Mathematics • Year 7 • Unit 3 • Lesson 9
Special Quadrilaterals — Mixed Challenge
Mix every special quadrilateral: rhombus, square, kite and trapezium. Combine the 360° angle sum with shape-specific properties (perpendicular diagonals, equal-angle pairs, parallel sides). Spot a property mix-up, then design your own shape puzzle.
1. Mixed problems
Show your working with a reason in brackets. 2 marks each
1.1 Rhombus ABCD has ∠A = 130°. Find ∠B, ∠C and ∠D.
1.2 Kite ABCD has ∠A = 130° (between the two long sides) and ∠C = 50° (between the two short sides). Find ∠B and ∠D.
1.3 Trapezium ABCD has AB ∥ DC. ∠A = 100° and ∠B = 90°. Find ∠C and ∠D.
1.4 Square ABCD has diagonals AC and BD meeting at M. State (i) the angle at M between the two diagonals, (ii) the angle between AC and side AB.
1.5 A quadrilateral with 4 equal sides has angles 60°, 120°, 60°, 120°. Name the shape using its most specific name and explain.
1.6 In trapezium ABCD with AB ∥ DC, the angles are 2x°, 3x°, 4x° and x°. Find x and state every angle. (Assume the labels go around the trapezium in order.)
2. Find the mistake
A Year 7 student tried to find ∠B and ∠D in kite ABCD with ∠A = 80° and ∠C = 40°. Their working is shown. Exactly one line is wrong. Spot it, explain why, then redo the working. 3 marks
Student's working — kite ABCD, ∠A = 80°, ∠C = 40°:
Line 1: In a kite, all four angles are equal because there are two axes of symmetry.
Line 2: ∠B = ∠D (only one pair of opposite angles is equal in a kite — the pair NOT at the symmetry vertices).
Line 3: 80 + ∠B + 40 + ∠B = 360.
Line 4: 2∠B = 240, so ∠B = ∠D = 120°.
(a) Which line contains the conceptual error?
(b) Explain in one or two sentences why that line is wrong.
(c) Confirm whether the final answer ∠B = ∠D = 120° is correct, and redo the working with the right reasoning.
Stuck? Kites have ONE axis of symmetry (through the equal-side vertices), not two. The "all four angles equal" line is the wrong claim — but the final number can still come out right.3. Open-ended challenge — design a "one-given" shape puzzle
This question has more than one correct answer. 4 marks
3.1 Pick ONE special quadrilateral (rhombus, square, kite or trapezium) and design a puzzle where ONE single given measurement is enough to find every other angle in the shape.
Requirements: (i) name the shape and draw it with all four vertices labelled; (ii) state the single given measurement; (iii) work out every other angle, with a property in brackets after each line; (iv) state which property of your shape made one number enough.
Suggestions: rhombus + 1 angle (use opposite-equal + co-interior); square + nothing (all 4 are 90°!); kite + 2 angles is normally needed, so think carefully; trapezium AB ∥ DC + 2 angles on one parallel side.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Rhombus ABCD, ∠A = 130°
∠C = 130° (opp. angles of parallelogram).
∠B = 180 − 130 = 50° (co-int. angles, AB ∥ DC).
∠D = 50° (opp. angles). Check: 130 + 50 + 130 + 50 = 360° ✓.
1.2 — Kite ABCD, ∠A = 130°, ∠C = 50°
∠B = ∠D (kite). 130 + ∠B + 50 + ∠B = 360 (∠ sum of quad). 2∠B = 180, so ∠B = ∠D = 90°.
Check: 130 + 90 + 50 + 90 = 360° ✓.
1.3 — Trapezium ABCD, AB ∥ DC, ∠A = 100°, ∠B = 90°
∠D = 180 − 100 = 80° (co-int. on leg AD).
∠C = 180 − 90 = 90° (co-int. on leg BC). Check: 100 + 90 + 90 + 80 = 360° ✓.
1.4 — Square ABCD diagonals
(i) Diagonals are perpendicular: angle at M between AC and BD = 90° (diagonals of square perpendicular).
(ii) Each diagonal bisects a 90° corner, so the angle between AC and AB = 45° (diagonals of square bisect corners).
1.5 — 4 equal sides, angles 60°, 120°, 60°, 120°
4 equal sides → rhombus. The angles satisfy the parallelogram properties (opposite equal, adjacent co-interior: 60 + 120 = 180 ✓). NOT a square (no right angles). Most specific name: rhombus.
1.6 — Trapezium with angles 2x, 3x, 4x, x
2x + 3x + 4x + x = 360 (∠ sum of quad). 10x = 360, so x = 36.
Angles = 72°, 108°, 144°, 36°. Check: 72 + 108 + 144 + 36 = 360° ✓. (Note: in a real trapezium with AB ∥ DC, the co-interior pairs on each leg must sum to 180° — students who solve and then notice 72 + 108 = 180 ✓ and 144 + 36 = 180 ✓ are confirming the shape works as labelled.)
2 — Find the mistake
(a) The error is on Line 1.
(b) A kite has only ONE axis of symmetry (through the vertices where the unequal-length sides meet). It does NOT have all four angles equal — that would be a rectangle or square.
(c) The final number ∠B = ∠D = 120° is correct, but the path got there by lucky guessing. Correct working:
Kite has one axis of symmetry through AC, so ∠B = ∠D (one pair of opposite angles equal in a kite).
∠A + ∠B + ∠C + ∠D = 360 (∠ sum of quad).
80 + ∠B + 40 + ∠B = 360.
2∠B = 240, so ∠B = ∠D = 120°. Check: 80 + 120 + 40 + 120 = 360° ✓.
3 — Open-ended challenge (sample solution)
Sample design. Rhombus ABCD with one given: ∠A = 50°.
∠C = ∠A = 50° (opp. angles of parallelogram; rhombus ⊂ parallelogram).
∠B = 180 − 50 = 130° (co-int. angles, AB ∥ DC).
∠D = ∠B = 130° (opp. angles).
Check: 50 + 130 + 50 + 130 = 360° ✓.
Why one number is enough: a rhombus's "opposite-angles-equal" property pairs ∠A with ∠C, and "co-interior-angles" forces the other two to be 180° − 50° = 130° each. Together these fix every angle from just one given.
Marking: 1 for choosing a shape + one given; 1 for using shape-specific properties; 1 for correct angles + check; 1 for the "why one number is enough" explanation. Accept any valid shape design.