Construct a triangle with sides 6 cm, 8 cm and 10 cm. Construct another triangle whose sides are of the corresponding sides of original triangle.

Let’s first construct ∆ABC with sides

Steps to draw ∆ABC

1. Draw base AB of side 6cm

2. With A as center, and 6cm as radius, C

Draw an arc

With B as center, and 10cm as radius,

Draw an arc.

3. Let C be the point where the two arc 8cm amd10cm

Intersect. Join AC & BC.

⇒ Thus, ∆ABC, is required triangle.

⇒Now, let’s make similar triangle with scale factor = .

Steps of construction

1.Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.

2. mark 5(the greater of 3 & 5 in ) points

On AX so that AA₁ = A₁A₂ = A₂A_3…...and so on.

3.join A₃B (3^rd point is smaller in)And draw a line through A₅ parallel to A₃B, to intersect AB extended at B^’_.

4.Draw a line through B^’ to the line BC to intersect AC extended at C^’.

Thus, ∆AB'C' is the required triangle.

Justification

Since scale factor is ,

We need to prove

By construction→

Also, BC is parallel to B’C

SO, they will make the same angle with AB

⇒ ∠ABC = ∠AB'C' (corresponding angles)…….(2)

Now, In ∆ABC & ∆AB’C’

∠A = ∠A (common)

∠ABC = ∠AB'C' (from 2)

⇒ ∆ABC~∆AB'C' (AA Similarity)

Since, corresponding sides of similar triangle are in the same ratio

So,

Thus, our construction is justified.