# 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.

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