Question

Draw a ∆ABC in which BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct another triangle whose sides are 3/4 times corresponding sides of ∆ABC.

Answer 1

• Draw a line segment BC = 7 cm.
• In ∆ABC, ∠B = 45°, 2A = 105°
  ∴ ∠C = 180° – (45° + 105°) = 30°
  At B draw and angle ∠B = 45° and draw ∠C = 30° intersecting each other at A to get ∆ABC

• At B draw any ray BX making an acute angle with BC on the side opposite to the vertex A
• Along BX mark 4 points B₁(B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
• Join B₄C
• From B₃ draw B<sub3C’ | | B₄C meeting BC at C’
• From C’ draw C’A’ | | AC meeting AB at A
  Then A’BC’ is required triangle, each of whose side is 3/4 of corresponding sides of ∆ABC.