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: Construct line bisectors, angle bisectors, perpendiculars, and triangles using only a straightedge and a pair of compasses.
Geometric construction is the art of drawing precise shapes, lines, and angles using only two classical tools:
- A Straightedge (Unmarked Ruler): Used solely to draw straight lines between two points.
- A Compass: Used to transfer fixed distances and draw circles or arcs of equal radius.
Interactive Compass: Perpendicular Bisector
Click the button or canvas to advance each step of the construction.
Step 1: Click to swing an arc from point A (radius > half of AB).
Core Principles of Construction
- Equidistance Principle: Any point on the perpendicular bisector of a segment \(AB\) is at the exact same distance from point \(A\) as it is from point \(B\). When two arcs of equal radius intersect at points \(C\) and \(D\), both points satisfy \(CA = CB\) and \(DA = DB\). Connecting \(C\) and \(D\) produces the perpendicular bisector.
- Angular Symmetry: An angle bisector is the locus of points equidistant from both arms (rays) of an angle. By creating equal lengths along both rays, we form an isosceles or rhombus framework where the diagonal splits the angle into two equal halves.
- Rigid Triangles (SSS Congruence): A triangle is uniquely fixed if all three side lengths are given. Compass arcs set each vertex at the exact specified distance from the others.
Fundamental Rule: When constructing perpendicular or angle bisectors, never change the compass setting between drawing pair arcs from the two reference points!
Key Formulas
1. Perpendicular Bisector & Midpoint
If \(M\) is the midpoint of segment \(AB\) with length \(d\):
\[AM = MB = \frac{d}{2}\]The line \(\ell_{\perp}\) passes through \(M\) and forms right angles: \(\ell_{\perp} \perp AB\) (i.e., \(90^\circ\)).
2. Angle Bisector
When ray \(BD\) bisects angle \(\angle ABC\):
\[\angle ABD = \angle DBC = \frac{1}{2} \angle ABC\]For instance, bisecting \(60^\circ\) yields two \(30^\circ\) angles; bisecting \(90^\circ\) yields two \(45^\circ\) angles.
3. Special Constructible Angles
Using compass and straightedge alone:
- \(60^\circ\): The interior angle of an equilateral triangle (radius equals base).
- \(90^\circ\): Perpendicular line or bisector of a straight line (\(180^\circ\)).
- \(30^\circ\): Bisect \(60^\circ\).
- \(45^\circ\): Bisect \(90^\circ\).
- \(75^\circ\): \(45^\circ + 30^\circ\) or bisecting between \(60^\circ\) and \(90^\circ\).
4. Circumcentre of a Right-Angled Triangle
The circumcentre (intersection of perpendicular bisectors) of a right-angled triangle lies exactly on the midpoint of the hypotenuse \(c\):
\[R = \frac{c}{2} = \frac{\sqrt{a^2 + b^2}}{2}\]Worked Examples
Problem: A surveyor in Nakuru marks a corner boundary of \(70^\circ\). She uses a straightedge and compass to construct the angle bisector. What is the size of each smaller angle?
Step-by-Step Solution:
- Identify the formula: An angle bisector divides the total angle into two equal parts: \(\theta_{\text{half}} = \frac{\theta}{2}\).
- Calculate:\[\theta_{\text{half}} = \frac{70^\circ}{2} = 35^\circ\]
Answer: Each angle measures \(35^\circ\).
Problem: A carpenter in Nairobi is preparing a wooden timber beam of length \(AB = 18\text{ cm}\). He constructs the perpendicular bisector of the beam, which intersects \(AB\) at point \(M\). He then marks a point \(P\) on the bisector such that \(PM = 12\text{ cm}\). What is the distance from \(P\) to vertex \(A\)?
Step-by-Step Solution:
- Find the midpoint distance \(AM\): Since the perpendicular bisector cuts \(AB\) exactly in half:\[AM = \frac{18}{2} = 9\text{ cm}\]
- Identify the geometric relationship: The line \(PM\) is perpendicular to \(AB\), forming right-angled triangle \(\triangle PMA\) with \(\angle PMA = 90^\circ\).
- Apply the Pythagorean theorem:\[PA = \sqrt{AM^2 + PM^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ cm}\]
Answer: The distance \(PA\) is \(15\text{ cm}\).
Problem: A student constructs a right-angled triangle with legs \(a = 9\text{ cm}\) and \(b = 12\text{ cm}\). She then constructs the perpendicular bisectors of all three sides to find the circumcentre and draw the circumscribed circle passing through all 3 vertices. What is the radius \(R\) of this circumcircle in centimeters?
Step-by-Step Solution:
- Calculate the hypotenuse \(c\):\[c = \sqrt{a^2 + b^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ cm}\]
- Recall the circumcentre theorem for right triangles: The intersection of the perpendicular bisectors of any right-angled triangle lies exactly at the midpoint of its hypotenuse.
- Calculate the circumradius \(R\):\[R = \frac{c}{2} = \frac{15}{2} = 7.5\text{ cm}\]
Answer: The radius of the circumcircle is \(7.5\text{ cm}\).
Common Mistakes
Misconception 1: Changing the compass radius between paired arcs
Why it feels right: Learners often believe that as long as two arcs cross each other, the intersection line will still bisect the segment.
Correction: If the compass width changes between swinging from \(A\) and swinging from \(B\), the intersection point is no longer equidistant from both endpoints, causing the line to tilt away from the true midpoint and \(90^\circ\) perpendicular.
Misconception 2: Setting the compass radius too small (less than half the segment)
Why it feels right: Setting a small radius seems easier and keeps pencil arcs compact on paper.
Correction: If \(r < \frac{1}{2}AB\), the two arcs will never meet because the sum of their radii is less than the total distance \(AB\). Always ensure \(r > \frac{1}{2}AB\).
Misconception 3: Measuring with a ruler instead of constructing
Why it feels right: It seems faster to measure an angle with a protractor or find a midpoint by dividing ruler numbers.
Correction: In pure geometric construction, measurements are subject to visual error. Compass constructions are exact theoretical proofs of equidistance and symmetry.
Real World
1. Roof Truss Construction (Carpentry in Kenya)
When local fundis build King-post roof trusses for homes or classrooms, the central vertical strut must meet the tie-beam at an exact right angle (\(90^\circ\)) right at the midpoint. Using the 3-4-5 rope method or arc swinging guarantees the roof ridge carries weight symmetrically without buckling.
2. Shamba and Road Boundary Surveying
Land surveyors use optical instruments to project perpendicular offsets and bisect angles when partitioning agricultural land (shambas). Bisecting an angle between two fencing lines ensures pathways or irrigation channels divide water access equally between neighbors.
3. Solar Panel Orientation
To maximize electricity production across seasons near the equator, solar panel frames are constructed with precise fixed inclination angles (such as \(15^\circ\) or \(30^\circ\)), easily generated by bisecting standard \(60^\circ\) and \(30^\circ\) geometric constructions.
Practice