Mathematics • Year 7 • Unit 3 • Lesson 19
Constructions — Mixed Challenge
Combine all three constructions, justify why they work, spot a plausible Year-7 mistake, and tackle an open-ended design puzzle.
1. Mixed problems
Each question mixes ideas from the lesson. Answer clearly. 2 marks each
1.1 Name the three constructions from Lesson 19 (one phrase each).
1.2 Bisecting an angle of 144° creates two angles of size ___° each.
1.3 A segment AB has length 12 cm. After perpendicular bisection, what is the length of each half? Where does the bisector cross AB?
1.4 An angle of x° is bisected, producing two angles of 27° each. Find x.
1.5 Why is a protractor NOT allowed in a classical construction? (1–2 sentences.)
1.6 Bisecting an angle THREE times in a row (bisect, then bisect a half, then bisect a quarter) produces a piece of size 1/8 of the original. If the original is 240°, what is the smallest piece produced?
2. Find the mistake
Another Year 7 student tried to construct the perpendicular bisector of AB. Their working is shown below. Exactly one step contains a mistake that breaks the construction. Spot it, explain why it's wrong, then describe the correct step. 3 marks
Student's working — perpendicular bisector of AB:
Step 1: Open the compass to about 3/4 of AB.
Step 2: Place compass on A; draw an arc above AB only (skipped the arc below).
Step 3: Same radius from B; draw an arc above AB.
Step 4: Join the upper crossing to the midpoint of AB by eye.
(a) Which step contains the mistake?
(b) Explain in one or two sentences why that step breaks the construction (think: where does the perpendicular bisector come from?).
(c) Write the corrected version of Step 2 (or Step 4, depending on which one you chose) and explain how the corrected step properly produces the perpendicular bisector.
Stuck? The construction needs TWO crossing points (one above, one below) so the line between them is a real straight-edge line — not eyeballed. Step 2 should draw arcs both ABOVE and BELOW.3. Open-ended challenge — build a list of constructible angles
This question has many correct answers. Show your work clearly. 4 marks
3.1 Starting with a straight angle (180°) and using ONLY the perpendicular and angle-bisector constructions (no protractor), how many DIFFERENT exact angles can you produce? List at least five different sizes you can construct, and for each one explain the chain of constructions used.
Example to start you off: 90° from perpendicular at a point on a straight line. 45° by bisecting 90°. 22.5° by bisecting 45°.
Bonus: Can you make 60° using these tools alone? (Hint: yes — using a special property of an equilateral triangle constructed with two equal-radius arcs from the endpoints of a segment.) Describe how.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Three constructions
(i) Perpendicular bisector of a segment. (ii) Angle bisector. (iii) Perpendicular from a point to a line.
1.2 — Bisecting 144°
Each angle = 144 ÷ 2 = 72°.
1.3 — Perpendicular bisector of 12 cm segment
Each half = 12 ÷ 2 = 6 cm. The bisector crosses AB at the midpoint and is perpendicular (90°) to it.
1.4 — Original angle from halves
x = 2 × 27 = 54°.
1.5 — Why no protractor
Classical constructions use ONLY a straight-edge (no measurements with it) and a pair of compasses (for arcs). Using a protractor would mean reading off degree measurements directly, which is not allowed. The whole point is to produce exact geometric figures from arcs and lines, not from measured numbers.
1.6 — Three halvings of 240°
240 ÷ 2 = 120; 120 ÷ 2 = 60; 60 ÷ 2 = 30°. The smallest piece produced is 30°.
2 — Find the mistake
(a) The mistake is in Step 2 (and Step 4 makes it worse by eyeballing).
(b) The construction needs the arcs from A to cross the arcs from B in TWO places (one above AB, one below). Drawing only the upper arc means you have only ONE crossing point and you can't draw a unique straight line through one point — you need TWO points for a line. The student then "joined to the midpoint by eye" — but the whole reason for the construction is to AVOID eyeballing.
(c) Corrected Step 2: "Place compass on A; draw an arc ABOVE AB AND an arc BELOW AB." Combined with Step 3 (arcs above and below from B), you now have TWO crossing points (one upper, one lower). Step 4 should be: "Use the straight-edge to join the upper crossing to the lower crossing." This gives the exact perpendicular bisector — no eyeballing, no measuring.
3 — Open-ended challenge (sample solutions)
Constructible angles from 180° using only perpendicular and angle bisection:
90° — perpendicular at a point on a line.
45° — bisect 90°.
22.5° — bisect 45°.
11.25° — bisect 22.5°.
67.5° — bisect 90° to get two 45° halves, then bisect one half (45 → 22.5), and add to the unbisected half (45 + 22.5).
135° — bisect a straight angle (180 → 90 + 90), then add 45° from a further bisection.
Many more by combining halves.
Bonus (60°): Pick a segment AB. With compass radius = length of AB, draw an arc from A and another arc from B. The arcs cross at a point C. Triangle ABC has all sides equal (length AB) — it's equilateral, so every angle = 60°. So ∠BAC = ∠ABC = ∠ACB = 60°. From there you can also construct 30° (bisect 60°), 15° (bisect 30°), 120° (60 + 60), etc.
Marking: 1 mark per pair of valid distinct angles with construction described (3 marks for 5+ different angles), 1 mark for any valid 60° construction in the bonus.