Mathematics • Year 8 • Unit 3 • Lesson 16
Quadrilateral Angles — Mixed Challenge
Pull everything from Lesson 16 together: the 360° rule, special quadrilateral angle properties, algebraic angle problems, and the diagonal-into-two-triangles proof. Six mixed problems, one "find the mistake", and one open-ended quadrilateral design challenge.
1. Mixed problems — choose the right move
Each question pulls a different idea from Lesson 16. Decide which approach applies before you start writing. Show your working. 3 marks each
1.1 Find the missing angle in a quadrilateral whose other three angles are 132°, 47° and 95°.
1.2 A rhombus has one angle of 110°. State the sizes of the other three angles and justify with the rhombus angle properties.
1.3 A quadrilateral has angles in the ratio 1 : 2 : 3 : 4. Find each angle.
1.4 A quadrilateral has angles 4x°, 4x°, 2x° and 2x°. Find x and the size of the largest angle.
1.5 A trapezium ABCD has AB parallel to DC. ∠A = 70° and ∠B = 80°. Find ∠C and ∠D using the co-interior angle rule.
1.6 A quadrilateral has angles (x + 30)°, (2x − 10)°, x° and (x + 20)°. Find x and list all four angles.
2. Find the mistake
A Year 8 student has tried to find the fourth angle of a quadrilateral with three known angles 130°, 90° and 70°. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find the fourth angle x when the other three are 130°, 90° and 70°:
Line 1: Sum of angles of a quadrilateral = 180°
Line 2: 130 + 90 + 70 + x = 180
Line 3: 290 + x = 180
Line 4: x = 180 − 290 = −110°
(a) Which line contains the original mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Revisit lesson § Card 10 — "Using 180° instead of 360° for the quadrilateral angle sum" is the top pitfall. A negative angle is also a strong warning sign.3. Open-ended challenge — Design a quadrilateral
This question has more than one valid answer. 4 marks
3.1 Design three different quadrilaterals, each meeting all of the conditions below:
- All four angles are whole numbers (integers).
- No angle is exactly 90°.
- No two angles are equal (the four angles are all different).
- The largest angle is at most 170°, the smallest at least 30°.
For each quadrilateral you design:
(i) List the four angles.
(ii) Show the check: the four angles sum to 360°.
(iii) Could the shape be a parallelogram, rhombus, rectangle, square, kite or trapezium? Justify (most will just be "general / irregular quadrilateral").
Bonus: at least one of your three quadrilaterals must have at least one angle that is reflex-close (greater than 150°).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 132°, 47°, 95°, x
x = 360 − (132 + 47 + 95) = 360 − 274 = 86°.
1.2 — Rhombus with one 110° angle
Opposite angle = 110° (opposite angles of a rhombus are equal). Adjacent angle = 180 − 110 = 70° (co-interior angles supplementary), and its opposite = 70°. Four angles: 110°, 70°, 110°, 70°. Check: 110 + 70 + 110 + 70 = 360 ✓
1.3 — Ratio 1 : 2 : 3 : 4
Parts total = 1 + 2 + 3 + 4 = 10. Each part = 360 ÷ 10 = 36°. Angles: 36°, 72°, 108°, 144°. Check: 36 + 72 + 108 + 144 = 360 ✓
1.4 — 4x°, 4x°, 2x°, 2x°
4x + 4x + 2x + 2x = 12x = 360, so x = 30°. Largest angle = 4(30) = 120°. Four angles: 120°, 120°, 60°, 60°.
1.5 — Trapezium ABCD with AB ∥ DC
Co-interior angles between parallel sides are supplementary. ∠A + ∠D = 180 → ∠D = 180 − 70 = 110°. ∠B + ∠C = 180 → ∠C = 180 − 80 = 100°. Check: 70 + 80 + 100 + 110 = 360 ✓
1.6 — (x+30)°, (2x−10)°, x°, (x+20)°
(x + 30) + (2x − 10) + x + (x + 20) = 360
5x + 40 = 360
5x = 320, x = 64.
Angles: 64+30 = 94°, 2(64)−10 = 118°, 64°, 64+20 = 84°. Check: 94 + 118 + 64 + 84 = 360 ✓
2 — Find the mistake
(a) The mistake is on Line 1 (which is then carried into Lines 2, 3 and 4).
(b) The student used 180° as the angle sum, but that is the rule for a triangle. A quadrilateral has four angles and splits into 2 triangles, so its angle sum is 2 × 180° = 360°.
(c) Corrected working:
Sum of angles of a quadrilateral = 360°
130 + 90 + 70 + x = 360
290 + x = 360
x = 360 − 290 = 70°.
Sanity check: a negative angle (−110°) is impossible inside a normal quadrilateral, which signals an error.
3 — Quadrilateral design (sample solutions)
There are many valid sets. Three good examples:
Quadrilateral 1: 60°, 80°, 100°, 120°. Sum = 360 ✓. All different, none = 90°, all between 30° and 170°. General irregular quadrilateral.
Quadrilateral 2: 40°, 70°, 110°, 140°. Sum = 360 ✓. Same checks pass. General irregular quadrilateral (could be drawn as an irregular trapezium if a pair of sides happens to be parallel — but the angles alone don't force this).
Quadrilateral 3 (with a near-reflex angle): 30°, 50°, 120°, 160°. Sum = 360 ✓. 160° is the bonus near-reflex angle. General irregular quadrilateral.
Why most are not special quadrilaterals: a square or rectangle needs four 90° angles (banned). A parallelogram/rhombus needs opposite angles equal (banned by "all different"). A kite needs one pair of opposite angles equal (banned). A trapezium needs two co-interior angles to be supplementary; this is possible but not forced by the angle list alone.
Marking: 1 mark per valid distinct quadrilateral that passes all four conditions with a 360° check (up to 3 marks). 1 bonus mark if at least one quadrilateral has an angle > 150°.