Geometrical Constructions
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Master the principles of geometric construction to accurately construct triangles, quadrilaterals, angle bisectors, and perpendicular bisectors using only a straightedge and a pair of compasses.
The Surveyor's Secret: Long before GPS and digital laser measures, land surveyors (such as those demarcation officers surveying shambas in the Rift Valley) laid out accurate boundaries using fixed-length ropes and pegs. In geometry, a pair of compasses acts as our fixed-length rope maintaining a constant radius, while a straightedge draws the straight boundary line connecting points.
Fundamental Geometric Principles
(a) Locus of Equidistant Points (Circles & Arcs): A circle is the set of all points that are an exact distance \(r\) from a fixed centre point \(O\). When two arcs of radii \(r_1\) and \(r_2\) drawn from centres \(A\) and \(B\) intersect at \(C\), point \(C\) is guaranteed to satisfy \(AC = r_1\) and \(BC = r_2\). This is the bedrock of constructing triangles given three side lengths (SSS).
(b) Perpendicular Bisector Principle: Any point \(P\) that lies on the perpendicular bisector of a segment \(\overline{AB}\) is equidistant from both endpoints: \[PA = PB\]By swinging arcs of equal radius \(r > \frac{1}{2}AB\) from both \(A\) and \(B\), their two intersection points define the unique perpendicular bisector line.
(c) The Triangle Inequality Law: A triangle can exist if and only if the sum of any two sides is strictly greater than the third side: \[a + b > c, \quad b + c > a, \quad \text{and} \quad a + c > b\]If \(a + b \le c\), the compass arcs will never meet!
Interactive Compass & Straightedge Studio
Adjust the side lengths below to see how compass arcs intersect to fix vertex \(C\). Notice what happens if the sides violate the triangle inequality!
Valid triangle. Click Animate to watch the compass arcs meet!
Key Formulas
Triangle Inequality Theorem:
\[a + b > c, \quad a + c > b, \quad b + c > a\]A triangle can be constructed if and only if every pair of side lengths sums to strictly more than the remaining side length.
Equidistant Locus for Perpendicular Bisector:
\[\text{For any point } P \text{ on the perpendicular bisector of } \overline{AB}: \quad PA = PB\]Equilateral Triangle Altitude (Pythagorean Relation):
\[h = \sqrt{s^2 - \left(\frac{s}{2}\right)^2} = \sqrt{s^2 - \frac{s^2}{4}} = \frac{s\sqrt{3}}{2} \approx 0.866 \times s\]Angle Bisector Property:
\[\text{If } \overline{BD} \text{ bisects } \angle ABC, \text{ then } \angle ABD = \angle DBC = \frac{1}{2} \angle ABC\]Standard Constructible Angles:
- \(60^\circ\): Equilateral triangle construction from a base segment.
- \(90^\circ\): Perpendicular bisector of a straight angle (\(180^\circ\)).
- \(45^\circ\): Bisection of a \(90^\circ\) angle.
- \(30^\circ\): Bisection of a \(60^\circ\) angle.
- \(15^\circ\): Bisection of a \(30^\circ\) angle.
- \(75^\circ\): \(60^\circ + 15^\circ\) or bisection between \(60^\circ\) and \(90^\circ\).
- \(120^\circ\): \(180^\circ - 60^\circ\) or two adjacent \(60^\circ\) angles.
- \(135^\circ\): \(90^\circ + 45^\circ\) or \(180^\circ - 45^\circ\).
Interior Angles of Polygons:
\[\text{Sum for Triangles} = 180^\circ, \qquad \text{Sum for Quadrilaterals} = (4-2) \times 180^\circ = 360^\circ\]Worked Examples
Example 1 (Easy): Constructing an Angle Bisector
Problem: Construct the angle bisector of \(\angle PQR = 76^\circ\) and calculate the measure of each resulting half-angle.
- Step 1: Mark equal radii on the arms: Place the compass needle at vertex \(Q\). Draw an arc cutting ray \(QP\) at \(X\) and ray \(QR\) at \(Y\). Here, \(QX = QY\).
- Step 2: Swing intersecting arcs: With the compass needle at \(X\), swing an arc in the interior of the angle. Keeping the same compass width, place the needle at \(Y\) and swing a second arc intersecting the first at point \(Z\).
- Step 3: Connect vertex to intersection: Draw the straight line from \(Q\) through \(Z\). Line \(QZ\) is the angle bisector.
- Step 4: Calculate the half-angle:\[\angle PQZ = \angle ZQR = \frac{76^\circ}{2} = 38^\circ\]
Example 2 (Medium): Constructing a Triangle given SSS and Finding its Altitude
Problem: A surveyor maps a triangular plot \(ABC\) where \(AB = 8\text{ cm}\), \(BC = 6\text{ cm}\), and \(AC = 10\text{ cm}\).
(a) Verify whether the triangle can be constructed.
(b) Construct the triangle and determine the length of the altitude from \(B\) to \(AC\).
- Step 1 (Check Triangle Inequality):\[8 + 6 = 14 > 10, \quad 8 + 10 = 18 > 6, \quad 6 + 10 = 16 > 8\]All conditions hold, so the triangle is valid. Furthermore, notice \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\), so \(\triangle ABC\) is a right-angled triangle with the right angle at \(B\).
- Step 2 (Construction Procedure):
- Draw base \(AC = 10\text{ cm}\) using a ruler.
- Set compass to \(8\text{ cm}\), place at \(A\), and draw an arc above \(AC\).
- Set compass to \(6\text{ cm}\), place at \(C\), and swing an arc intersecting the previous arc at \(B\).
- Join \(AB\) and \(BC\).
- Step 3 (Calculate Altitude from \(B\) to \(AC\)):
Using area equality \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\):\[\text{Area} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 8 \times 6 = 24\text{ cm}^2\]Using hypotenuse \(AC = 10\text{ cm}\) as base and altitude \(h\):\[\frac{1}{2} \times 10 \times h = 24 \implies 5h = 24 \implies h = \frac{24}{5} = 4.8\text{ cm}\]
Example 3 (Hard): Constructing a Rhombus from Given Diagonals
Problem: A decorative Kiondo woven pattern requires constructing a rhombus \(KLMN\) whose diagonals measure \(d_1 = KM = 12\text{ cm}\) and \(d_2 = LN = 16\text{ cm}\).
(a) Outline the step-by-step compass construction.
(b) Calculate the side length of the rhombus.
- Step 1 (Geometric Property): The diagonals of a rhombus are perpendicular bisectors of each other. Thus, \(KM\) and \(LN\) meet at \(O\) at right angles (\(90^\circ\)), where \(KO = OM = 6\text{ cm}\) and \(LO = ON = 8\text{ cm}\).
- Step 2 (Construction Steps):
- Draw diagonal \(KM = 12\text{ cm}\).
- Construct the perpendicular bisector of \(KM\) by swinging arcs greater than \(6\text{ cm}\) from \(K\) and \(M\). Draw the bisector line passing through midpoint \(O\).
- Set compass radius to \(\frac{16}{2} = 8\text{ cm}\). With needle at \(O\), mark points \(L\) and \(N\) on the perpendicular bisector on opposite sides of \(KM\).
- Join \(K\) to \(L\), \(L\) to \(M\), \(M\) to \(N\), and \(N\) to \(K\) to complete rhombus \(KLMN\).
- Step 3 (Calculate Side Length \(s\)):
In right-angled \(\triangle KOL\):\[s = \sqrt{KO^2 + LO^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}\]Every side of the rhombus is exactly \(10\text{ cm}\).
Common Mistakes
Example: Trying to construct a triangle with sides \(3\text{ cm}\), \(4\text{ cm}\), and \(8\text{ cm}\).
Real World
1. Land Surveying and Title Boundary Demarcation in Kenya
When the Ministry of Lands or local county surveyors demarcate parcels of land (shambas) across Nakuru, Uasin Gishu, or Kiambu, they use triangulation. By measuring accurate baseline distances along boundaries and using bearings (angles) or side lengths, surveyors create precise master title deeds. Accurate geometric constructions prevent boundary disputes between neighbours.
2. Roof Truss and Bridge Engineering
Civil engineers designing steel roof trusses for open-air markets, warehouses in Nairobi's Industrial Area, or footbridges across rivers use triangular frameworks (King-post and Pratt trusses). Triangles are the only rigid polygon that cannot deform without altering side lengths. Drafting these trusses to scale requires accurate angle constructions (\(30^\circ, 45^\circ, 60^\circ\)).
3. Sports Pitch Marking (Nyayo & Kasarani Stadiums)
Groundskeepers marking football pitches must lay out exact right angles (\(90^\circ\)) at the four pitch corners and locate the exact penalty spot and centre circle. They utilize the 3-4-5 rope construction technique (an ancient compass-and-straightedge application of the Pythagorean theorem) and swing measured ropes to trace the centre circle.
Practice