Question

Construct a ∆ABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45° Then construct a triangle whose sides are 3/5 of the corresponding sides of ∆ABC.

Answer 1

Steps of Construction:

• Draw a line segment AB = 5 cm.
• At A, draw ∠BAY = 45°
• From A draw an arc AC = 6 cm meeting A at C
• Join BC. Thus, AABC obtained.
• Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.

• Along AX mark 5 points A₁( A₂, A₃, A₄ and A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₆
• Join A₅B
• From A₃ draw A₃B’ || A₅B meeting AB at B’
• From B’ draw B’C’ || BC meeting at C’.
  Then AB’C is required triangle, each of whose side is 35 of corresponding sides of ∆ABC.

Justification:

Since

B’C || BC ‘

Therefore, AB’C ~ ∆ABC

AB′/AB = B′C′/BC = AC′/AC = 3/5