Question

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

Answer 1

In ΔABC, 

∠A + ∠B + ∠C = 180°   ...[Angle sum property of a triangle]

⇒ 105° + 45° + ∠C = 180°

⇒ ∠C = 30°

Steps of constructions:

1. Draw BC = 7 cm.
2. Construct ∠CBY = 45° and ∠BCZ = 30°.
3. Rays BY and CZ intersect at A.
4. ΔABC is given.
5. From B, draw a ray BX below BC making acute angle with BC.
6. Along it mark 4 points B₁, B₂, B₃, B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
7. Join B₄C. Make ∠BB₄C at B₃ such that the ray intersects BC at C'.
   ∴ ∠BB₄C = ∠BB₃C".
   So, B₄C || B₃C".
8. From C', make ∠BC'A' = ∠BCA so that C'A' || CA.
   Thus, A'BC' is the required triangle.