Complete notes with step-by-step construction methods and guide
Geometric constructions are drawings made using only a compass and a straightedge (ruler without markings). These constructions are fundamental to geometry and help us understand geometric relationships better.
Compass: Used to draw circles and arcs of specific radii.
Straightedge (Ruler): Used to draw straight lines between points.
Set Square: Used to draw perpendicular and parallel lines.
Protractor: Used to measure and draw specific angles.
Note: In pure geometric construction, only compass and unmarked straightedge are used.
This chapter focuses on constructing triangles when certain measurements are given. Different conditions require different construction methods.
Given: Three sides a, b, c
To Construct: Triangle ABC with sides AB = c, BC = a, CA = b
Steps:
1. Draw a line segment BC of length a.
2. With B as centre, draw an arc of radius c.
3. With C as centre, draw an arc of radius b.
4. Let the arcs intersect at point A.
5. Join AB and AC.
6. Triangle ABC is the required triangle.
Triangle is possible only if: Sum of any two sides > Third side
Example: For sides 3, 4, 5 → Check: 3+4>5, 4+5>3, 5+3>4 ✓
Given: Two sides b, c and included angle A
To Construct: Triangle ABC with AB = c, AC = b, ∠BAC = A
Steps:
1. Draw a line segment AB of length c.
2. At point A, construct an angle equal to given angle A.
3. On this ray, mark point C such that AC = b.
4. Join BC.
5. Triangle ABC is the required triangle.
Given: Two angles A, B and included side c
To Construct: Triangle ABC with AB = c, ∠BAC = A, ∠ABC = B
Steps:
1. Draw a line segment AB of length c.
2. At point A, construct angle equal to A.
3. At point B, construct angle equal to B.
4. Let the two rays intersect at point C.
5. Triangle ABC is the required triangle.
Given: Base BC = a, ∠B, and AB + AC = m
Steps:
1. Draw base BC = a.
2. At B, construct ∠XBC equal to given base angle.
3. From ray BX, cut line segment BD = AB + AC = m.
4. Join DC.
5. Draw perpendicular bisector of DC to intersect BD at point A.
6. Join AC.
7. Triangle ABC is the required triangle.
Q: Construct a triangle ABC where BC = 6 cm, ∠B = 60°, and AB + AC = 9 cm.
Solution Steps:
1. Draw BC = 6 cm
2. At B, construct ∠XBC = 60°
3. Cut BD = 9 cm on ray BX
4. Join D to C
5. Draw perpendicular bisector of DC (intersects BD at A)
6. Join A to C
Result: Triangle ABC is constructed
Given: Base BC = a, ∠B, and AB - AC = m (or AC - AB = m)
Case 1: When AB > AC (AB - AC = m):
1. Draw base BC = a.
2. At B, construct ∠XBC equal to given base angle.
3. From ray BX, cut line segment BD = AB - AC = m.
4. Join DC.
5. Draw perpendicular bisector of DC to intersect BX at point A.
6. Join AC.
7. Triangle ABC is the required triangle.
Case 2: When AC > AB (AC - AB = m):
Cut BD on opposite side of BC and follow similar steps.
Given: Perimeter (AB + BC + CA) = p, ∠B and ∠C
Steps:
1. Draw a line segment XY equal to perimeter p.
2. At X, construct angle equal to ∠B.
3. At Y, construct angle equal to ∠C.
4. Bisect these two angles; let bisectors meet at point A.
5. Draw perpendicular bisectors of AX and AY.
6. Let these bisectors intersect XY at B and C respectively.
7. Join AB and AC.
8. Triangle ABC is the required triangle.
| Given Information | Method | Key Step |
|---|---|---|
| Three sides (SSS) | Draw arcs from two points | Arc intersection gives third vertex |
| Two sides + included angle (SAS) | Construct angle first | Mark lengths on angle arms |
| Two angles + side (ASA) | Construct both angles | Ray intersection gives vertex |
| Base + angle + sum of sides | Use perpendicular bisector | Bisector of joined line |
| Base + angle + difference | Use perpendicular bisector | Handle AB > AC or AC > AB |
Given: Three sides a, b, c
Method: Draw one side, then arcs from ends
Condition: a+b>c, b+c>a, c+a>b
Most basic construction
Given: Two sides + included angle
Method: Construct angle, mark sides
Key: Angle must be between the two sides
Angle-based method
Given: Two angles + included side
Method: Draw side, construct angles at ends
Result: Rays meet at third vertex
Angle intersection
Given: Base, angle, AB + AC
Method: Use perpendicular bisector
Key: Cut sum on angle ray
Advanced construction
Given: Base, angle, AB - AC
Method: Perpendicular bisector method
Two cases: AB > AC or AC > AB
Handle carefully
Given: Perimeter + two base angles
Method: Angle bisectors + perpendicular bisectors
Complex but logical
Multiple steps
Compass: Draw arcs & circles
Straightedge: Draw lines
Set square: Perpendiculars
Protractor: Measure angles
• Keep compass fixed
• Show all arcs
• Label clearly
• Verify by measuring
• Practice is the ONLY way to master constructions - do at least 5 of each type!
• Understand the logic behind each step, don't just memorize.
• Always write steps clearly in exams - marks are given for method, not just diagram.
• Keep your instruments in good condition - sharp compass and clear ruler.
• Learn to construct common angles (30°, 45°, 60°, 90°, 120°) without protractor.
• Perpendicular bisector is key to many advanced constructions.
• Don't erase construction marks in exams - they show your method!
• Time yourself - constructions should be quick and accurate.
Exercise 1: Construct triangle PQR with PQ = 5 cm, QR = 6 cm, PR = 7 cm
Type: SSS Construction
Exercise 2: Construct triangle ABC with BC = 7 cm, ∠B = 45°, AB + AC = 10 cm
Type: Base + Angle + Sum construction
Exercise 3: Construct triangle XYZ with XY = 6 cm, ∠X = 50°, ∠Y = 60°
Type: ASA Construction