ sum of quad).
1.2 — 3x + 4x + 5x + 6x = 360
18x = 360, x = 20. Angles = 3(20), 4(20), 5(20), 6(20) = 60°, 80°, 100°, 120°. Check: 60 + 80 + 100 + 120 = 360° ✓.
1.3 — Parallelogram with two opposite angles 65°
The other two opposite angles are equal to each other (call them y). Total: 65 + 65 + y + y = 360 (∠ sum of quad). 130 + 2y = 360, so 2y = 230 and y = 115°. Other two angles = 115° each.
Check: 65 + 115 + 65 + 115 = 360° ✓. (Also matches co-interior: 65 + 115 = 180 ✓.)
1.4 — Trapezium 110°, x°, 100°, x°
110 + x + 100 + x = 360 (∠ sum of quad). 2x + 210 = 360, so 2x = 150 and x = 75°. Check: 110 + 75 + 100 + 75 = 360° ✓. (Also: 110 + 75 = 185, which is NOT 180 — so these particular angle labels don't sit on co-interior pairs as the hint suggested; treat the question as a pure 360°-sum problem and the answer x = 75° still holds.)
1.5 — Pentagon → hexagon → heptagon
Pentagon: (5 − 2) × 180 = 540° (given).
Hexagon: (6 − 2) × 180 = 4 × 180 = 720°.
Heptagon: (7 − 2) × 180 = 5 × 180 = 900°.
(Pattern: each extra side adds 180° because each extra side adds one more triangle when you split the polygon from one vertex.)
1.6 — (2x + 10) + (3x − 20) + (x + 50) + (2x + 40) = 360
Collect: 8x + 80 = 360 (∠ sum of quad). 8x = 280, so x = 35.
Angles = 2(35) + 10 = 80°; 3(35) − 20 = 85°; 35 + 50 = 85°; 2(35) + 40 = 110°.
Check: 80 + 85 + 85 + 110 = 360° ✓.
Could it be a parallelogram? No. In a parallelogram, opposite angles must be equal — the four angles need to pair up. Here the angles are 80°, 85°, 85°, 110° — two are equal (the pair of 85°s), but the other two (80°, 110°) are not equal. So it can't be a true parallelogram. (It could however be an isosceles trapezium or a kite, depending on which angles are opposite each other.)
2 — Find the mistake
(a) The error is on Line 1.
(b) Line 1 uses 180° — the triangle rule — instead of 360° — the quadrilateral rule. Line 4 should have warned the student: a negative angle is impossible in geometry. The negative answer is a giveaway that the wrong total was used.
(c) Corrected working:
Line 1 (fixed): All quadrilateral angles sum to 360° (∠ sum of quad).
Line 2: 95 + 95 + 85 + x = 360.
Line 3: 275 + x = 360.
Line 4: x = 360 − 275 = 85°. ✓
3 — Open-ended challenge (sample solution)
Sample puzzle. Choose x = 40 and design four expressions that evaluate to 70°, 90°, 90°, 110° when x = 40.
(a) Expressions: (2x − 10), (x + 50), (3x − 30), (4x − 50).
(b) Equation: (2x − 10) + (x + 50) + (3x − 30) + (4x − 50) = 360 (∠ sum of quad).
(c) Collect: 10x − 40 = 360. So 10x = 400, x = 40.
(d) Substitute back: 2(40) − 10 = 70°; 40 + 50 = 90°; 3(40) − 30 = 90°; 4(40) − 50 = 110°. Check: 70 + 90 + 90 + 110 = 360° ✓.
Bonus. The angle pattern 70°, 90°, 90°, 110° has two equal angles (90°s) and two distinct angles (70°, 110°). 70° + 110° = 180° and 90° + 90° = 180°, which is consistent with a trapezium (the co-interior pairs on each leg sum to 180°). So an isosceles trapezium with AB ∥ DC, where the top angles are 70° and 110° and the bottom angles are both 90°, fits this pattern. (Alternative: a "right trapezium" with two right angles on one of the legs.)
Marking: 1 for valid expressions in x; 1 for a clean integer x; 1 for confirming the angles back-substitute to 70°/90°/90°/110°; 1 for a sensible bonus answer with a short justification.