Construct a triangle PQR with sides QR = 7 cm, PQ = 6.5 cm and ∠ PQR = 60°. Now construct another triangle, whose sides are times the corresponding sides of the given triangle.

Steps of construction:

Step 1. Construct a line PQ of length 6.5 cm.

Step 2:

Draw angle of 60° at Q.

Step 3:

From Q draw an arc of 7 cm. The point intersects at R. Join PR.

Step 4:

Draw QX.

Step 5:

Mark the greater of i.e. 7 arcs in equal distance on QX.

Step 6:

Join Q₇P and then draw a line through Q₅ parallel to Q₇P.

Step 7:

From R₁ draw a line parallel to RQ.

Thus, R₁QP₁ is the required triangle.

Justification:

Since the scale factor is ,

We need to prove,

By construction,

… (1)

Also, R₁Q₁ is parallel to RQ.

So, this will make same angle with QP.

∴ ∠R₁P₁Q = ∠RPQ …. (2)

Now,

In Δ R₁P₁Q and ΔRPQ

∠Q = ∠Q (common)

∠R₁P₁Q = ∠RPQ (from 2)

Δ R₁P₁Q ∼ ΔRPQ

Since corresponding sides of similar triangles are in same ratio.

From (1)

Hence construction is justified.