Q. 95.0( 3 Votes )
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.
Draw angle of 60° at Q.
From Q draw an arc of 7 cm. The point intersects at R. Join PR.
Mark the greater of i.e. 7 arcs in equal distance on QX.
Join Q7P and then draw a line through Q5 parallel to Q7P.
From R1 draw a line parallel to RQ.
Thus, R1QP1 is the required triangle.
Since the scale factor is ,
We need to prove,
Also, R1Q1 is parallel to RQ.
So, this will make same angle with QP.
∴ ∠R1P1Q = ∠RPQ …. (2)
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.
Hence construction is justified.
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