Constructing Triangles (Class 9)
In Class 9 Geometry, triangle constructions go beyond the basic SSS, SAS, and ASA methods learned in earlier classes. Students now construct triangles with indirect measurements — given the base, base angles, and the sum or difference of the other two sides.
These constructions require careful use of a compass and straightedge. Each construction must be accompanied by a proper justification explaining why the method works.
The three main construction types in NCERT Class 9 are based on different sets of given data: (1) base, base angles, and sum of other two sides; (2) base, base angles, and difference of other two sides; (3) perimeter and two base angles.
What is Constructing Triangles (Class 9)?
Construction Type 1: Given base BC, a base angle, and sum of other two sides (AB + AC).
Construction Type 2: Given base BC, a base angle, and difference of other two sides (AB − AC or AC − AB).
Construction Type 3: Given the perimeter (AB + BC + CA) and both base angles.
Important:
- All constructions use only compass and straightedge.
- Each construction must include a written justification.
- The triangle inequality must be satisfied for a valid triangle to exist.
Constructing Triangles (Class 9) Formula
Key Relationships:
1. Sum of two sides > third side (triangle inequality)
AB + AC > BC
2. Angle sum property:
∠A + ∠B + ∠C = 180°
3. Perimeter relationship:
- Perimeter = AB + BC + CA
- If perimeter and BC are known: AB + CA = Perimeter − BC
Derivation and Proof
Construction 1: Given base, one base angle, and sum of other two sides
Given: Base BC, ∠B, and AB + AC = s.
Steps:
- Draw the base BC of given length.
- At B, construct ∠XBC = ∠B.
- From B, cut BD = s (= AB + AC) on ray BX.
- Join DC.
- Construct the perpendicular bisector of DC, meeting BD at A.
- Join AC. Triangle ABC is the required triangle.
Justification:
- A lies on the perpendicular bisector of DC, so AD = AC.
- BD = BA + AD = BA + AC = s. ✓
- ∠ABC = ∠B (by construction). ✓
Construction 2: Given base, one base angle, and difference of other two sides
Case (a): AB > AC.
Given: Base BC, ∠B, and AB − AC = d.
Steps:
- Draw BC and construct ∠XBC = ∠B at B.
- From B, cut BD = d on ray BX.
- Join DC.
- Construct the perpendicular bisector of DC, meeting BX at A.
- Join AC. Triangle ABC is the required triangle.
Justification:
- A lies on the perpendicular bisector of DC, so AD = AC.
- BD = BA − AD = BA − AC = d. ✓
Construction 3: Given perimeter and two base angles
Given: Perimeter = AB + BC + CA, ∠B, ∠C.
Steps:
- Draw a line segment XY = perimeter.
- At X, construct ∠LXY = ∠B. At Y, construct ∠MYX = ∠C.
- Construct the bisector of ∠LXY and the bisector of ∠MYX. Let them meet at A.
- Construct the perpendicular bisector of AX meeting XY at B.
- Construct the perpendicular bisector of AY meeting XY at C.
- Join AB and AC. Triangle ABC is the required triangle.
Types and Properties
Summary of Construction Types:
1. Base + Base angle + Sum of other two sides
- Given: BC, ∠B, AB + AC
- Key idea: Mark the total length on one ray, then use the perpendicular bisector to locate the vertex.
- Condition: AB + AC > BC (triangle inequality).
2. Base + Base angle + Difference of other two sides
- Given: BC, ∠B, |AB − AC|
- Key idea: Mark the difference on the ray, then use the perpendicular bisector.
- Two sub-cases depending on which side is longer.
3. Perimeter + Two base angles
- Given: AB + BC + CA, ∠B, ∠C
- Key idea: Draw the full perimeter as a segment, use angle bisectors and perpendicular bisectors.
Solved Examples
Example 1: Example 1: Construct with sum of sides
Problem: Construct ▵ABC where BC = 7 cm, ∠B = 45°, and AB + AC = 13 cm.
Solution:
- Draw BC = 7 cm.
- At B, construct ∠XBC = 45°.
- From B, mark D on BX such that BD = 13 cm.
- Join DC.
- Draw the perpendicular bisector of DC, meeting BD at A.
- Join AC.
Verification:
- Measure AB + AC. It equals 13 cm. ✓
- Measure ∠B. It is 45°. ✓
Answer: ▵ABC is constructed with the given conditions.
Example 2: Example 2: Construct with difference of sides (AB > AC)
Problem: Construct ▵ABC where BC = 8 cm, ∠B = 60°, and AB − AC = 3 cm.
Solution:
- Draw BC = 8 cm.
- At B, construct ∠XBC = 60°.
- From B, mark D on BX such that BD = 3 cm (on the ray BX, since AB > AC).
- Join DC.
- Draw the perpendicular bisector of DC, meeting BX at A.
- Join AC.
Verification:
- Measure AB − AC. It equals 3 cm. ✓
- Measure ∠B. It is 60°. ✓
Answer: ▵ABC is constructed.
Example 3: Example 3: Construct with difference (AC > AB)
Problem: Construct ▵PQR where QR = 6 cm, ∠Q = 60°, and PR − PQ = 2 cm (PR > PQ).
Solution:
- Draw QR = 6 cm.
- At Q, construct ∠XQR = 60°.
- Since PR > PQ, mark D on the extension of QX below Q such that QD = 2 cm.
- Join DR.
- Draw the perpendicular bisector of DR, meeting QX at P.
- Join PR.
Verification:
- Measure PR − PQ. It equals 2 cm. ✓
Answer: ▵PQR is constructed.
Example 4: Example 4: Construct given perimeter and base angles
Problem: Construct ▵ABC where perimeter = 15 cm, ∠B = 50°, ∠C = 70°.
Solution:
- Draw XY = 15 cm.
- At X, construct ∠LXY = 50°. At Y, construct ∠MYX = 70°.
- Bisect ∠LXY and ∠MYX. Let bisectors meet at A.
- Draw the perpendicular bisector of AX, meeting XY at B.
- Draw the perpendicular bisector of AY, meeting XY at C.
- Join AB and AC.
Verification:
- Measure AB + BC + CA = 15 cm. ✓
- Measure ∠B = 50°, ∠C = 70°. ✓
Answer: ▵ABC is constructed.
Example 5: Example 5: Sum construction with 30-degree angle
Problem: Construct ▵XYZ where YZ = 5 cm, ∠Y = 30°, and XY + XZ = 10 cm.
Solution:
- Draw YZ = 5 cm.
- At Y, construct ∠PYZ = 30°.
- From Y, cut YD = 10 cm on ray YP.
- Join DZ.
- Draw the perpendicular bisector of DZ, meeting YD at X.
- Join XZ.
Answer: ▵XYZ with the given measurements is constructed.
Example 6: Example 6: Verify angle sum after construction
Problem: Construct ▵ABC with BC = 6 cm, ∠B = 75°, AB + AC = 11 cm. Measure ∠A and ∠C and verify the angle sum property.
Solution:
- Construct the triangle as in Construction Type 1.
- Measure angles: ∠A ≈ 62°, ∠C ≈ 43°.
- Sum = 75° + 62° + 43° = 180°. ✓
Answer: The angle sum property is verified.
Example 7: Example 7: When construction is not possible
Problem: Can you construct ▵ABC with BC = 10 cm, ∠B = 50°, and AB + AC = 8 cm?
Solution:
Check triangle inequality:
- AB + AC = 8 cm, but BC = 10 cm.
- AB + AC must be greater than BC for a triangle to exist.
- 8 < 10, so the condition is violated.
Answer: The construction is not possible because AB + AC < BC violates the triangle inequality.
Example 8: Example 8: Perimeter construction with equal base angles
Problem: Construct ▵PQR where perimeter = 18 cm, ∠Q = 65°, ∠R = 65°.
Solution:
- Draw XY = 18 cm.
- At X, construct ∠LXY = 65°. At Y, construct ∠MYX = 65°.
- Bisect both angles. The bisectors meet at P.
- Perpendicular bisector of PX meets XY at Q.
- Perpendicular bisector of PY meets XY at R.
- Join PQ and PR.
Note: Since ∠Q = ∠R, the triangle is isosceles with PQ = PR.
Answer: ▵PQR is an isosceles triangle with perimeter 18 cm.
Real-World Applications
Applications:
- Engineering Drawing: Constructing triangular frameworks when only indirect measurements are available.
- Architecture: Designing triangular roof sections, gables, and arches with specific perimeter and angle constraints.
- Navigation: Triangulation problems where the sum or difference of distances from a point to two fixed points is known.
- Proof Skills: These constructions develop logical reasoning and the ability to write mathematical justifications — a core skill in higher mathematics.
Key Points to Remember
- Class 9 triangle constructions involve indirect measurements (sum/difference of sides, perimeter).
- Type 1: Base + base angle + sum of other two sides.
- Type 2: Base + base angle + difference of other two sides (two sub-cases).
- Type 3: Perimeter + two base angles.
- The perpendicular bisector is the key tool in all three types.
- The triangle inequality must be satisfied: sum of two sides > third side.
- Every construction must include a written justification.
- The equidistant property of the perpendicular bisector is used for justification.
- These constructions are part of NCERT Class 9 Chapter 11.
Practice Problems
- Construct triangle ABC where BC = 6 cm, angle B = 60 degrees, and AB + AC = 11 cm.
- Construct triangle PQR where QR = 7 cm, angle Q = 75 degrees, and PQ - PR = 2.5 cm (PQ > PR).
- Construct a triangle with perimeter 16 cm and base angles 45 degrees and 60 degrees.
- Construct triangle XYZ where YZ = 5.5 cm, angle Y = 50 degrees, and XZ - XY = 1.5 cm.
- Can a triangle be constructed with BC = 9 cm, angle B = 40 degrees, and AB + AC = 7 cm? Justify your answer.
- Construct triangle DEF with perimeter 14 cm and base angles D = 55 degrees and F = 70 degrees. Measure the third angle.
Frequently Asked Questions
Q1. What are the three types of triangle constructions in Class 9?
The three types are: (1) given base, base angle, and sum of other two sides; (2) given base, base angle, and difference of other two sides; (3) given perimeter and two base angles.
Q2. Why is the perpendicular bisector important in these constructions?
The perpendicular bisector of a segment ensures any point on it is equidistant from both endpoints. This property is used to locate the vertex such that the sum or difference condition is satisfied.
Q3. What is the justification for Construction Type 1?
Point A lies on the perpendicular bisector of DC, so AD = AC. Since BD = BA + AD = BA + AC = given sum, the construction is correct.
Q4. When is a construction not possible?
A construction is not possible when the triangle inequality is violated. For instance, if the given sum of two sides is less than or equal to the base, no triangle can be formed.
Q5. Is justification compulsory in CBSE exams?
Yes. NCERT Class 9 requires a written justification for every construction. In CBSE exams, marks are allocated for both the construction and its justification.
Q6. What tools are allowed for these constructions?
Only a compass (for drawing arcs) and a straightedge or ruler (for drawing straight lines) are allowed. Protractors may be used to draw the given angle if it is not a standard constructible angle.
Q7. How do you handle the case when the difference of sides is negative?
If AB − AC is negative (meaning AC > AB), mark the difference on the opposite side of the ray. The perpendicular bisector method then locates vertex A correctly.
Q8. Can these methods construct any triangle?
Yes, as long as the given data satisfies the triangle inequality and angle sum property. If the conditions are contradictory, the construction will not yield a valid triangle.
