Mathematics • Year 7 • Unit 3 • Lesson 20
Geometry Synthesis — Mixed Challenge
The grand finale. Six problems sweeping every Unit 3 topic. Find a plausible classmate mistake. Design your own multi-topic question that uses three different facts.
1. Mixed problems — six topics, six questions
Each question covers a different area of Unit 3. Show all working with named reasons. 2 marks each
1.1 (Triangle algebra) A triangle has angles 2x, 3x, 4x. Find each angle.
1.2 (Polygon) A regular polygon has each interior angle equal to 144°. How many sides does it have?
1.3 (Parallel lines) Two parallel lines are crossed by a transversal. A pair of co-interior angles are (3x + 20)° and (x + 60)°. Find x.
1.4 (Similarity) △ABC ∼ △DEF. AB = 5, BC = 7, AC = 8, DE = 15. Find EF, DF and the scale factor.
1.5 (Quadrilateral) A parallelogram has one angle of 75°. Find the other three angles, with reasons.
1.6 (Construction) Bisecting a 144° angle twice produces three pieces. State the three angle sizes and check they sum to 144°.
2. Find the mistake
Another Year 7 student tried to solve a synthesis problem combining a parallelogram with a triangle. 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
Problem: A parallelogram ABCD has ∠A = 80°. Diagonal AC makes an angle of 35° with side AB. Find ∠ACB.
Line 1: ∠A + ∠B = 180° (co-int ∠s, AD ∥ BC).
Line 2: ∠B = 180 − 80 = 100°.
Line 3: In △ABC: ∠BAC + ∠ABC + ∠ACB = 360° (∠ sum of △).
Line 4: 35 + 100 + ∠ACB = 360.
Line 5: ∠ACB = 360 − 35 − 100 = 225°. Answer: 225°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (think: triangle vs quadrilateral angle sum).
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Triangles sum to 180°, not 360°. Also: 225° is impossible for an interior angle (interior angles are between 0° and 180°). Sanity-check should have caught it!3. Open-ended challenge — design a 3-topic synthesis problem
This question asks you to invent your own multi-topic problem. Be thorough. 4 marks
3.1 Design a synthesis problem from Unit 3 that uses at least THREE different topics in its solution. Choose three topics from this list and use them in your problem:
• Angle sum of triangle (180°)
• Angle sum of quadrilateral (360°)
• Polygon angle sum (n − 2) × 180°
• Parallel-line angles (alternate, corresponding, co-interior)
• Similarity and scale factor
• Congruence (SSS, SAS, AAS, RHS)
• Isosceles triangle base-angle property
• Properties of a special quadrilateral (square, parallelogram, kite, etc.)
Your problem must include:
(i) A clear description of the figure (or labelled sketch).
(ii) Given values (at least one angle or length).
(iii) A question that requires the THREE chosen topics.
(iv) A FULLY WORKED solution with every step's named reason.
(v) State which three topics you used.
Bonus: Make your problem use FOUR different topics.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Triangle 2x, 3x, 4x
2x + 3x + 4x = 180 (∠ sum of △). 9x = 180, x = 20°.
Angles: 2(20) = 40°, 3(20) = 60°, 4(20) = 80°.
1.2 — Regular polygon with 144° interior angle
Each angle = (n − 2) × 180 / n = 144.
Solve: (n − 2) × 180 = 144n → 180n − 360 = 144n → 36n = 360 → n = 10.
It is a regular decagon.
1.3 — Co-interior with algebra
Co-interior angles add to 180°: (3x + 20) + (x + 60) = 180.
4x + 80 = 180. 4x = 100. x = 25.
Check: 3(25) + 20 = 95°, 25 + 60 = 85°. 95 + 85 = 180 ✓.
1.4 — Similar triangles
SF = DE ÷ AB = 15 ÷ 5 = 3.
EF = BC × 3 = 7 × 3 = 21.
DF = AC × 3 = 8 × 3 = 24.
1.5 — Parallelogram
In a parallelogram, opposite angles are equal and adjacent angles are co-interior on parallel sides (so they add to 180°).
Given ∠A = 75°. So ∠C (opposite) = 75°.
∠B = 180 − 75 = 105° (co-int ∠s, AD ∥ BC). ∠D (opposite ∠B) = 105°.
Check: 75 + 105 + 75 + 105 = 360 ✓.
1.6 — Bisecting 144° twice
First bisection: 144 → 72° + 72°.
Bisect ONE 72° half: 72 → 36° + 36°.
Pieces: 36°, 36°, 72°. Check: 36 + 36 + 72 = 144 ✓.
2 — Find the mistake
(a) The mistake is on Line 3 (and Lines 4–5 propagate it).
(b) The angle sum of a TRIANGLE is 180°, not 360°. The student used the wrong total. (360° is the angle sum of a quadrilateral.) That gave an impossible answer (225° for an interior angle), which a sanity-check would have flagged.
(c) Corrected working:
Line 1: ∠A + ∠B = 180° (co-int ∠s, AD ∥ BC). (unchanged)
Line 2: ∠B = 180 − 80 = 100°. (unchanged)
Line 3 (fixed): In △ABC: ∠BAC + ∠ABC + ∠ACB = 180° (∠ sum of △).
Line 4 (fixed): 35 + 100 + ∠ACB = 180.
Line 5 (fixed): ∠ACB = 180 − 35 − 100 = 45°.
The fix: triangles sum to 180°, quadrilaterals to 360°. Sanity-check answers — interior angles must be between 0° and 180°.
3 — Open-ended challenge (sample solution)
Sample problem: An isosceles triangle △ABC has AB = AC, with apex ∠BAC = 40°. A line through B is parallel to AC, meeting an extension of CA at point D. Find ∠ABD.
Worked solution:
Step 1: Base angles of isosceles △ABC are equal. ∠ABC = ∠ACB = (180 − 40) ÷ 2 = 70° (∠ sum of △, isosceles base angles).
Step 2: Since BD ∥ AC and AB is a transversal: ∠ABD = ∠BAC = 40° (alt ∠s, BD ∥ AC).
Wait — actually we want the angle at the vertex from D-side. Let me redo: ∠DBC and ∠ACB are co-interior on parallel lines BD and AC (transversal BC). So ∠DBC + ∠ACB = 180° → ∠DBC = 180 − 70 = 110°. Then ∠ABD = ∠DBC − ∠ABC = 110 − 70 = 40° — which matches the alternate-angle result (alt ∠s, BD ∥ AC, transversal AB).
Three topics used: (1) ∠ sum of △ (and isosceles property), (2) alternate angles on parallel lines, (3) angle subtraction/co-interior.
Bonus (4 topics): Add: "What scale factor enlarges △ABD by 3?" — adds similarity.
Marking: 1 for clear figure; 1 for given values; 1 for fully worked solution citing 3+ topics; 1 for naming the topics used. Bonus: 1 extra if a fourth topic is woven in.