Question

Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?

Answer 1

A ΔP'Q'R' whose sides are `3/5`of the corresponding sides of ΔPQR can be drawn as follows:

Step 1

Draw a ΔPQR with side QR = 6 cm, PQ = 5 cm and ∠PQR = 60°.

Step 2

Draw a ray QX making an acute angle with QR on the opposite side of vertex P.

Step 3

Locate 5 points (as 5 is greater in 3 and 5), Q₁, Q₂, Q₃, Q₄ and Q₅ on line segment QX.

Step 4

Join Q₅R and draw a line through Q₃, parallel to Q₅R intersecting QR at R'.

Step 5

Draw a line through R' parallel to PR intersecting PQ at P'.

Δ P'Q'R' is the required triangle.