Question

Construct a triangle ABC with sides BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are `3/4` times the corresponding sides of ∆ABC.

Answer 1

Steps of construction:

1. Draw line BC = 7 cm

2. At B, construct ∠CBY = 45^° and at C, construct 180^° −(105^° + 45^°) = 30^°.

3. Mark the point of intersection of ∠B = 45^° and ∠C = 30^° as A. Thus, ∆ABC is obtained.

4. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

5. Locate 4 (3 < 4) points B₁, B₂, B₃ and B₄ on BX such that  BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

6. Join B₄C and draw a line through B3 parallel to B₄C to intersect BC at C'.

7. Draw a line through C′ parallel to the line CA to intersect BA at A′

Then, ΔA′BC′ is the required triangle similar to the ΔABC