Question

4. Construct ∆ ABC, in which BC = 3.2 cm, angle ACB = 45° and perimeter of A ABC
is 10 cm.???​

Answer 1

10 = AB + 3.2 + AC ∴ AB + AC = 10 – 3.2 ∴ AB + AC = 6.8 cm Now, In ∆ABC BC = 3.2 cm, ∠ACB = 45° and AB + AC = 6.8 cm ….(i) As shown in the rough figure draw j seg BC = 3.2 cm Draw a ray CT making an angle of 45° with CB Take a point D on ray CT, such that CD = 6.8 cm Now, CA + AD = CD [C - A - D] ∴ CA + AD = 6.8 cm …(ii) Also, AB + AC = 6.8 cm ….(iii) [From (i)] ∴ CA + AD = AB + AC [From (ii) and (iii)] ∴ AD = AB