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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

  1. AB > AC and difference AB – AC is given
  2. 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 :

  1. Draw base BC  and at point B make an angle XBC  equal to the given angle.
  2. Cut a line segment BD equal to AB – AC from ray BX
  3. Join DC. Draw perpendicular bisector PQ of line DC.
  4. 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 :

  1. Draw base BC  and at point B make an angle XBC equal to the given angle.
  2. Cut a line segment BD equal to AB – AC from ray BX
  3. Join DC. Draw perpendicular bisector PQ of line DC.
  4. 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

-ev-

  1. Chapter 11 Class 9 Constructions
  2. Concept wise
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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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