Question

Construct a ΔABC 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 ΔABC.

Answer 1

Steps of construction:

1. Draw AB = 5 cm. With A as the centre, draw ∠BAC= 45°. Join BC, ∠ABC is thus formed.

2. Draw AX such that ∠BAX is an acute angle.

3. Cut % equals arcs AA₁, A₁A₂, A₂A₂, A₂A₄ and A₄A_3.

4. Join A3 to B and draw a line through A3 parallel to A3B which meets AB at B

Here, AB' = `3/5` AB

5. Now draw a line through B' parallel to BC which joins AC at C'

Here, BC' = 3/5 BC and AC = `3/5` AC

Thus, AB'C is the required triangle.