Draw a triangle PQR such that PQ = 5 cm, P = $120^{\circ}$ and PR = 6 cm. Construct another triangle
Whose sides are $\frac{3}{4}$ times the corresponding sides of $Δ$ PQR?

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Solution

A $Δ$ PQ'R' whose sides are $\frac{3}{4}$ of the corresponding sides of $Δ$ PQR can be drawn as follows.
Step1. Draw a $Δ$ PQR with side PQ = 5 cm, PR = 6 cm and P = $120^{\circ}$
Step2. Draw a ray PX making an acute angle with PR on the opposite side of vertex Q.
Step3. Locate 4 points (as 4 is greater in 3 and 4), $P_{1}, P_{2}, P_{3}, P_{4}$ , on line segment PX.
Step4. Join $P_{4} R$ and draw a line through $P_{3}$ , parallel to $P_{4} R$ intersecting PR at R'.
Step5. Draw a line through R' parallel to QR intersecting PQ at Q'. $Δ$ PQ'R' is the required triangle.

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