Question

Construct a ∆ABC, with BC = 7 cm, $\angle$B = 60º and AB = 6 cm. Construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of ∆ABC. [CBSE 2008, '09, '15]

Answer 1

Steps of Construction

Step 1. Draw a line segment BC = 7 cm.

Step 2. At B, draw $\angle$XBC = 60º.

Step 3. With B as centre and radius 6 cm, draw an arc cutting the ray BX at A.

Step 4. Join AC. Thus, ∆ABC is the required triangle.

Step 5. Below BC, draw an acute angle $\angle$YBC.

Step 6. Along BY, mark four points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

Step 7. Join CB₄.

Step 8. From B₃, draw B₃C' || CB₄ meeting BC at C'.

Step 9. From C', draw A'C' || AC meeting AB in A'.


Here, ∆A'BC' is the required triangle whose sides are $\frac{3}{4}$ times the corresponding sides of ∆ABC.