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  1. Chapter 11 Class 9 Constructions
  2. Serial order wise
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Construction 11.5 : To construct a triangle, given its base, a base angle and the difference of the other two sides. There can be two cases AB > AC and difference AB – AC is given AC > AB and difference AC – AB is given Case 1: AB > AC Given the base BC, a base angle ∠B and the difference AB – AC, we need to construct Δ ABC Steps of Construction: Draw base BC  and at point B make an angle XBC  equal to the given angle. Cut a line segment BD equal to AB – AC from ray BX Join DC. Draw perpendicular bisector PQ of line DC. Let PQ intersect BX a point A. Join AC. Then, Δ ABC is the required triangle. Justification We need to prove that AB – AC = BD. Given Base BC and ∠B In Δ ACD,   PQ is the perpendicular bisector of CD ∴ AD = AC Now, BD = AB – AD BD  = AB – AC Thus, our construction is justified Case 2: AC > AB Given the base BC, a base angle ∠B and the difference AC – AB, we need to construct Δ ABC Steps of Construction: Draw base BC  and at point B make an angle XBC equal to the given angle. Cut a line segment BD equal to AB – AC from ray BX Join DC. Draw perpendicular bisector PQ of line DC. Let PQ intersect BX a point A. Join AC. Then, Δ ABC is the required triangle. Justification We need to prove that AC – AB = BD. Given Base BC and ∠B In Δ ACD,   PQ is the perpendicular bisector of CD ∴ AD = AC Now, BD = AB – AD BD  = AB – AC Thus, our construction is justified

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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