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Construction 11.1 Deleted for CBSE Board 2025 Exams
Construction 11.2 Important Deleted for CBSE Board 2025 Exams
Construction 11.3 Deleted for CBSE Board 2025 Exams
Construction 11.4 Deleted for CBSE Board 2025 Exams You are here
Construction 11.5 Important Deleted for CBSE Board 2025 Exams
Construction 11.6 Important Deleted for CBSE Board 2025 Exams
Last updated at April 16, 2024 by Teachoo
Construction 11.4 : To construct a triangle, given its base, a base angle and sum of other two sides. Given base BC, a base angle ∠B and the sum AB + AC, we need to construct Δ ABC Steps of Construction: Draw base BC 2. Now, let’s draw ∠ B Construct angle B from point B. Let the ray be BX 3. Open the compass to length AB + AC. From point B as center, cut an arc on ray BX. Let the arc intersect BX at D 4. Join CD Now, we will draw perpendicular bisector of CD 6. Mark point A where perpendicular bisector intersects BD Join AC ∴ Δ ABC is the required triangle Check Construction 11.2, Class 9 on how to draw perpendicular bisector Justification We need to prove that AB + AC = BD. Let perpendicular bisector intersect CD at point R Thus, AR is the perpendicular bisector of CD ∴ CR = DR & ∠ ARC = ∠ ARD = 90° Now, In Δ ADR and Δ ACR AR = AR ∠ ARD = ∠ ARC DR = CR ∴ Δ ADR ≅ Δ ACR ⇒ AC = AD Now, BD = AB + AD BD = AB + AC Thus, our construction is justified (Common) (From (2)) (From (1)) (SAS Congruency) (CPCT) (From (3))