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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 8 · 8.16 Constructions

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.Centre protractor on vertex; align baseline with one arm; read where the other arm crosses the scale
2.Obtuse
3.Draw a ray. Open compass to any radius. From vertex draw an arc cutting the ray at P. From P with same radius draw another arc cutting the first. Join.
4.Open compass to the segment's length; mark off the same length on the target line
5.90°. Use the standard perpendicular construction.
6.Two arcs from A and B, radius = AB; intersection point C; join AC and BC.
7.Arcs of radius > half AB from A and B; line through intersections is perpendicular bisector.
8.Six circles arranged around a central circle, each touching the centre
9.Set compass to radius; step around the circle 6 times — divides circle exactly into 6 equal arcs
10.First construct 60°, then bisect it
Silver
11.Draw base 7 cm; from endpoints draw arcs of 5 and 6; intersection is third vertex
12.40°
13.Build a perpendicular (90°), then bisect
14.Equilateral triangle (all sides equal)
15.Step the radius around the circle 6 times, join consecutive points
16.By construction: each arc was drawn with the same radius from A and B, so intersection points are equidistant.
17.Build a side, then perpendiculars at each end, mark off 5 cm, join.
18.Bisect 30° (which itself comes from bisecting 60°)
19.60° comes from equilateral triangle; 20° = trisection of 60°, which is famously impossible by classical methods.
20.Arc = part of a circle's circumference; full circle = entire boundary
Gold
21.Height 4 cm
22.Hyp = 5
23.Midpoint = 5 cm from each end; bisector at 90° as constructed
24.Standard construction using the diagonal of a particular rectangle
25.Either start with 60° and add 15° (bisect 30°) — or start with 45° (bisect 90°) and add 30°
26.6 surrounding circles around 1 central; angle between adjacent radii from centre = 60°
27.(a) 43≈6.934\sqrt{3} \approx 6.93 cm (b) 8 cm
28.Method: pick a point on the line; create an angle at the point; copy the angle at the external point
29.Area ≈10.83\approx 10.83 cm²
30.Cannot do exactly with compass-and-straightedge alone. Can construct 45° + 2° approximation.
Platinum
31.Find the centroid by bisecting two sides; their intersection is the centre
32.Standard "circumscribed-circle" pentagon construction
33.It demonstrates the hexagonal close-packing in 2D: a perfect tiling of equilateral triangles
34.Area ≈26.83\approx 26.83 cm²
35.For an equilateral triangle, the altitude, median, and angle bisector from a vertex are all the same line.
36.Yes — 30° can be constructed (60° bisected), so 90° → three 30°s is possible
37.Side ≈4.62\approx 4.62 cm
38.Sides: 4.5, 6, 7.5 cm
39.Start with an inscribed hexagon, then bisect each side
40.Use chord-bisecting techniques to position the parallel sides

Pack B — Answers

Bronze
1.Same.
2.Reflex
3.Same.
4.Same.
5.Same.
6.Same.
7.Same.
8.Same.
9.Same.
10.Same.
Silver
11.Same.
12.55°
13.Build 60° + 15° (bisect 30°)
14.Isosceles
15.Same.
16.Same.
17.Same.
18.Bisect 45° (from perpendicular bisection)
19.Same.
20.Same.
Gold
21.Height 3 cm
22.Hyp = 13
23.Same.
24.Same.
25.Same.
26.Same.
27.(a) 53≈8.665\sqrt{3} \approx 8.66 (b) 10
28.Same.
29.Area ≈15.59\approx 15.59 cm²
30.Same.
Platinum
31.Same.
32.Same.
33.Same.
34.Area = 24 cm² (right triangle)
35.Same.
36.Same.
37.Side ≈5.77\approx 5.77 cm
38.6, 8, 10 cm
39.Same.
40.Same.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) AC = BC = AB = 8 cm (by construction) (b) 43≈6.934\sqrt{3} \approx 6.93 cm (c) 163≈27.7116\sqrt{3} \approx 27.71 cm²
Full working
(a) Compass arcs from A and B of radius 8 → intersection C. By construction, AC = BC = 8. Plus AB = 8 → all equal.

(b) Height h=82−42=48=43h = \sqrt{8^2 - 4^2} = \sqrt{48} = 4\sqrt{3}.

(c) Area = (1/2)(8)(43)=163(1/2)(8)(4\sqrt{3}) = 16\sqrt{3}.
2Problem 2
Answer
(a) See working (b) Halves of AB (c) Geometric proof via arcs of equal radius
Full working
(a) Open compass to > 6 cm. Centre A, draw arcs above and below the line. Centre B (same radius), draw arcs intersecting the first two. Join the two intersection points.

(b) Midpoint M is constructed; AM = MB = 6 cm. Verify with ruler.

(c) Each intersection point P satisfies PA = PB (same compass setting). The locus of all such points is the perpendicular bisector — points equidistant from A and B form exactly this line.
3Problem 3
Answer
(a) 6 cm (b) Inscribed hexagon construction (c) ≈ 23.4 m²
Full working
(a) Scale 1:50 → 3 m = 6 cm on plan.

(b) Draw a circle of radius 6 cm. Step the radius around the circumference 6 times. Join consecutive marks.

(c) Real area =332(3)2=2732≈23.4= \tfrac{3\sqrt{3}}{2}(3)^2 = \tfrac{27\sqrt{3}}{2} \approx 23.4 m².
4Problem 4
Answer
(a) 60° + 15° or 45° + 30° (b) Check: should read 75°
Full working
Method 1: Construct 60° (equilateral triangle). Bisect to get 30°. Bisect again to get 15°. Add 15° to 60°: total 75°.

Method 2: Construct 90° (perpendicular). Bisect to get 45°. Construct 30° (bisect 60°). Add 30° to 45°: total 75°.

(b) Measure with protractor — should read 75°.
5Problem 5
Answer
(a) Standard procedure (b) 8 cm (c) Hexagonal symmetry — radii equal sides of inscribed hexagon
Full working
(a) Steps:
1. Draw the central circle C₀ of radius 4 cm.
2. Mark a point P₁ on its circumference.
3. With centre P₁, radius 4 cm, draw circle C₁.
4. Mark P₂ where C₀ and C₁ intersect (next position around).
5. With centre P₂, draw C₂. Repeat for P₃, P₄, P₅, P₆.
6. The sixth circle's endpoint coincides with P₁ — close-up.

(b) The 6 surrounding circles form a hexagon; their centres lie on a circle of radius 4 cm (= radius of C₀). The outer boundary (where the 6 outermost points lie) has radius 2×4=82 \times 4 = 8 cm.

(c) The 6 surrounding centres form an equilateral hexagon — six 60° rotations around the central point. Six 60°s = 360°, so the construction is rotationally symmetric and closes exactly.
6Problem 6
Answer
(a) 525\sqrt{2} cm (b) 50 cm² (c) ≈ 63.7%
Full working
(a) Inscribed square: diagonal = diameter = 10. Side =10/2=52= 10/\sqrt{2} = 5\sqrt{2}.

(b) Side² = 50 cm².

(c) Circle area ≈π×25≈78.5\approx \pi \times 25 \approx 78.5. Square/Circle =50/78.5≈0.637=63.7%= 50/78.5 \approx 0.637 = 63.7\%.
7Problem 7
Answer
(a) Standard bisection (b) 50° (c) Bisect 50° to get 25°
Full working
(a) From the vertex, draw an arc cutting both arms at points P and Q. From P and Q (same radius), draw arcs intersecting at R. Vertex-through-R is the bisector.

(b) Half-angle = 50°.

(c) Bisect 50° using the same procedure: result 25°.
8Problem 8
Answer
(a) Start with square; bisect each arc; result has 8 vertices (b) ≈ 4.59 cm (c) ≈ 101.8 cm²
Full working
(a) Steps:
1. Draw circle radius 6 cm.
2. Construct a square inscribed in the circle (using horizontal and vertical diameters).
3. Bisect each of the four arcs between adjacent square vertices.
4. The 4 new bisection points + 4 square vertices = 8 evenly-spaced points → octagon.

(b) Each side: s=2rsin⁡(180°/8)=12sin⁡(22.5°)≈4.59s = 2 r \sin(180°/8) = 12 \sin(22.5°) \approx 4.59 cm.

(c) Area = 2r22≈2×36×1.414≈101.82 r^2 \sqrt{2} \approx 2 \times 36 \times 1.414 \approx 101.8 cm². (Or use general regular-polygon formula.)
9Problem 9
Answer
(a) 4+7=11 > 9 ✓ (b) Standard SSS construction (c) ≈ 3.32 cm
Full working
(a) 4+7=11>9 ✓; 4+9=13>7 ✓; 7+9=16>4 ✓. Triangle exists.

(b) Draw base 9 cm. From one endpoint, arc radius 4. From other, arc radius 7. Intersection = third vertex.

(c) Use Heron's: s=10s = 10, Area =10×6×3×1=180≈13.42= \sqrt{10 \times 6 \times 3 \times 1} = \sqrt{180} \approx 13.42. Height from the 7 cm side: h=2A/b=26.83/7≈3.83h = 2A/b = 26.83/7 \approx 3.83. Hmm — recompute. Heron: s=(4+7+9)/2=10s = (4+7+9)/2 = 10. A=10×6×3×1=180≈13.42A = \sqrt{10 \times 6 \times 3 \times 1} = \sqrt{180} \approx 13.42. Wait: but the heights depend on which side is the base. Height to the 9 cm side: 2A/9≈2.982A/9 \approx 2.98. Height to the 7 cm side: 2A/7≈3.832A/7 \approx 3.83. **Correct answer: ≈3.83\approx 3.83 cm**.
10Problem 10
Answer
(a) Copy an angle (b) Corresponding angles equal (c) Standard construction
Full working
(a) Draw a transversal from P to ℓ\ell, hitting at A. Construct the angle at A on the line. Copy this angle at P. The new ray from P is parallel.

(b) The angle copied at P matches the angle at A — by corresponding-angle property of parallel lines, the new ray must be parallel to ℓ\ell.

(c) Parallelogram with sides 6 and 4: draw the 6 cm side as AB. At A, construct a 60° angle (or any chosen angle) and mark D at distance 4 cm along this ray. Through B and D, construct lines parallel to AD and AB respectively — their intersection is C.
11Problem 11
Answer
(a) Standard hexagon construction (b) 333\sqrt{3} cm ≈ 5.20 (c) 6 cm (d) ≈ 1.155
Full working
(a) Draw circle radius 6 cm. Step radius around 6 times. Join consecutive marks.

(b) Apothem = s3/2=33s\sqrt{3}/2 = 3\sqrt{3} ≈ 5.20 cm.

(c) Circumradius = side = 6 cm.

(d) Ratio = 6/(33)=2/3≈1.1556/(3\sqrt{3}) = 2/\sqrt{3} \approx 1.155.
12Problem 12
Answer
(a) Standard constructions (b) Side of hexagon = circumradius of triangle = 232\sqrt{3} cm (c) Triangle area: 939\sqrt{3} ≈ 15.59. Hexagon area: 18318\sqrt{3} ≈ 31.18. Hexagon area = 2 × triangle area.
Full working
(a) (i) Equilateral triangle: arcs of 6 from endpoints. (ii) Circumscribed circle: find centroid (intersection of medians); radius = 2/32/3 of median length. Median = 333\sqrt{3}. Circumradius = 232\sqrt{3} cm. (iii) Step this radius around the circle 6 times.

(b) Hexagon side = circumradius = 232\sqrt{3} cm.

(c) Triangle area = (3/4)×36=93≈15.6(\sqrt{3}/4) \times 36 = 9\sqrt{3} \approx 15.6 cm². Hexagon area = (33/2)(23)2=(33/2)(12)=183≈31.2(3\sqrt{3}/2)(2\sqrt{3})^2 = (3\sqrt{3}/2)(12) = 18\sqrt{3} \approx 31.2 cm². The hexagon is exactly **twice** the triangle's area — a striking geometric fact.