Corrigé
8.16 Constructions
Pack A — Réponses
| # | Question | Réponse |
|---|---|---|
| 1 | Describe how to measure an angle using a protractor. | Centre protractor on vertex; align baseline with one arm; read where the other arm crosses the scale |
| 2 | A protractor reading is 130°. Identify whether the angle is acute, right, obtuse, or reflex. | Obtuse |
| 3 | Construct an angle of 60° using a compass on a starting line. Describe the steps. | 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 | Describe how to copy a given line segment using a compass and straight-edge. | Open compass to the segment's length; mark off the same length on the target line |
| 5 | A right angle is what number of degrees? Describe how to construct a perpendicular through a point on a line. | 90°. Use the standard perpendicular construction. |
| 6 | Construct an equilateral triangle on a given segment AB using compass and straightedge. Describe the construction. | Two arcs from A and B, radius = AB; intersection point C; join AC and BC. |
| 7 | Sketch and describe how to bisect a line segment AB using compass and ruler. | Arcs of radius > half AB from A and B; line through intersections is perpendicular bisector. |
| 8 | The "seed of life" is a classical construction. Describe it. | Six circles arranged around a central circle, each touching the centre |
| 9 | Sketch the inscribed regular hexagon in a circle. Why does the construction work? | Set compass to radius; step around the circle 6 times — divides circle exactly into 6 equal arcs |
| 10 | Use compass and ruler to construct a 30° angle. Describe. | First construct 60°, then bisect it |
| 11 | Construct a triangle with sides 5 cm, 6 cm, 7 cm using compass and ruler. | Draw base 7 cm; from endpoints draw arcs of 5 and 6; intersection is third vertex |
| 12 | Bisect an angle of 80°. Describe the steps and find the resulting angle. | 40° |
| 13 | Construct a 45° angle. | Build a perpendicular (90°), then bisect |
| 14 | Construct a triangle with sides 6 cm, 6 cm, 6 cm. What type of triangle is this? | Equilateral triangle (all sides equal) |
| 15 | Construct a regular hexagon inscribed in a circle of radius 4 cm. | Step the radius around the circle 6 times, join consecutive points |
| 16 | A perpendicular bisector of segment AB passes through the midpoint and is at 90° to AB. Why is every point on this bisector equidistant from A and B? | By construction: each arc was drawn with the same radius from A and B, so intersection points are equidistant. |
| 17 | Construct a square of side 5 cm. | Build a side, then perpendiculars at each end, mark off 5 cm, join. |
| 18 | Construct a 15° angle. Describe. | Bisect 30° (which itself comes from bisecting 60°) |
| 19 | Why can a 60° angle be constructed but not a 20° angle with compass and straightedge only? | 60° comes from equilateral triangle; 20° = trisection of 60°, which is famously impossible by classical methods. |
| 20 | A construction involves "marking and arc". Describe the difference between drawing an arc and a full circle. | Arc = part of a circle's circumference; full circle = entire boundary |
| 21 | Construct an isosceles triangle with base 6 cm and equal sides 5 cm each. Calculate the height from apex to base. | Height 4 cm |
| 22 | A right-angled triangle has legs 3 and 4. Construct it and verify the hypotenuse using Pythagoras. | Hyp = 5 |
| 23 | Construct the perpendicular bisector of a 10 cm segment. Verify that midpoint is 5 cm from each endpoint and the bisector is at 90°. | Midpoint = 5 cm from each end; bisector at 90° as constructed |
| 24 | Construct a regular pentagon inscribed in a circle. (Outline the steps for the standard ruler-and-compass construction.) | Standard construction using the diagonal of a particular rectangle |
| 25 | A 75° angle can be constructed as 60° + 15° (or 45° + 30°). Describe both methods. | Either start with 60° and add 15° (bisect 30°) — or start with 45° (bisect 90°) and add 30° |
| 26 | Construct the seed-of-life flower (7 interlocking circles). Describe the procedure and the angles created at the centres. | 6 surrounding circles around 1 central; angle between adjacent radii from centre = 60° |
| 27 | A regular hexagon has side 8 cm. Find (a) the radius of the inscribed circle (apothem) and (b) the radius of the circumscribed circle. | (a) $4\sqrt{3} \approx 6.93$ cm (b) 8 cm |
| 28 | Construct a parallel to a given line through an external point. Describe. | Method: pick a point on the line; create an angle at the point; copy the angle at the external point |
| 29 | Construct a triangle with sides 5, 5, 5 cm. Find the area without measuring (use Pythagoras to find the height). | Area $\approx 10.83$ cm² |
| 30 | A protractor reading of 47° is given. Construct (without protractor!) an angle of 47° approximately — list approximate methods and their limits. | Cannot do exactly with compass-and-straightedge alone. Can construct 45° + 2° approximation. |
| 31 | Construct an equilateral triangle on a given segment AB, then construct its **circumscribed circle** (passing through all three vertices). | Find the centroid by bisecting two sides; their intersection is the centre |
| 32 | Describe how to construct a regular pentagon with side 5 cm using only ruler and compass. | Standard "circumscribed-circle" pentagon construction |
| 33 | Verify the "seed of life" structure: six surrounding circles in a hexagonal pattern. Why is this geometric arrangement special? | It demonstrates the hexagonal close-packing in 2D: a perfect tiling of equilateral triangles |
| 34 | Construct a triangle with sides 7 cm, 8 cm, 9 cm. Calculate its area using Heron's formula. | Area $\approx 26.83$ cm² |
| 35 | A complex construction: starting from an equilateral triangle ABC, construct the perpendicular from one vertex to the opposite side. What is the relationship between this perpendicular and the median? | For an equilateral triangle, the altitude, median, and angle bisector from a vertex are all the same line. |
| 36 | Trisect a 90° angle — is this possible using only compass and straightedge? | Yes — 30° can be constructed (60° bisected), so 90° → three 30°s is possible |
| 37 | Construct a regular hexagon with apothem 4 cm. Find the side length. | Side $\approx 4.62$ cm |
| 38 | Construct a triangle similar to a given triangle with sides 3, 4, 5 cm, but with a linear scale factor of 1.5. | Sides: 4.5, 6, 7.5 cm |
| 39 | Construct an inscribed regular dodecagon (12-gon) in a circle. Outline the procedure. | Start with an inscribed hexagon, then bisect each side |
| 40 | In a circle of radius 5 cm, construct an isosceles trapezium with parallel sides 6 cm and 8 cm. | Use chord-bisecting techniques to position the parallel sides |
Pack B — Réponses
| # | Question | Réponse |
|---|---|---|
| 1 | Describe how to measure an angle using a protractor. | Same. |
| 2 | A protractor reading is 130°. Identify whether the angle is acute, right, obtuse, or reflex. | Reflex |
| 3 | Construct an angle of 60° using a compass on a starting line. Describe the steps. | Same. |
| 4 | Describe how to copy a given line segment using a compass and straight-edge. | Same. |
| 5 | A right angle is what number of degrees? Describe how to construct a perpendicular through a point on a line. | Same. |
| 6 | Construct an equilateral triangle on a given segment AB using compass and straightedge. Describe the construction. | Same. |
| 7 | Sketch and describe how to bisect a line segment AB using compass and ruler. | Same. |
| 8 | The "seed of life" is a classical construction. Describe it. | Same. |
| 9 | Sketch the inscribed regular hexagon in a circle. Why does the construction work? | Same. |
| 10 | Use compass and ruler to construct a 30° angle. Describe. | Same. |
| 11 | Construct a triangle with sides 5 cm, 6 cm, 7 cm using compass and ruler. | Same. |
| 12 | Bisect an angle of 80°. Describe the steps and find the resulting angle. | 55° |
| 13 | Construct a 45° angle. | Build 60° + 15° (bisect 30°) |
| 14 | Construct a triangle with sides 6 cm, 6 cm, 6 cm. What type of triangle is this? | Isosceles |
| 15 | Construct a regular hexagon inscribed in a circle of radius 4 cm. | Same. |
| 16 | A perpendicular bisector of segment AB passes through the midpoint and is at 90° to AB. Why is every point on this bisector equidistant from A and B? | Same. |
| 17 | Construct a square of side 5 cm. | Same. |
| 18 | Construct a 15° angle. Describe. | Bisect 45° (from perpendicular bisection) |
| 19 | Why can a 60° angle be constructed but not a 20° angle with compass and straightedge only? | Same. |
| 20 | A construction involves "marking and arc". Describe the difference between drawing an arc and a full circle. | Same. |
| 21 | Construct an isosceles triangle with base 6 cm and equal sides 5 cm each. Calculate the height from apex to base. | Height 3 cm |
| 22 | A right-angled triangle has legs 3 and 4. Construct it and verify the hypotenuse using Pythagoras. | Hyp = 13 |
| 23 | Construct the perpendicular bisector of a 10 cm segment. Verify that midpoint is 5 cm from each endpoint and the bisector is at 90°. | Same. |
| 24 | Construct a regular pentagon inscribed in a circle. (Outline the steps for the standard ruler-and-compass construction.) | Same. |
| 25 | A 75° angle can be constructed as 60° + 15° (or 45° + 30°). Describe both methods. | Same. |
| 26 | Construct the seed-of-life flower (7 interlocking circles). Describe the procedure and the angles created at the centres. | Same. |
| 27 | A regular hexagon has side 8 cm. Find (a) the radius of the inscribed circle (apothem) and (b) the radius of the circumscribed circle. | (a) $5\sqrt{3} \approx 8.66$ (b) 10 |
| 28 | Construct a parallel to a given line through an external point. Describe. | Same. |
| 29 | Construct a triangle with sides 5, 5, 5 cm. Find the area without measuring (use Pythagoras to find the height). | Area $\approx 15.59$ cm² |
| 30 | A protractor reading of 47° is given. Construct (without protractor!) an angle of 47° approximately — list approximate methods and their limits. | Same. |
| 31 | Construct an equilateral triangle on a given segment AB, then construct its **circumscribed circle** (passing through all three vertices). | Same. |
| 32 | Describe how to construct a regular pentagon with side 5 cm using only ruler and compass. | Same. |
| 33 | Verify the "seed of life" structure: six surrounding circles in a hexagonal pattern. Why is this geometric arrangement special? | Same. |
| 34 | Construct a triangle with sides 7 cm, 8 cm, 9 cm. Calculate its area using Heron's formula. | Area = 24 cm² (right triangle) |
| 35 | A complex construction: starting from an equilateral triangle ABC, construct the perpendicular from one vertex to the opposite side. What is the relationship between this perpendicular and the median? | Same. |
| 36 | Trisect a 90° angle — is this possible using only compass and straightedge? | Same. |
| 37 | Construct a regular hexagon with apothem 4 cm. Find the side length. | Side $\approx 5.77$ cm |
| 38 | Construct a triangle similar to a given triangle with sides 3, 4, 5 cm, but with a linear scale factor of 1.5. | 6, 8, 10 cm |
| 39 | Construct an inscribed regular dodecagon (12-gon) in a circle. Outline the procedure. | Same. |
| 40 | In a circle of radius 5 cm, construct an isosceles trapezium with parallel sides 6 cm and 8 cm. | Same. |
Problèmes — Solutions détaillées
1
**Constructing an equilateral triangle.** Use ruler and compass to construct an equilateral triangle on the segment AB = 8 cm. Then: (a) Verify the construction works (all sides equal). (b) Find the height from one vertex to the opposite side. (c) Calculate the area.
Réponse
(a) AC = BC = AB = 8 cm (by construction) (b) $4\sqrt{3} \approx 6.93$ cm (c) $16\sqrt{3} \approx 27.71$ cm²
(a) Compass arcs from A and B of radius 8 → intersection C. By construction, AC = BC = 8. Plus AB = 8 → all equal.
(b) Height $h = \sqrt{8^2 - 4^2} = \sqrt{48} = 4\sqrt{3}$.
(c) Area = $(1/2)(8)(4\sqrt{3}) = 16\sqrt{3}$.
2
**Perpendicular bisector.** Construct the perpendicular bisector of a 12 cm segment AB. (a) Describe the construction. (b) Verify the midpoint is 6 cm from each end. (c) Justify why the bisector is the locus of points equidistant from A and B.
Réponse
(a) See working (b) Halves of AB (c) Geometric proof via arcs of equal radius
(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.
3
**Hexagonal patio plan.** Design a regular hexagonal patio with side 3 m using compass-and-ruler-only methods on a 1:50 scale plan. (a) State the side length on the plan. (b) Describe the construction. (c) Find the area of the patio (real).
Réponse
(a) 6 cm (b) Inscribed hexagon construction (c) ≈ 23.4 m²
(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 $= \tfrac{3\sqrt{3}}{2}(3)^2 = \tfrac{27\sqrt{3}}{2} \approx 23.4$ m².
4
**Constructing 75°.** Construct an angle of 75° using only ruler and compass. (a) Describe two different methods. (b) Verify with a protractor (in principle).
Réponse
(a) 60° + 15° or 45° + 30° (b) Check: should read 75°
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°.
5
**Seed of life.** Starting with one central circle of radius 4 cm, construct six interlocking circles forming the "seed of life". (a) Describe the procedure step-by-step. (b) Find the radius of the surrounding hexagon's circumcircle. (c) Why does the construction close exactly after 6 circles?
Réponse
(a) Standard procedure (b) 8 cm (c) Hexagonal symmetry — radii equal sides of inscribed hexagon
(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 $2 \times 4 = 8$ 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.
6
**Constructing a square from a circle.** A circle has radius 5 cm. Construct a square such that all four corners touch the circle. (a) Find the side length of the inscribed square. (b) Find the area of the square. (c) What fraction of the circle is the square?
Réponse
(a) $5\sqrt{2}$ cm (b) 50 cm² (c) ≈ 63.7%
(a) Inscribed square: diagonal = diameter = 10. Side $= 10/\sqrt{2} = 5\sqrt{2}$.
(b) Side² = 50 cm².
(c) Circle area $\approx \pi \times 25 \approx 78.5$. Square/Circle $= 50/78.5 \approx 0.637 = 63.7\%$.
7
**Angle bisection.** Bisect a 100° angle using compass and straightedge. (a) Describe the procedure. (b) Find the resulting half-angle. (c) Use the bisection again to find a 25° angle.
Réponse
(a) Standard bisection (b) 50° (c) Bisect 50° to get 25°
(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°.
8
**Constructing a regular octagon.** Inscribe a regular octagon in a circle of radius 6 cm. (a) Outline the construction (starting from a square). (b) Find the side length of the octagon. (c) Find the area.
Réponse
(a) Start with square; bisect each arc; result has 8 vertices (b) ≈ 4.59 cm (c) ≈ 101.8 cm²
(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: $s = 2 r \sin(180°/8) = 12 \sin(22.5°) \approx 4.59$ cm.
(c) Area = $2 r^2 \sqrt{2} \approx 2 \times 36 \times 1.414 \approx 101.8$ cm². (Or use general regular-polygon formula.)
9
**Triangle from three lengths.** Construct a triangle with sides 4 cm, 7 cm, 9 cm. (a) Verify the triangle inequality. (b) Describe the construction. (c) Find the height from the 7 cm side to the opposite vertex.
Réponse
(a) 4+7=11 > 9 ✓ (b) Standard SSS construction (c) ≈ 3.32 cm
(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: $s = 10$, Area $= \sqrt{10 \times 6 \times 3 \times 1} = \sqrt{180} \approx 13.42$. Height from the 7 cm side: $h = 2A/b = 26.83/7 \approx 3.83$. Hmm — recompute. Heron: $s = (4+7+9)/2 = 10$. $A = \sqrt{10 \times 6 \times 3 \times 1} = \sqrt{180} \approx 13.42$. Wait: but the heights depend on which side is the base. Height to the 9 cm side: $2A/9 \approx 2.98$. Height to the 7 cm side: $2A/7 \approx 3.83$. **Correct answer: $\approx 3.83$ cm**.
10
**Constructing a parallel line.** Through point P (not on line $\ell$), construct a line parallel to $\ell$ using only compass and ruler. (a) Describe the method. (b) Justify why the new line is parallel. (c) Use this to construct a parallelogram with side 6 cm and adjacent side 4 cm.
Réponse
(a) Copy an angle (b) Corresponding angles equal (c) Standard construction
(a) Draw a transversal from P to $\ell$, 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 $\ell$.
(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.
11
**Apothem and circumradius.** A regular hexagon has side 6 cm. (a) Construct the hexagon. (b) Find the inscribed-circle radius (apothem). (c) Find the circumscribed-circle radius. (d) Compute the ratio of the two radii.
Réponse
(a) Standard hexagon construction (b) $3\sqrt{3}$ cm ≈ 5.20 (c) 6 cm (d) ≈ 1.155
(a) Draw circle radius 6 cm. Step radius around 6 times. Join consecutive marks.
(b) Apothem = $s\sqrt{3}/2 = 3\sqrt{3}$ ≈ 5.20 cm.
(c) Circumradius = side = 6 cm.
(d) Ratio = $6/(3\sqrt{3}) = 2/\sqrt{3} \approx 1.155$.
12
**Combining constructions.** Construct (i) an equilateral triangle of side 6 cm, (ii) inscribe its circumscribed circle, and (iii) inscribe a regular hexagon in the same circle. (a) Describe each step. (b) Find the side length of the hexagon. (c) Compare the area of the triangle and the hexagon.
Réponse
(a) Standard constructions (b) Side of hexagon = circumradius of triangle = $2\sqrt{3}$ cm (c) Triangle area: $9\sqrt{3}$ ≈ 15.59. Hexagon area: $18\sqrt{3}$ ≈ 31.18. Hexagon area = 2 × triangle area.
(a) (i) Equilateral triangle: arcs of 6 from endpoints. (ii) Circumscribed circle: find centroid (intersection of medians); radius = $2/3$ of median length. Median = $3\sqrt{3}$. Circumradius = $2\sqrt{3}$ cm. (iii) Step this radius around the circle 6 times.
(b) Hexagon side = circumradius = $2\sqrt{3}$ cm.
(c) Triangle area = $(\sqrt{3}/4) \times 36 = 9\sqrt{3} \approx 15.6$ cm². Hexagon area = $(3\sqrt{3}/2)(2\sqrt{3})^2 = (3\sqrt{3}/2)(12) = 18\sqrt{3} \approx 31.2$ cm². The hexagon is exactly **twice** the triangle's area — a striking geometric fact.