Bank Guides Notes Mocks Board Questions Maharashtra State Board Class 9 Mathematics Constructions of Triangles 18 Maharashtra State Board Class 9 Mathematics practice questions with answers and worked solutions.
About this chapter Difficulty: 12 moderate · 6 hard. Most-asked subtopics: Construction Given Base, Base Angle and Sum of Remaining Sides (6), Construction Given Base, Base Angle and Difference of Remaining Sides (6), Construction Given Two Angles and Perimeter (5). Updated 23 August 2026. Balbharati textbook · 4.1 SolvedEx.1 Moderate Add #1 · Construction Given Base, Base Angle and Sum of Remaining Sides
Construct △ A B C \triangle ABC △ A B C in which B C = 6.3 BC = 6.3 B C = 6.3 cm, ∠ B = 75 ∘ \angle B = 75^\circ ∠ B = 7 5 ∘ and A B + A C = 9 AB + AC = 9 A B + A C = 9 cm. Balbharati textbook · Ex 4.1 Q1 Moderate Add #2 · Construction Given Base, Base Angle and Sum of Remaining Sides
Construct △ P Q R \triangle PQR △ P QR , in which Q R = 4.2 QR = 4.2 QR = 4.2 cm, m ∠ Q = 40 ∘ m\angle Q = 40^\circ m ∠ Q = 4 0 ∘ and P Q + P R = 8.5 PQ + PR = 8.5 P Q + P R = 8.5 cm. Balbharati textbook · Ex 4.1 Q2 Moderate Add #3 · Construction Given Base, Base Angle and Sum of Remaining Sides
Construct △ X Y Z \triangle XYZ △ X Y Z , in which Y Z = 6 YZ = 6 Y Z = 6 cm, X Y + X Z = 9 XY + XZ = 9 X Y + X Z = 9 cm, ∠ X Y Z = 50 ∘ \angle XYZ = 50^\circ ∠ X Y Z = 5 0 ∘ . Balbharati textbook · Ex 4.1 Q3 Moderate Add #4 · Construction Given Base, Base Angle and Sum of Remaining Sides
Construct △ A B C \triangle ABC △ A B C , in which B C = 6.2 BC = 6.2 B C = 6.2 cm, ∠ A C B = 50 ∘ \angle ACB = 50^\circ ∠ A C B = 5 0 ∘ , A B + A C = 9.8 AB + AC = 9.8 A B + A C = 9.8 cm. Balbharati textbook · Ex 4.1 Q4 Moderate Add #5 · Construction Given Base, Base Angle and Sum of Remaining Sides
Construct △ A B C \triangle ABC △ A B C , in which B C = 3.2 BC = 3.2 B C = 3.2 cm, ∠ A C B = 45 ∘ \angle ACB = 45^\circ ∠ A C B = 4 5 ∘ and perimeter of △ A B C \triangle ABC △ A B C is 10 cm. Done a few? Here’s your next move
Balbharati textbook · 4.2 SolvedEx.1 Moderate Add #6 · Construction Given Base, Base Angle and Difference of Remaining Sides
Construct △ A B C \triangle ABC △ A B C , such that B C = 7.5 BC = 7.5 B C = 7.5 cm, ∠ A B C = 40 ∘ \angle ABC = 40^\circ ∠ A B C = 4 0 ∘ , A B − A C = 3 AB - AC = 3 A B − A C = 3 cm. Balbharati textbook · 4.2 SolvedEx.2 Moderate Add #7 · Construction Given Base, Base Angle and Difference of Remaining Sides
Construct △ A B C \triangle ABC △ A B C , in which side B C = 7 BC = 7 B C = 7 cm, ∠ B = 40 ∘ \angle B = 40^\circ ∠ B = 4 0 ∘ and A C − A B = 3 AC - AB = 3 A C − A B = 3 cm. Balbharati textbook · Ex 4.2 Q1 Moderate Add #8 · Construction Given Base, Base Angle and Difference of Remaining Sides
Construct △ X Y Z \triangle XYZ △ X Y Z , such that Y Z = 7.4 YZ = 7.4 Y Z = 7.4 cm, ∠ X Y Z = 45 ∘ \angle XYZ = 45^\circ ∠ X Y Z = 4 5 ∘ and X Y − X Z = 2.7 XY - XZ = 2.7 X Y − X Z = 2.7 cm. Balbharati textbook · Ex 4.2 Q2 Moderate Add #9 · Construction Given Base, Base Angle and Difference of Remaining Sides
Construct △ P Q R \triangle PQR △ P QR , such that Q R = 6.5 QR = 6.5 QR = 6.5 cm, ∠ P Q R = 40 ∘ \angle PQR = 40^\circ ∠ P QR = 4 0 ∘ and P Q − P R = 2.5 PQ - PR = 2.5 P Q − P R = 2.5 cm. Balbharati textbook · Ex 4.2 Q3 Moderate Add #10 · Construction Given Base, Base Angle and Difference of Remaining Sides
Construct △ A B C \triangle ABC △ A B C , such that B C = 6 BC = 6 B C = 6 cm, ∠ A B C = 100 ∘ \angle ABC = 100^\circ ∠ A B C = 10 0 ∘ and A C − A B = 2.5 AC - AB = 2.5 A C − A B = 2.5 cm. Balbharati textbook · 4.3 SolvedEx.1 Hard Add #11 · Construction Given Two Angles and Perimeter
Construct △ A B C \triangle ABC △ A B C such that A B + B C + C A = 11.3 AB + BC + CA = 11.3 A B + B C + C A = 11.3 cm, ∠ B = 70 ∘ \angle B = 70^\circ ∠ B = 7 0 ∘ , ∠ C = 60 ∘ \angle C = 60^\circ ∠ C = 6 0 ∘ . Balbharati textbook · Ex 4.3 Q1 Hard Add #12 · Construction Given Two Angles and Perimeter
Construct △ P Q R \triangle PQR △ P QR , in which ∠ Q = 70 ∘ \angle Q = 70^\circ ∠ Q = 7 0 ∘ , ∠ R = 80 ∘ \angle R = 80^\circ ∠ R = 8 0 ∘ and P Q + Q R + P R = 9.5 PQ + QR + PR = 9.5 P Q + QR + P R = 9.5 cm. Balbharati textbook · Ex 4.3 Q2 Hard Add #13 · Construction Given Two Angles and Perimeter
Construct △ X Y Z \triangle XYZ △ X Y Z , in which ∠ Y = 58 ∘ \angle Y = 58^\circ ∠ Y = 5 8 ∘ , ∠ X = 46 ∘ \angle X = 46^\circ ∠ X = 4 6 ∘ and perimeter of triangle is 10.5 cm. Balbharati textbook · Ex 4.3 Q3 Hard Add #14 · Construction Given Two Angles and Perimeter
Construct △ L M N \triangle LMN △ L M N , in which ∠ M = 60 ∘ \angle M = 60^\circ ∠ M = 6 0 ∘ , ∠ N = 80 ∘ \angle N = 80^\circ ∠ N = 8 0 ∘ and L M + M N + N L = 11 LM + MN + NL = 11 L M + M N + N L = 11 cm. Balbharati textbook · Prob Q1 Moderate Add #15 · Construction Given Base, Base Angle and Sum of Remaining Sides
Construct △ X Y Z \triangle XYZ △ X Y Z , such that X Y + X Z = 10.3 XY + XZ = 10.3 X Y + X Z = 10.3 cm, Y Z = 4.9 YZ = 4.9 Y Z = 4.9 cm, ∠ X Y Z = 45 ∘ \angle XYZ = 45^\circ ∠ X Y Z = 4 5 ∘ . Balbharati textbook · Prob Q2 Hard Add #16 · Construction Given Two Angles and Perimeter
Construct △ A B C \triangle ABC △ A B C , in which ∠ B = 70 ∘ \angle B = 70^\circ ∠ B = 7 0 ∘ , ∠ C = 60 ∘ \angle C = 60^\circ ∠ C = 6 0 ∘ , A B + B C + A C = 11.2 AB + BC + AC = 11.2 A B + B C + A C = 11.2 cm. Balbharati textbook · Prob Q3 Hard Add #17 · Construction Given Perimeter and Ratio of Sides
The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4 2 : 3 : 4 2 : 3 : 4 . Construct the triangle. Balbharati textbook · Prob Q4 Moderate Add #18 · Construction Given Base, Base Angle and Difference of Remaining Sides
Construct △ P Q R \triangle PQR △ P QR , in which P Q − P R = 2.4 PQ - PR = 2.4 P Q − P R = 2.4 cm, Q R = 6.4 QR = 6.4 QR = 6.4 cm and ∠ P Q R = 55 ∘ \angle PQR = 55^\circ ∠ P QR = 5 5 ∘ . Studying this with friends? Send them this chapter's questions. It's free for them too.
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